MCQ 11 Mark
The wickets taken by a bowler in a one day cricket match are $4, 5, 6, 3, 4, 0, 3, 2, 3, 5.$ The mode of the data is $...........$
AnswerMode of the set of data is the observation which occurs the most.
$4, 6$ occurs $2$ times each, $6, 2$ and $00$ occurs $11$ time each,
where as $3$ occurs $3$ times.
Thus, the number $33$ occurs the maximum number of times
i.e., $3.$ Therefore, mode of the data is $33.$
View full question & answer→MCQ 21 Mark
The average of the first five odd prime numbers is:
AnswerRequired average $=\frac{3 + 5 + 7 + 11 + 13}{5}$
$=\frac{39}{5}$
$=7.8$
View full question & answer→MCQ 31 Mark
Mode of the distribution is that value of the variate for which the $..........$ is $..........$
- ✓
- B
- C
frequency, arithmetic mean
- D
frequency, arithmetic mean
AnswerMode of the distribution is that value of the variate for which the frequency is maximum.
View full question & answer→MCQ 41 Mark
The average age of $6$ students is $11$ years. If two more students of age $14$ and $16$ years join, their average will become
- A
$13$ years
- ✓
$12$ years
- C
$12\dfrac{1}{2}$ years
- D
$12\dfrac{1}{2}$ years
AnswerCorrect option: B. $12$ years
$\Rightarrow$ The average age of $66$ students is $11$ years.
$\Rightarrow$ Sum of age of $66$ students $= 6 \times 11 = 66$
$\Rightarrow$ When two more students of age $14$ and $16 $added to $6$ students then, total students will become $8$
$\Rightarrow$ Sum of age of $8$ students $= 66 + 14 + 16 = 96$
$\Rightarrow$ Required average $ =\dfrac{96}{8}=12\ \text{years}$
View full question & answer→MCQ 51 Mark
If the mean of $x + 2, 2x+ 3, 3x + 4, 4x + 5$ is $x + 2$ then $x$ is equal to:
AnswerMean of the given distribution is,
$=\frac{(\text{x}+2)+(2\text{x}+3)+(3\text{x}+4)+(4\text{x}+5)}{4}$
$= \text{x} +2$
$= 4(x + 2) + (2x + 3) + (3x + 4) + (4x + 5)$
$= x + 2, ($given$)$
$=\frac{10\text{x}+14}{4} = \text{x}+2$
$= 10x + 14 = 4x + 8$
$\Rightarrow x = -1$
View full question & answer→MCQ 61 Mark
Choose the correct answer.
Consider the first $10$ positive integers. If we multiply each number by $-1$ and then add $1$ to each number, the variance of the numbers so obtained is:
- ✓
$8.25$
- B
$6.5$
- C
$3.87$
- D
$3.87$
AnswerCorrect option: A. $8.25$
Since, the first $10$ positive integers are $1, 2, 3, 4, 5, 6, 7, 8, 9$ and $10.$
On multiplying each number by $-1,$ we get $-1, -2, -3, -4, -5, -6, -7, -8, -9, -10$ On adding $1$ in each number.
We get $0, -1, -2, -3, -4, -5, -6, -7, -8, -9.$
$\therefore\ \sum\text{x}_\text{i}=-\frac{9\times10}{2}=-45$
and $\sum\text{x}^2_\text{i}=0^2+(-1)^2+(-2)^2+\ ....\ +(9)^2=\frac{9\times10\times19}{6}=285$
$\text{SD}=\sqrt{\frac{285}{10}-\Big(\frac{-45}{10}\Big)^2}$
$=\sqrt{\frac{285}{10}-\frac{2025}{100}}$
$=\sqrt{\frac{2850-2025}{100}}$
$=\sqrt{8.25}$
Now, $\text{variance}=(\text{SD})^2$
$=\big(\sqrt{8.25}\big)^2$
$=8.25$
View full question & answer→MCQ 71 Mark
The standard deviation of first $10$ natural numbers is:
- A
$5.5$
- B
$3.87$
- C
$2.97$
- ✓
$2.87$
AnswerCorrect option: D. $2.87$
We know that the standard deviation of first $n$ natural number is $\sqrt{\frac{\text{n}^2-1}{12}}.$
$\therefore$ Standard deviation of first $10$ natural numbers
$=\sqrt{\frac{10^2-1}{12}}$
$=\sqrt{\frac{99}{12}}$
$=\sqrt{8.25}$
$=2.87$
Hence, the correct answer is option $(d).$
View full question & answer→MCQ 81 Mark
Let $a, b, c, d, e$ be the observations with mean $m$ and standard deviation $s.$ The standard deviation of the observations $a + k, b + k, c + k, d + k, e + k$ is:
- ✓
$s$
- B
$ks$
- C
$s + k$
- D
$s + k$
AnswerThe given observations are $a, b, c, d, e.$
$\text{Mean}=\text{m}=\frac{\text{a+b+c+d+e}}{5}$
$\Rightarrow\sum\text{x}_\text{i}=\text{a}+\text{b}+\text{c}+\text{d}+\text{e}=5\text{m}\ ...(1)$
Standard deviation, $\text{s}=\sqrt{\frac{\sum\text{x}_\text{i}^2}{5}-\text{m}^2}$
Now, consider the observations $a + k, b + k, c + k, d + k, e + k.$
New mean $=\frac{(\text{a+k})+(\text{b+k})+(\text{c+k})+(\text{d+k})+(\text{e+k})}{5}$
$=\frac{\text{a+b+c+d+e+5k}}{5}$
$=\frac{5\text{m}+5\text{k}}{5}$
$=\text{m}+\text{k}$
$\therefore$ New standard deviation
$=\sqrt{\frac{\sum(\text{x}_\text{i}+\text{k})^2}{5}-(\text{m}+\text{k})^2}$
$=\sqrt{\frac{\sum(\text{x}_\text{i}^2+\text{k}^2+2\text{x}_\text{i}\text{k})}{5}-(\text{m}^2+\text{k}^2+2\text{mk})}$
$=\sqrt{\frac{\sum\text{x}_\text{i}^2}{5}+\frac{\sum\text{k}^2}{5}+\frac{\sum2\text{x}_\text{i}\text{k}}{5}-(\text{m}^2+\text{k}^2+2\text{mk})}$
$=\sqrt{\frac{\sum\text{x}_\text{i}^2}{5}-\text{m}^2+\frac{5\text{k}^2}{5}-\text{k}^2+\frac{2\text{k}\sum\text{x}_\text{i}}{5}-2\text{mk}}$
$=\sqrt{\frac{\sum\text{x}_\text{i}^2}{5}-\text{m}^2+\frac{2\text{k}\times5\text{m}}{5}-2\text{mk}} [$Using $(1)]$
$=\sqrt{\frac{\sum\text{x}_\text{i}^2}{5}-\text{m}^2}$
$=\text{s}$
Hence, the correct answer is option $(a).$
View full question & answer→MCQ 91 Mark
The mean of $9$ observations is $36.$ If the mean of the first $5$ observations is $32$ and that of the last $5$ observations is $39,$ then the fifth observation is $............$
AnswerMean of $9$ observations $= 36$
$\Rightarrow$ Sum of these $9$ observations $= 324$
Sum of first five observations $= 32 \times 5 = 160$
Sum of last five observations $= 39 \times 5 = 195$
Fifth observation $=$ Sum of first five observations $+$ Sum of last five observations $-$ Sum of all $9$ observations
$= 160 + 195 - 324$
$= 31$
View full question & answer→MCQ 101 Mark
If $n = 10, \overline{\text{X}}=12$ and $\sum\text{x}_\text{i}^2=1530,$ then the coefficient of variation is:
- A
$36\%$
- B
$41\%$
- ✓
$25\%$
- D
$25\%$
AnswerCorrect option: C. $25\%$
Standard deviation is expressed in the following manner:
$\sigma=\sqrt{\frac{1}{\text{n}}\sum_\text{i}\text{x}_\text{i}^2-(\overline{\text{X}})^2}$
$=\sqrt{\frac{1530}{10}-(12)^2}$
$=\sqrt9$
$=3$
$\text{CV}=\frac{\sigma}{\overline{\text{X}}}\times100$
$=\frac{3}{12}\times100$
$=25\%$
View full question & answer→MCQ 111 Mark
The difference between the maximum and the minimum obervations in data is called the $...........$
AnswerIn arithmetic, the range of a set of data is the difference between the largest and smallest values.
So, difference between minimum and maximum values is called range.
View full question & answer→MCQ 121 Mark
The mean of $8$ numbers is $25$ if each number is multiplied by $2$ the new mean will be:
AnswerMean of $8$ numbers$=25$
$\therefore\ \text{A.M}=\frac{\sum \text{x}}{\text{n}}$
$\Rightarrow 25=\dfrac{\sum \text{x}}{8}$
$\Rightarrow∑\text{x}=25×8=200$
If each number is multiply by $2$ then new sum
$=200\times 2=400$
$\therefore \text{New mean}=8400=50$
View full question & answer→MCQ 131 Mark
Let $x_1, x_2, ..., x_n$ be $n$ observations. Let $y_i = ax_i + by_i + b$ for $i = 1, 2, 3, ..., n,$ where $a$ and $b$ are constants. If the mean of $x_i's$ is $48$ and their standard deviation is $12,$ the mean of $y_i's$ $55$ and standard deviation of $y_i's$ is $15,$ the values of $a$ and $b$ are:
- ✓
$a = 1.25, b = -5$
- B
$a = -1.25, b = 5$
- C
$a = 2.5, b = -5$
- D
$a = 2.5, b = -5$
AnswerCorrect option: A. $a = 1.25, b = -5$
It is given that $y_i = ax_i + b$ for $i = 1, 2, 3, ..., n$, where $a$ and $b$ are constants.
$\overline{\text{x}_\text{i}}=48$ and $\sigma_{\text{x}_\text{i}}=12$
$\overline{\text{y}_\text{i}}=55$ and $\sigma_{\text{y}_\text{i}}=15$
$\text{y}_\text{i}=\text{ax}_\text{i}+\text{b}$
$\Rightarrow\frac{\sum\text{y}_\text{i}}{\text{n}}=\frac{\sum(\text{ax}_\text{i}+\text{b})}{\text{n}}$
$\Rightarrow\frac{\sum\text{y}_\text{i}}{\text{n}}=\text{a}\frac{\sum\text{x}_\text{i}}{\text{n}}+\frac{\sum\text{b}}{\text{n}}$
$\Rightarrow\overline{\text{y}_\text{i}}=\text{a}\overline{\text{x}_\text{i}}+\text{b}$
$\Rightarrow55=48\text{a}+\text{b}\ ...(1)$
Now,
Standard deviation of $y_i =$ Standard deviation of $ax_i + b$
$\Rightarrow\sigma_{\text{y}_\text{i}}=\text{a}\times\sigma_{\text{x}_\text{i}}$
$\Rightarrow15=12\text{a}$
$\Rightarrow\text{a}=\frac{15}{12}=1.25$
Putting $a = 1.25$ in $(1),$ we get
$b = 55 - 48 \times 1.25$
$= 55 - 60$
$= -5$
Thus, the values of $a$ and $b$ are $1.25$ and $-5,$ respectively.
Hence, the correct answer is option $(a).$
View full question & answer→MCQ 141 Mark
Let set $M = \{ x, 2x, 4x \}$ for any number $x.$ If the average $($arithmetic mean$)$ of the numbers in set $M$ is $14,$ find the value of $x:$
AnswerGiven set $M = \{x, 2x, 4x\}$
Average $($arithmetic mean$)$ of the numbers in set $M$ is $14.$
Value of $x$ will be,
$\Rightarrow \frac{\text{x}\ +\ 2\text{x}\ +\ \text{4x}}{3}$
$\Rightarrow7\text{x}=42$
$\Rightarrow\text{x}=6$
View full question & answer→MCQ 151 Mark
Given the list of numbers $\{1, 6, 3, 9, 16, 11, 2, 9, 5, 712, 13, 8\}$ what is the median?
AnswerGiven list is $\{1, 6, 3, 9, 16, 11, 2, 9, 5, 712, 13, 8\}$ Arrange given set of integers in ascending order.
Then, we have $\{1, 2, 3, 5, 6, 8, 9, 11, 13, 16, 712\}$
The middle number of the set is $8.$
Therefore the median is $8.$
View full question & answer→MCQ 161 Mark
The age of $13$ school students are listed below. Find the median: $12, 9, 8, 13, 15, 14, 6, 18, 7, 11, 9, 14, 10$
AnswerThe median of a set of data is the middlemost number in the set.
So, first arrange the data in order.
$6, 7, 8, 9, 10, 10, 11, 12, 13, 14, 14, 15, 18$
The median is $11.$
View full question & answer→MCQ 171 Mark
The mean of the squares of the first $n$ natural numbers is:
- A
$\displaystyle {\text{n}}^{2}+1\text{n}2+1$
- B
$\displaystyle \frac{\text{n}^{4}+1}{\text{n}}$
- ✓
$\displaystyle \frac{\left ( \text{n}+1 \right )\left ( 2\text{n}+1 \right )}{6}$
- D
$\displaystyle \frac{\left ( \text{n}+1 \right )\left ( 2\text{n}+1 \right )}{5}$
AnswerCorrect option: C. $\displaystyle \frac{\left ( \text{n}+1 \right )\left ( 2\text{n}+1 \right )}{6}$
The first natural numbers are $1,2,3,......n$
Their square are $1\ 1^2,2^2,3^2......\text{n}^2$
$\text{Mean}=\dfrac{1^2+2^2+3^2+........+\text{n}^2}{\text{n}}$
$=\dfrac{\text{n}(\text{n}+1)(2\text{n}+1)}{6\text{n}}$
$\therefore$ Mean $=n\ 12 + 22 + 32 +........+ n^2$
$\therefore$ square of $n$ natural numbers is $ =\dfrac{\text{n}(\text{n}+1)(2\text{n}+1)}{6}$
$\text{Mean}=\dfrac{(\text{n}+1)(2\text{n}+1)}{6}$
View full question & answer→MCQ 181 Mark
Kavita obtained $16, 14, 18$ and $20$ marks $($out of $25)$ in maths in weekly test in the month of Jan $2000;$ then mean marks of Kavita is:
AnswerNo. of test in the month Jan $2000 = 4$ Total Marks obtained in $4$ test
$= 16 + 14 + 18 + 20$
$=68$
$\therefore \text{A.M}=\frac{\sum \text{x}}{\text{n}}=\frac{68}{4}=17$
View full question & answer→MCQ 191 Mark
Choose the correct answer. Let $a, b, c, d, e$ be the observations with mean $m$ and standard deviation $s.$ The standard deviation of the observations $a + k, b + k, c + k, d + k, e + k$ is:
- ✓
$s$
- B
$ks$
- C
$s + k$
- D
$s - k$
AnswerGiven observations are $a, b, c d$ and $e.$
$\text{Mean}=\text{m}=\frac{\text{a}+\text{b}+\text{c}+\text{d}+\text{e}}{5}$
$\sum\text{x}_\text{i}={\text{a}+\text{b}+\text{c}+\text{d}+\text{e}}={5}\text{m}$
Now, $\text{mean}=\frac{\text{a}+\text{k}+\text{b}+\text{k}+\text{c}+\text{k}+\text{d}+\text{k}+\text{e}+\text{k}}{5}$
$=\frac{(\text{a}+\text{b}+\text{c}+\text{d}+\text{e})+5\text{k}}{5}=\text{m}+\text{k}$
$\therefore\ \text{SD}=\sqrt{\frac{{\sum(\text{x}_\text{i}^2+\text{k}^2+2\text{k}\text{x}_\text{i})}}{\text{n}}-(\text{m}^2+\text{k}^2+2\text{mk})}$
$=\sqrt{\frac{\sum\text{x}_\text{i}^2}{\text{n}}-\text{m}^2+\frac{2\text{k}\sum\text{x}_\text{i}}{\text{n}}-2\text{mk}}$
$=\sqrt{\frac{\sum\text{x}_\text{i}^2}{\text{n}}-\text{m}^2+2\text{km}-2\text{mk}}$ $\Big[\because\ \frac{\sum\text{x}_\text{i}}{\text{n}}=\text{m}\Big]$
$=\sqrt{\frac{\sum\text{x}_\text{i}^2}{\text{n}}-\text{m}^2}$
$=\text{s}$
View full question & answer→MCQ 201 Mark
A company produces on an average $4000$ items per month for the first $3$ months. How many items it must produce on an average per month over the next $9$ months, to average $4375$ items per month over the whole?
- ✓
$4500$
- B
$4600$
- C
$4670$
- D
$4680$
AnswerCorrect option: A. $4500$
Total production has to be $4375 \times 12 = 52500$
First three months production is $4000 \times 3 = 12000$
Total production has to be in remaining $9$ months $= 52500 - 12000 = 40500$
Average production per month in remaining $9$ months $= \frac{40500}{9} = 4500$
View full question & answer→MCQ 211 Mark
If the standard deviation of a variable $X$ is $\sigma,$ then the standard deviation of variable $\frac{\text{aX+b}}{\text{c}}$ is:
- A
$\text{a}\ \sigma$
- B
$\frac{\text{a}}{\text{c}}\sigma$
- ✓
$\Big|\frac{\text{a}}{\text{c}}\Big|\sigma$
- D
$\Big|\frac{\text{a}}{\text{d}}\Big|\sigma$
AnswerCorrect option: C. $\Big|\frac{\text{a}}{\text{c}}\Big|\sigma$
$\text{Y}=\frac{\text{aX+b}}{\text{c}}$
$\overline{\text{Y}}=\frac{\sum\text{y}_\text{i}}{\text{n}}=\frac{\frac{\text{a}\sum\text{X}+\text{nb}}{\text{c}}}{\text{n}}$
$=\frac{\text{a}\sum\text{X}}{\text{nc}}+\frac{\text{nb}}{\text{nc}}$
$=\frac{\text{a}\overline{\text{X}}}{\text{c}}+\frac{\text{b}}{\text{c}}$
$\text{Var}(\text{X})=\frac{\sum\big(\text{x}_\text{i}-\overline{\text{X}}\big)^2}{\text{n}}$
$=\sigma^2$
$\text{Var}(\text{Y})=\frac{\sum\big(\text{y}_\text{i}-\overline{\text{Y}}\big)^2}{\text{n}}$
$=\frac{\sum\Big(\frac{\text{aX}}{\text{c}}+\frac{\text{b}}{\text{c}}-\frac{\text{a}}{\text{c}}\overline{\text{X}}-\frac{\text{b}}{\text{c}}\Big)}{\text{n}}$
$=\frac{\sum\Big(\frac{\text{aX}}{\text{c}}-\frac{\text{a}}{\text{c}}\overline{\text{X}}\Big)^2}{\text{n}}$
$=\Big(\frac{\text{a}}{\text{c}}\Big)^2\frac{\sum\big(\text{x}_1-\overline{\text{X}}\big)^2}{\text{n}}$
$=\Big(\frac{\text{a}}{\text{c}}\Big)^2\sigma^2$
$\text{SD}(\sigma)=\sqrt{\Big(\frac{\text{a}}{\text{c}}\Big)^2\sigma^2}$
$=\Big|\frac{\text{a}}{\text{c}}\Big|\sigma$
View full question & answer→MCQ 221 Mark
Choose the correct answer. The standard deviation of some temperature data in $^\circ C$ is $5.$ If the data were converted into $^\circ F,$ the variance would be:
AnswerGiven that $\sigma_\text{c}=5$
We know that $\text{C}=\frac{5}{9}(\text{F}-32)$
$\Rightarrow\text{F}=\frac{9\text{C}}{5}+32$
$\therefore\ \sigma_\text{F}=\frac{9}{5}\sigma_\text{c}=\frac{9}{5}\times5=9$
$\therefore\ \sigma^2_{\text{F}}=(9)^2=81$
View full question & answer→MCQ 231 Mark
A measure of central location which splits the data set into two equal groups is called the:
AnswerMedian is the middle most value of a series.
So it divides a series of observations into two equal parts where $50\%$ of the observations are below.
The median value and other $50\%$ are above the median value.
View full question & answer→MCQ 241 Mark
The average of $2, 4, 6, 8, 10$ is $.........$
Answer$\text{ Average} = \displaystyle \frac{2 + 4 + 6 + 8 + 10}{5} = 6$
View full question & answer→MCQ 251 Mark
If $i < m < n,$ then the median of the list $i, m, n$ is $..........$
AnswerThe median of a set of data is the middlemost number in the set.
So, the median of the list $i, m, n$ is $m.$
View full question & answer→MCQ 261 Mark
Find the mean of the first five multiples of $7.$
AnswerThe first five multiples of $7$ are $7, 14, 21, 28$ and $35.$
$\text{Required mean }= \dfrac{7+14+21+28+35}{7}=\dfrac{105}{7}=15$
View full question & answer→MCQ 271 Mark
Find the mean of: $9, 11, 12, 4$ and $7$
Answer$\text{ Mean} = \frac{9+11+12+4+7}{5}$
$ \text{Mean} = \dfrac{43}{5}= 8.6$
View full question & answer→MCQ 281 Mark
The average of monthly salary of fifteen employees in a company is $Rs. 9450$. If the supervisors salary is added, the average salary increase by $Rs. 650$ What is the salary of the supervisor?
- A
$Rs.19,850$
- ✓
$Rs.20,050$
- C
$Rs. 20,250$
- D
$Rs. 20,205$
AnswerCorrect option: B. $Rs.20,050$
Average salary of $15$ employees $= Rs. 9450$
Sum of the salaries of $15$ employees $= 15 \times 9450 = 141750$
New average after adding salary of supervisor $= 9450 + 650 = 10100$
Sum of salaries of $16$ employees $= 10100 \times 16 = 1616600$
Let the salary of the supervisor $= x$
Thus. $x + 141750 = 161600$
$x = 19,850$
View full question & answer→MCQ 291 Mark
Variance of the distribution $73, 77, 81, 85, ...,113$ is:
View full question & answer→MCQ 301 Mark
In a triangle, the side lengths are $a = 5, b = 3$ and $c = 2.$ Find the length of the median drawn to the side $c:$
AnswerMedian of length drawn to the side, $\text{c}= \frac{1}{2}\sqrt{\left (2(\text{a}^{2}+\text{b}^{2})-\text{x}\text{c}^{2}\right)}$
$= \frac{1}{2}\sqrt{\left (2(5^{2}+3^{2})-2^{2}\right)} $
$= \frac{1}{2}\sqrt{\left (2(34)-2^{2}\right)}$
$= \frac{1}{2}\sqrt{\left (68-4\right)} $
$= \frac{1}{2}\sqrt{64} $
$= \frac{8}{2}$
$=4$
View full question & answer→MCQ 311 Mark
The mean deviation of the series $a, a + d, a + 2d, ..., a + 2n$ from its mean is:
- A
$\frac{(\text{n}+1)\text{d}}{2\text{n}+1}$
- B
$\frac{\text{n}\text{d}}{2\text{n}+1}$
- ✓
$\frac{\text{n}(\text{n}+1)\text{d}}{2\text{n}+1}$
- D
$\frac{\text{n}(\text{n}+1)\text{}}{2\text{n}+1}$
AnswerCorrect option: C. $\frac{\text{n}(\text{n}+1)\text{d}}{2\text{n}+1}$
| $x_i$ |
$\Big|\text{x}_\text{i}-\overline{\text{X}}\Big|=\big|\text{x}_\text{i}-(\text{a}+\text{nd})\big|$ |
| $a$ |
$nd$ |
| $a + d$ |
$(n - 1)d$ |
| $a + 2d$ |
$(n - 2)d$ |
| $a + 3d$ |
$(n - 3)d$ |
| $:$ |
$:$ |
| $:$ |
$:$ |
| $a + (n + 1)d$ |
$d$ |
| $a + nd$ |
$0$ |
| $a + (n + 1)d$ |
$d$ |
| $:$ |
$:$ |
| $:$ |
$:$ |
| $a + 2^{nd}$ |
$nd$ |
| $\sum\text{x}_\text{i}=(2\text{n}+1)(\text{a}+\text{nd})$ |
$\sum\Big|\text{x}_\text{i}-\overline{\text{X}}\Big|=\text{n}(\text{n}+1)\text{d}$ |
Therefore are $2n + 1$ terms.
$\Rightarrow N = 2n + 1$
$\sum\text{x}_\text{i}=\text{a}+\text{a}+\text{d}+\text{a}+2\text{d}+\text{a}+3\text{d}+...+\text{a}+2^\text{nd}$
$=(2\text{n}+1)\text{a}+\text{d}(1+2+3+...+2\text{n}) [a + a + a + ...(2n + 1)$ times $= (2n + 1)a]$
$=(2\text{n}+1)\text{a}+\frac{2\text{n}(2\text{n}+1)\text{d}}{2}$ $\Big[$Sum of the first $n$ natural numbers is $\frac{\text{n}(\text{n}+1)}{2},$ but here we are considering$\Big]$
$=(2\text{n}+1)\text{a}+(2\text{n}+1)^\text{nd}$
$=(2\text{n}+1)(\text{a}+\text{nd})$
$\overline{\text{X}}=\frac{(2\text{n}+1)(\text{a}+\text{nd})}{(2\text{n}+1)}$
$=\text{a}+\text{nd}$
$\sum\big|\text{x}_\text{i}-\overline{\text{X}}\big|=\text{nd}+(\text{n}-1)\text{d} (\text{n}-2)\text{d}\\+...+\text{d}+0+\text{d}+2\text{d}+3\text{d}+...+\text{nd}$
$=\text{d}(\text{n}+(\text{n}-1)+(\text{n}-2)+...+1)\\ \ +0+\text{d}(1+2+3+....+\text{n})$
$=\frac{\text{dn}(\text{n}+1)}{2}+\frac{\text{dn}(\text{n}+1)}{2}$ $\Big\{\because1+2+3+....+\text{n}=\frac{\text{n}(\text{n}+1)}{2}\Big\}$
$=\text{n}(\text{n}+1)\text{d}$
Mean deviation about the mean $=\frac{\sum\Big|\text{x}_\text{i}-\overline{\text{X}}\Big|}{\text{N}}$
$=\frac{\text{n}(\text{n}+1)\text{d}}{(2\text{n}+1)}$ View full question & answer→MCQ 321 Mark
The mean of first five prime numbers is:
AnswerThe first five prime numbers are $2, 3, 5, 7, 11$
$\text{Mean}=\frac{\text{sum of the terms}}{\text{no. of terms}}$
$\text{Mean}=\frac{2+3+5+7+11}{5}$
$=\frac{28}{5}$
$=5.6$
View full question & answer→MCQ 331 Mark
Choose the correct answer. The following information relates to a sample of size $60 \sum\text{x}^2=18000$ and $\sum\text{x}=960,$ then the variance is:
AnswerWe know that variance $(\sigma^2)\frac{\sum\text{x}_\text{i}^2}{\text{N}}-\Big(\frac{\sum\text{x}_\text{i}}{\text{N}}\Big)^2$
$=\frac{18000}{60}-\Big(\frac{960}{60}\Big)^2$
$=300-256$
$=44$
View full question & answer→MCQ 341 Mark
In a class of $100$ students there are $70$ boys whose average marks in a subject are $75$ If the average marks of whole class is $72$ then what is the average marks of the girls?
AnswerTotal students $= 100$ Average marks $=72$
Total marks of the class $= 72 \times 100 = 7200$
Total marks of the boys $= 70 \times 75 = 5250$
Total marks of the girls $= 7200 = 5250 = 1950$
Average marks of the girls $ = \dfrac{1950}{30}=65$
hence, option $A$ is correct.
View full question & answer→MCQ 351 Mark
The sum of the squares deviations for $10$ observations taken from their mean $50$ is $250.$ The coefficient of variation is:
- ✓
$10\%$
- B
$40\%$
- C
$30\%$
- D
$50\%$
AnswerCorrect option: A. $10\%$
We have:
$\overline{\text{X}}=50,\ \text{n}=10$
$\sum\limits^{10}_{\text{i}=1}\Big(\text{x}_\text{i}-\overline{\text{X}}\Big)^2=250$
$\therefore\text{SD}=\sqrt{\text{Variance of X}}$
$=\sqrt{\frac{\sum\limits^{10}_{\text{i}=1}\Big(\text{x}_\text{i}-\overline{\text{X}}\Big)^2}{\text{n}}}$
$=\sqrt{\frac{250}{10}}$
$=5$
Using $\text{CV}=\frac{\sigma}{\overline{\text{X}}}\times100$
$\Rightarrow\text{CV}=\frac{5}{50}\times100$
$=10\%$
View full question & answer→MCQ 361 Mark
The mean age of $30$ student is $9$ years. If the age of their teacher is included, it becomes $10$ years. The age of teacher $($in years$)$ is:
AnswerGiven : Average age of $30$ students $= 9$ years.
Total age of $3030$ students $= 9 \times 30 = 270$ years.
Teachers age included.
So, average age of $30$ students $+$ one teacher $= 10$ years.
$\Rightarrow$ Total age of $30$ students $+$ one teacher $= 10 \times 31 = 310$ years.
$\therefore$ age of teacher $= 310 - 270 = 40$ years.
View full question & answer→MCQ 371 Mark
Means of a set of $60$ values is $23,$ if $4$ is added to each these values the the new mean is:
AnswerNew mean $= x\sim = 23 + 4 = 27$
View full question & answer→MCQ 381 Mark
The following observations have been arranged in ascending order. If the median of the data is $78,$ find the value of $x. 44, 47, 63, 65, x + 13, 87, 93, 99, 110.$
AnswerThe series in ascending order is: $44, 47, 63, 65, x + 13, 87, 93, 99, 110.$
The series has $9$ observations.
hence, the middle observation will be the median of the series.
Here, $x + 13$ is the middle observation
Therefore, $x + 13 = 78$
$x = 65$
View full question & answer→MCQ 391 Mark
Median of $15, 28, 72, 56, 44, 32, 31, 43$ and $51$ is $43:$
AnswerThe terms are: $15, 28, 72, 56, 44, 32, 31, 43$ and $51.$
Arranging them in ascending order: $15, 28, 31, 32, 43, 44, 51, 56, 72$
Since the total number of terms is odd that is $9,$
therefore the median will be the middle term that is the $5^{th}$ term which is $43.$
View full question & answer→MCQ 401 Mark
The attendance of a class of $45$ boys for $10$ days is given as $40, 30, 35, 45, 44, 41, 38, 44$ and $41$ then the mean attendance of a class is:
AnswerIn this question one day attendance not givenGiven attendance as per Answer.
are $40, 42, 30, 35, 45, 44, 41, 38, 44$ and $41$ Then mean
$=\frac{40+42+30+35+45+44+41+38+44+41}{10}$
$=\frac{400}{10}$
$=40$
View full question & answer→MCQ 411 Mark
The mean of $864, 874, 884, 1000$ and $1008$ is:
- A
$928$
- B
$1010$
- ✓
$926$
- D
$927$
AnswerFormula,
$\cfrac{\sum {\text{x }} }{N}=\cfrac{864+874+884+1000+1008}{5}$
$=\frac{4630}{5}$
$=926$
View full question & answer→MCQ 421 Mark
The daily sale of milk $($in litres$)$ in a ration shop for eight days is as follows$-$
$60, 40, 10, 40, 4, 70, 30$ and $10.$ The average daily sale is:
AnswerBy definition of average,
$=\cfrac{60+40+10+40+4+70+30+10}{8}$
$=\cfrac{264}{8}$
$=33$
View full question & answer→MCQ 431 Mark
A set $F,$ which contains the elements $4, 5, 11, 13, 16, 18,$ and $x.$ If both the median and average $($arithmetic mean$)$ of Set $F$ equal $11,$ what must be the value of $x\ ?$
AnswerGiven, set $f = 4, 5, 11, 13, 16, 18, x$ Average of set
$f = 11$
$\Rightarrow \dfrac{4+5+11+13+16+18+\text{x}}{7}=11$
$\Rightarrow \frac{67+\text{x}}{7}=11$
$\Rightarrow 67+{\text{x}}=77$
$\Rightarrow \text{x}=77-67$
$=10$
View full question & answer→MCQ 441 Mark
$...........$ is the most frequently observed data value:
AnswerThe mode is the value thats repeated the maximum number of times in the data set.
A worked example: Marks obtained in an examination is
given as $5, 9, 7, 12, 15, 7, 5, 7, 7, 8, 7$
We identify the number that is repeated the maximum number of times as: $7 ($repeated $5$ times$).$
Thus the mode for this data set is $7.$
View full question & answer→MCQ 451 Mark
Mean of $10$ values is $32.6.$ If another values is included the mean becomes $31.$ The included value is:
AnswerIncluded value $= 31 \times 11 − 32.6 \times 10 = 15$
View full question & answer→MCQ 461 Mark
The mean of the cubes of the first $n$ natural numbers is:
- A
$ \displaystyle \frac{\text{n}\left (\text{ n}+1 \right )^{2}}{2}$
- ✓
$ \displaystyle \frac{\text{n}\left ( \text{n}+1 \right )^{2}}{4}$
- C
$ \displaystyle \frac{\text{n}\left ( \text{n}+1 \right )\left ( \text{n}+2 \right )}{7}$
- D
$ \displaystyle \frac{\text{n}\left ( \text{n}+1 \right )\left ( \text{n}+2 \right )}{8}$
AnswerCorrect option: B. $ \displaystyle \frac{\text{n}\left ( \text{n}+1 \right )^{2}}{4}$
First $n$ natural numbers are $1,2,3,4,......,n$
$\therefore \text{Mean}=\dfrac{1^3+2^3+3^3+.......+\text{n}^3}{\text{n}}$
Sum of the cubes of $n$ natural numbers $ =\left(\frac{\text{n}(\text{n}+1)}{2}\right)^2$
$\therefore \text{Mean}=\left(\dfrac{\text{n}(\text{n}+1)}{2}\right)^2\times\frac{1}{\text{n}}$
$=\dfrac{\text{n}(\text{n}+1)^2}{4}$
View full question & answer→MCQ 471 Mark
let $x_1, x_2, ...,x_n$ be $n$ observations and $\overline{\text{X}}$ be their arithmetic mean. The standard deviation is given by:
- A
$\sum\limits^\text{n}_{\text{i}=1}\Big(\text{x}_\text{i}-\overline{\text{X}}\Big)^2$
- B
$\frac{1}{\text{n}}\sum\limits^\text{n}_{\text{i}=1}\Big(\text{x}_\text{i}-\overline{\text{X}}\Big)^2$
- ✓
$\sqrt{\frac{1}{\text{n}}\sum^\text{n}_{\text{i}=1}\Big(\text{x}_\text{i}-\overline{\text{X}}\Big)^2}$
- D
$\sqrt{\frac{1}{\text{n}}\sum^\text{n}_{\text{i}=1}\Big(\text{x}_\text{i}-\overline{\text{X}}\Big)}$
AnswerCorrect option: C. $\sqrt{\frac{1}{\text{n}}\sum^\text{n}_{\text{i}=1}\Big(\text{x}_\text{i}-\overline{\text{X}}\Big)^2}$
It is given that $x_1, x_2, ...,x_n$ be $n$ observations and $\overline{\text{X}}$ be their arithmetic mean.
The standard deviation is given observations is $\sqrt{\frac{1}{\text{n}}\sum^\text{n}_{\text{i}=1}\Big(\text{x}_\text{i}-\overline{\text{X}}\Big)^2}.$
Also,
$\sqrt{\frac{1}{\text{n}}\sum^\text{n}_{\text{i}=1}\Big(\text{x}_\text{i}-\overline{\text{X}}\Big)^2}=\sqrt{\frac{1}{\text{n}}\sum^\text{n}_{\text{i}=1}\text{x}_\text{i}^2-\overline{\text{X}}^2}$
Hence, the correct answers are options $(c)$ and $(d).$
Disclaimer: For option $(c)$ to be the only correct answer, option $(d)$ should be different from the given value.
View full question & answer→MCQ 481 Mark
The daily sale of kerosene $($in litres$)$ in a ration shop for six days is as follows: $75, 120, 12, 50, 70.5$ and $140.5$ The average daily sale is:
Answer$\text{Mean}=\frac{75+120+12+50+70.5+140.5}{6}= 78$
The average daily sale is therefore the mean $= 78.$
View full question & answer→MCQ 491 Mark
Choose the correct answer. Mean deviation for n observations $x_1, x_2, ...,x_n$ from their mean $x$ is given by:
- A
$\sum\limits^{\text{n}}_{\text{i}=1}(\text{x}_\text{i}-\bar{\text{x}})$
- ✓
$\frac{1}{\text{n}}\sum\limits^{\text{n}}_{\text{i}=1}|\text{x}_\text{i}-\bar{\text{x}}|$
- C
$\sum\limits^{\text{n}}_{\text{i}=1}(\text{x}_\text{i}-\bar{\text{x}})^2$
- D
$\sum\limits^{\text{n}}_{\text{i}=1}(\text{x}_\text{i}-\bar{\text{x}})^2$
AnswerCorrect option: B. $\frac{1}{\text{n}}\sum\limits^{\text{n}}_{\text{i}=1}|\text{x}_\text{i}-\bar{\text{x}}|$
$\text{MD}=\frac{1}{\text{n}}\sum\limits^{\text{n}}_{\text{i}=1}|\text{x}_\text{i}-\bar{\text{x}}|$
View full question & answer→MCQ 501 Mark
The average marks of boys in a class is $52$ and that of girls is $42.$ The average marks of boys and girls combined is $50.$ The percentage of boys in the class is $-$
AnswerLet the number of boys and girls be $x$ and $y.$
$\therefore 52\text{x}+42\text{y}=50(\text{x}+{y})$
$\Rightarrow 52\text{x}+42\text{y}=50\text{x}+50\text{y}$
$\Rightarrow 2\text{x}=8{\text{y}} = \text{x }{ =4y} $
$ \therefore$Total number of students in class
$=x + y = 4y + y = 5y$
$ \therefore$ Required $\%$ of boys
$=\frac { 4\text{y} }{ 5\text{y} } \times 100=80\%$
View full question & answer→MCQ 511 Mark
The most frequently occurring data value in a data set is the $...........$
AnswerMode is the highest occurring figure in a series.
It is the value in a series of observation that repeats maximum number of times and,
which represents the whole series as most of the values in the series revolves around this value.
View full question & answer→MCQ 521 Mark
Choose the correct answer. Let $x_1, x_2, ...,x_n$ be n observations and $\bar{\text{x}}$ be their arithmetic mean. The formula for the standard deviation is given by:
- A
$\sum(\text{x}_\text{i}-\bar{\text{x}})^2$
- B
$\frac{\sum(\text{x}_\text{i}-\bar{\text{x}})^2}{\text{x}}$
- ✓
$\sqrt{\frac{\sum(\text{x}_\text{i}-\bar{\text{x}})^2}{\text{n}}}$
- D
$\sqrt{\frac{\sum(\text{x}_\text{i}-\bar{\text{x}})^2}{\text{n}}}$
AnswerCorrect option: C. $\sqrt{\frac{\sum(\text{x}_\text{i}-\bar{\text{x}})^2}{\text{n}}}$
The formula for $\text{S.D}=\sigma=\sqrt{\frac{\sum(\text{x}_\text{i}-\bar{\text{x}})^2}{\text{n}}}$
View full question & answer→MCQ 531 Mark
Choose the correct answer. The mean deviation of the data $3, 10, 10, 4, 7, 10, 5$ from the mean is:
AnswerCorrect option: B. $2.57$
Observations are fiven by $3, 10, 10, 4, 7, 10,$ and $5$
$\therefore\ \bar{\text{x}}=\frac{3+10+10+4+7+10+5}{7}=\frac{49}{7}=7$
$\text{MD}=\frac{\sum\text{d}_\text{i}}{\text{n}}=\frac{18}{7}=2.57$
|
$x_i$
|
$\text{d}_\text{i}=|\text{x}_\text{i}-\bar{\text{x}}|$
|
| $3$ |
$4$ |
| $10$ |
$3$ |
| $10$ |
$3$ |
| $4$ |
$3$ |
| $7$ |
$0$ |
| $10$ |
$3$ |
| $5$ |
$2$ |
|
Total
|
$\sum\text{d}_\text{i}=18$
|
View full question & answer→MCQ 541 Mark
The sum $\displaystyle \sum _{\text{ r}=1 }^{ 10 }{ \left( {\text{ r} }^{ 2 }+1 \right) \times \left( \text{r}\ ! \right) }$ is equal to:
- A
$(11)!$
- ✓
$10 \times (11)!$
- C
$101 \times (10)!$
- D
$101 \times (10)!$
AnswerCorrect option: B. $10 \times (11)!$
$ \sum(\text{r}^2+1)\text{r}!=\sum[\text{r}(\text{r }+1) -(\text{r}-1)\text{r}!$
$ =\sum\limits^{10}_\text{r=1}[\text{r}(\text{r }+1)!-(\text{r}-1)\text{r}!]$
$= (1\times 2!−0\times 1!)+(2\times 3!−1\times 2!)+......+(10\times 11!−9\times 10!)$
$=10\times 11! $
View full question & answer→MCQ 551 Mark
If the first and the second letters of the word $\text{MISJUDGEMENTS}$ are interchanged with the last and the second last letters respectively, and similarly the third and the fourth letters are interchanged with the third and the fourth letters from the last respectively , and so on,then what will be the fifth letter to the right of the third letter from the left end?
Answer
| $M$ |
$I$ |
$S$ |
$J$ |
$U$ |
$D$ |
$G$ |
$E$ |
$M$ |
$E$ |
$N$ |
$T$ |
$S$ |
| $1$ |
$2$ |
$3$ |
$4$ |
$5$ |
$6$ |
$7$ |
$8$ |
$9$ |
$10$ |
$11$ |
$12$ |
$13$ |
The alphabet series is shown in the above diagram with positions.
Now, with the required interchange in positions,
as the question says, we get following letters interchanged
$1^{st} \rightarrow 13^{th}$
$2^{nd} \rightarrow 12^{th}$
$3^{rd} \rightarrow 11^{th}$
$4^{th} \rightarrow 10^{th}$
$5^{th} \rightarrow 9^{th}$
$6^{th} \rightarrow 8^{th}$
$7^{th} \rightarrow 7^{th}$
Using above mentioned interchange we get complete reversal of the letters of this word.
which is: $\text{STNEMEGDUJSIM}$
Here Third letter from the left end is $N.$
According to the question,
$\Rightarrow (5 + 3)^{th}$ letter from the left
$\Rightarrow 8^{th}$ letter from the left Hence, $8^{th}$ letter from the left end is $D.$ View full question & answer→MCQ 561 Mark
The mean of $x, x + 3, x + 4, x + 8$ and $x + 10:$
- A
$x + 4$
- B
$x + 8$
- C
$x + 3$
- ✓
$x + 5$
AnswerCorrect option: D. $x + 5$
By definition
$\text{Average} =\cfrac{\text{x}+(\text{x}+3)+(\text{x}+4)+((\text{x}+8)+(\text{x}+10)}{5}$
$=\cfrac{5\text{x}+25}{5}$
$=(\text{x}+5)$
View full question & answer→MCQ 571 Mark
The average can be found only in $............$ variables:
AnswerThe average can be found only in quantitative variables.
Example: A quantitative variable is something that
can be measured and written out as a number.
So, we can find the average marks of $2020$ students in $1212$ class but,
we can not find the average of the
cleverness of students.
View full question & answer→MCQ 581 Mark
Choose the correct answer. Coefficient of variation of two distributions are $50$ and $60,$ and their arithmetic means are $30$ and $25$ respectively. Difference of their standard deviation is:
AnswerHere, $\text{CV}_1=50,\ \text{CV}_2=60,\ \bar{\text{x}}_1=30$ and $\bar{\text{x}}_2=25$
$\therefore\ \text{CV}_1=\frac{\sigma_1}{\bar{\text{x}}_1}\times100$
$\Rightarrow50=\frac{\sigma_1}{30}\times100$
$\therefore\ \sigma_1=\frac{30\times50}{100}=15$
and $\text{CV}_2=\frac{\sigma_2}{\bar{\text{x}}_2}\times100$
$\Rightarrow60=\frac{\sigma_2}{25}\times100$
$\therefore\ \sigma^2=\frac{60\times25}{100}=15$
Now, $\sigma_1-\sigma_2=15-15=0$
View full question & answer→MCQ 591 Mark
If the mode of five observations, in order, $0, 2, 3, m, 5$ is $3$ then $m = ...........$
AnswerIf the mode of five observation, in order, $0, 2, 3, m, 5$ is $3$, then $m=3.$
View full question & answer→MCQ 601 Mark
For dealing with qualitative data the best average is:
AnswerMedian is the middle most value.
Also for even number of observations, median is the average of to middle values.
Hence, it divides the whole series into two equal halv.
It gives the more accurate and best average for qualitative data.
View full question & answer→MCQ 611 Mark
If the arithmetic mean of $7, 8, x, 11, 14$ is $x,$ then $x:$
AnswerWe have,
$= \frac{7+8+\text{x}+11+14}{5}$
$= \text{x} = 40+\text{x} = 5\text{x} = \text{x} = 10$
View full question & answer→MCQ 621 Mark
The mean of $100$ observations is $50$ and their standard deviation is $5.$ The sum of all squares of all the observations is:
- A
$50,000$
- B
$250,000$
- ✓
$252500$
- D
$25000$
AnswerCorrect option: C. $252500$
Let $\overline{\text{x}}$ and $\sigma$ be the mean and standard deviation of $100$ observations, respectively.
$\therefore\overline{\text{x}}=50,\ \sigma=5$ and $n = 100$
$\text{Mean},\ \overline{\text{x}}=50$
$\Rightarrow\frac{\sum\text{x}_\text{i}}{100}=50$
$\Rightarrow\sum\text{x}_\text{i}=5000\ ...(1)$
Now,
Standard deviation, $\sigma=5$
$\Rightarrow\sqrt{\frac{\sum\text{x}_\text{i}^2}{100}-\Big(\frac{\sum\text{x}_\text{i}}{100}\Big)^2}=5$
$\Rightarrow\frac{\sum\text{x}_\text{i}^2}{100}-\Big(\frac{5000}{100}\Big)^2=25 [$From $(1)]$
$\Rightarrow\frac{\sum\text{x}_\text{i}^2}{100}=25+2500=2525$
$\Rightarrow{\sum\text{x}_\text{i}^2}=252500$
Thus, the sum of all squares of all the observations is $252500.$
Hence, the correct answer is option $(c).$
View full question & answer→MCQ 631 Mark
The following data has been arranged in ascending order. If their median is $63,$ find the value of $x.34, 37, 53, 55, x, x + 2, 77, 83, 89$ and $100.$
AnswerThe series in ascending order is: $34, 37, 53, 55, x, x + 2, 77, 83, 89$ and
The series has $10$ numbers, even numbers.
Hence, the median will be the mean of the two middle numbers:
median $=$ mean of $5^{th}$ and $6^{th}$ terms
$63=\frac{\text{x}\ + \ \text{x}\ +\ 2}{2}$
$1260=2\text{x}+2$
$2\text{x}=124$
$\text{x}=62$
View full question & answer→MCQ 641 Mark
If $150$ is the mean of $200$ observations and $100$ is the mean of some $300$ other observations, find the mean of the combination:
AnswerMean of $200$ observations $= 150$
Sum of $200$ observations $= 200 \times 150 = 30000$
Mean of $300$ observations $= 100$
Sum of $300$ observations $= 300 \times 100 = 30000$
Total Sum $= 30000 + 30000 = 60000$
Number of observations $= 100 + 200 = 500$
$\text{Mean} = \frac{\text{Sum}}{\text{Number of observations}}$
$= \frac{60000}{500}$
$= 120$
View full question & answer→MCQ 651 Mark
if $x_1, x_2, x_3, x_4, x_5$ are five consecutive odd numbers, then their average is:
AnswerThe five consecutive odd numbers are $\text{x}_1+ \text{x}_1+2, \text{x}_1 + 4,\text{x}_1 +6,\text{x}_1 +8$
$\therefore \text{mean}=\frac{\text{x}_1 \ + \ \text{x}_1 \ +\ 2+\text{x}_1 \ +\ 4\ +\ \text{x}_1 \ +\ 6\text{x}_1 \ +\ 8}{6}$
$=\frac{5\text{x}_1\ +\ 20}{5}$
$=\text{x}_1\ +\ 4$
$=\text{x}_3$
View full question & answer→MCQ 661 Mark
The standard deviation of the observations $6, 5, 9, 13, 12, 8, 10$ is:
- A
$6$
- B
$\sqrt6$
- C
$\frac{52}{7}$
- ✓
$\sqrt{\frac{52}{7}}$
AnswerCorrect option: D. $\sqrt{\frac{52}{7}}$
The given observations are $6, 5, 9, 13, 12, 8, 10.$
Now,
$\sum\text{x}_\text{i}= 6 + 5 + 9 + 13 + 12 + 8 + 10 = 63$
$\sum\text{x}_\text{i}^2=36 + 25 + 81 + 169 + 144 + 64 + 100 = 619$
$\therefore$ Standard deviation of the observations, $\sigma$
$=\sqrt{\frac{1}{\text{N}}\sum\text{x}_\text{i}^2-\Big(\frac{1}{\text{N}}\sum\text{x}_\text{i}\Big)^2}$
$=\sqrt{\frac{1}{7}\times619-\Big(\frac{1}{7}\times63\Big)^2}$
$=\sqrt{\frac{619}{7}-81}$
$=\sqrt{\frac{619-567}{7}}$
$=\sqrt{\frac{52}{7}}$
Hence, the correct answer is option $(d).$
View full question & answer→MCQ 671 Mark
The average of a collection of $20$ measurements was calculated to be $56 \ cm.$ But later it was found that a mistake occured in one of the measurements which was recorded as $64 \ cm$ but should have been $61 \ cm.$ What is the correct average?
- A
$39.55\ cm$
- B
$29.55 \ cm$
- C
$55.85 \ cm$
- ✓
$55.85 \ cm$
AnswerCorrect option: D. $55.85 \ cm$
Incorrect total of $20$ measurement $= 20 \times 56 = 1120$
Correct total $= 1120 - 64 + 61 = 1117$
$\therefore \text{Correct average} = \displaystyle \frac{1117}{20} = 55.85\ \text{cm}$
View full question & answer→MCQ 681 Mark
If the mean of five observations $x ,x + 2, x + 4, x + 6$ and $x + 8$ is $11,$ then the mean of last three obsevations is:
AnswerGiven observations $x, x + 2, x + 4, x + 6, x + 8$
$\Rightarrow\frac{5\text{x}+ 20}{5}=11$
$\Rightarrow x = 7$
So, the observations are $7, 9, 11, 13, 15$
Req. mean $=\frac{11+13+15}{3}=13$
View full question & answer→MCQ 691 Mark
The standard deviation of the data:
| $x$ |
$1$ |
$a$ |
$a^2$ |
....
|
$a^n$ |
| $f$ |
${ }^n C_0$ |
${ }^n C_1$ |
${ }^n C_2$ |
....
|
${ }^n C_2$ |
is, - A
$\Big(\frac{1+\text{a}^2}{2}\Big)^\text{n}-\Big(\frac{1+\text{a}}{2}\Big)^\text{n}$
- B
$\Big(\frac{1+\text{a}^2}{2}\Big)^{2\text{n}}-\Big(\frac{1+\text{a}}{2}\Big)^\text{n}$
- C
$\Big(\frac{1+\text{a}^2}{2}\Big)^{2\text{n}}-\Big(\frac{1+\text{a}^2}{2}\Big)^\text{n}$
- ✓
$\Big(\frac{1+\text{a}^2}{2}\Big)^{2\text{n}}-\Big(\frac{1+\text{a}^2}{2}\Big)^\text{n}$
AnswerCorrect option: D. $\Big(\frac{1+\text{a}^2}{2}\Big)^{2\text{n}}-\Big(\frac{1+\text{a}^2}{2}\Big)^\text{n}$
|
$x_i$
|
$f_i$
|
$f_ix_i$
|
${x_i}^2$
|
${f_ix_i}^2$
|
|
1
|
${ }^n C_0$ |
${ }^n C_0$ |
1
|
1
|
|
a
|
${ }^n C_1$ |
$a{ }^n C_1$ |
$a^2$ |
$a^2{ ~}^n C_1$ |
|
a
|
${ }^n C_2$ |
$a^2{ ~}^n C_2$ |
$a^4$ |
$a^4{ ~}^n C_2$ |
|
a
|
${ }^n C_3$ |
$a^3{~}^n C_3$ |
$a^6$ |
$a^6{ ~}^n C_3$ |
|
$:$
$:$
$:$
|
$:$
$:$
$:$
|
$:$
$:$
$:$
|
$:$
$:$
$:$
|
$:$
$:$
$:$
|
|
$a^n$
|
${ }^n C_n$ |
$a^n{~ }^n C_n$ |
$a^{2n}$ |
$a^{2n}{~ }^n C_n$ |
|
|
$\sum\limits_{\text{i}=1}^{\text{n}}\text{f}_\text{i}=2^\text{n}$
|
$\sum\limits_{\text{i}=1}^{\text{n}}\text{f}_\text{i}\text{x}_\text{i}=(1+\text{a})^\text{n}$
|
|
$\sum\limits_{\text{i}=1}^{\text{n}}\text{f}_\text{i}\text{x}_\text{i}^2=(1+\text{a}^2)^\text{n}$ |
Number of terms, $\text{N}=\sum\limits_{\text{i}=1}^\text{n}\text{f}_\text{i}=2^\text{n}$
$\sum\limits_{\text{i}=1}^\text{n}\text{f}_\text{i}\text{x}_\text{i}=^\text{n}\text{C}_0+\text{a}^\text{n}\text{C}_1+\text{a}^2{^\text{ n}\text{C}_2+...+\text{a}^\text{n}{^\text{ n}\text{C}_\text{n}}}=(1+\text{a})^\text{n}$
$\overline{\text{X}}=\frac{\sum\limits_{\text{i}=1}^\text{n}\text{f}_\text{i}\text{x}_\text{i}}{\text{N}}$
$=\frac{(1+\text{a})^\text{n}}{2^\text{n}}$
$\sum\limits_{\text{i}=1}^\text{n}\text{f}_\text{i}\text{x}_\text{i}^2=(1+\text{a}^2)^\text{n}$
$\sigma^2=\text{Variance}(\text{X})=\frac{1}{\text{N}}\sum\limits_{\text{i}=1}^\text{n}\text{f}_\text{i}\text{x}_\text{i}^2-\Bigg(\frac{\sum\limits_{\text{i}=1}^\text{n}\text{f}_\text{i}\text{x}_\text{i}}{\text{N}}\Bigg)^2$
$=\frac{(1+\text{a}^2)^\text{n}}{2^\text{n}}-\Big[\frac{(1+\text{a}^2)^\text{n}}{2^\text{n}}\Big]^2$
$=\Big[\frac{1+\text{a}^2}{2}\Big]^\text{n}-\Big[\frac{1+\text{a}}{2}\Big]^2\text{n}$
$\sigma=\sqrt{\text{Variance}(\text{X})}$
$=\sqrt{\Big[\frac{1+\text{a}^2}{2}\Big]^\text{n}-\Big[\frac{1+\text{a}}{2}\Big]^2\text{n}}$ View full question & answer→MCQ 701 Mark
Of the three numbers, the first is twice the second, and the second is twice the third. The average of the reciprocal of the numbers is $\frac{7}{72}$. What are the three numbers?
- ✓
$24, 12, 6$
- B
$8, 4, 2$
- C
$12, 6, 3$
- D
$12, 6, 3$
AnswerCorrect option: A. $24, 12, 6$
Let the third number be $x$ Then Second number $= 2x$
and first number $= 4x$
Sum of the reciprocals of these $3$ numbers $ =\frac{1}{\text{4x}}+\frac{1}{2\text{x}}+\frac{1}{\text{x}}=\frac{1+2+4}{4\text{x}}$
$= \frac{7}{\text{4x}}$
Given, $\frac{7}{4\text{x}}=3\times \frac{7}{72}$
$=4\text{x}=24= \text{x}=6 $
Therefore, the three numbers are,
$4 \times 6, 2 \times 6, 6$
i.e. $24, 12, 6.$
View full question & answer→MCQ 711 Mark
The weight of a body, calculated as the average of seven different experiments is $53.735g.$The average of the first three experiments is $54.005g.$ The fourth was greater than the fifth by $0.0040.004 g$ and the average of sixth and seventh was $0.010g$ less than the average of the first three. Find the weight of the body in the fourth experiment.
- A
$52.071g$
- ✓
$53.072g$
- C
$51.450g$
- D
$51.450g$
AnswerCorrect option: B. $53.072g$
View full question & answer→MCQ 721 Mark
The mean of $100$ numbers is $45.$ The mean of the last $99$ numbers is $44.$ The first number is
AnswerThe mean of $100$ numbers is $45$
$\therefore$ Sum of all $100$ numbers $= 100 \times 45 = 4500$
The mean of last $99$ numbers is $44$
$\therefore$ Sum of all last $99$ numbers $= 99 \times 44 = 4356$
$\Rightarrow$ The first number $= 4500 - 4356 = 144$
View full question & answer→MCQ 731 Mark
Which of the following is not changed for the observations $31, 48, 50, 60, 25, 8, 3x, 26, 32? ($where $x$ lies between $10$ and $15):$
View full question & answer→MCQ 741 Mark
Find the mean of first six natural numbers:
AnswerFirst six natural numbers $= 1, 2, 3, 4, 5, 6$
$\text{ Mean} = \frac{\text{Sum}}{\text{Number of observations}} $
$\text{Mean} = \frac{1 + 2 + 3 + 4 + 5 +6}{6} = \dfrac{21}{6}$
$ =3.5$
View full question & answer→MCQ 751 Mark
On Thursday, $20$ of the $25$ students in a chemistry class took a test and their average $($arithmetic mean$)$ was $80.$ On Friday, the other $5$ students took the test and their average $($arithmetic mean$)$ was $90.$ What was the average for the entire class?
AnswerAverage $= \frac{20\left ( 80 \right )\ +\ 5\left ( 90 \right )}{25}$
$=\frac{1600\ +\ 450}{25}$
$=\frac{2050}{25}$
$=82$
View full question & answer→MCQ 761 Mark
The mean deviation of the numbers $3, 4, 5, 6, 7$ from the mean is:
Answer$\text{Mean}(\overline{\text{X}})=\frac{3+4+5+6+7}{5}$
$=\frac{25}{5}$
$=5$
Taking the absolute value of deviation of each term from the mean, we get:
$\text{MD}=\frac{|(3-5)|+|(4-5)|+|(5-5)|+|(6-5)|+|(7-5)|}{5}$
$=\frac{2+1+0+1+2}{5}$
$=\frac{6}{5}$
$=1.2$
View full question & answer→MCQ 771 Mark
When there are $2$ observations in the middle, median is calculated by $.........$
AnswerCorrect option: D. Both $(B)$ and $(C) $
Median is the middle most value of a series.
So when the series has odd number of elements then,
median can be calculated easily but when the series has even number of elements then,
The series has two middle values, so
median is calculated either by taking out the average of both the
value or the median is the $\frac{(\text{N}+1)}{ 2}^{th}$ element of the series.
View full question & answer→MCQ 781 Mark
The average age of a group of eight is same as it was $3$ years ago when a young member is substituted for an old member the incoming member is younger to the outgoing member by:
- A
$11$ years
- ✓
$24$ years
- C
$25$ years
- D
$28$ years
AnswerCorrect option: B. $24$ years
let presently the member be $ x_1, x_2, x_3....x_8 $
So the average age $ =\dfrac{\text{x}_1+\text{x}_2+\text{x}_3+.......+\text{x}_8}{8}......1$
Now the average age of all the members $3$ years ago
$ =\dfrac{\text{x}_1+\text{x}_2+\text{x}_3+.......+\text{x}_8-24}{8}$
if $x_1$ the younger member is replaced by the older member $y_1$ then,
$=\dfrac{\text{x}_1+\text{x}_2+\text{x}_3+.......+\text{x}_8}{8}$
$=\dfrac{\text{x}_1+\text{x}_2+\text{x}_3+.......+\text{x}_8-24}{8}$
$\Rightarrow \text{x}1=\text{y}1+24$
$\Rightarrow \text{x}_1-\text{y}_1=24 $
View full question & answer→MCQ 791 Mark
The mean of $x, y, z$ is $y,$ then $x + z =:$
View full question & answer→MCQ 801 Mark
Choose the correct answer. Standard deviations for first $10$ natural numbers is:
- A
$5.5$
- B
$3.87$
- C
$2.17$
- ✓
$2.97$
AnswerCorrect option: D. $2.97$
We know that $SD$ of first $n$ natural numbers $\sqrt{\frac{\text{n}^2-1}{12}}$
Here, $\text{n}=10$
$\therefore\ \text{SD}=\sqrt{\frac{(10)^2-1}{12}}$
$=\sqrt{\frac{99}{12}} $
$=\sqrt{8.25}$
$=2.97$
View full question & answer→MCQ 811 Mark
Given the following data set, what is the value of median $(2\ 4\ 3\ 6\ 1\ 8\ 9\ 2\ 5\ 7).$
AnswerMedian is the middle most value of a series.
So when the series has odd number of elements then median can be calculated easily but,
when the series has even number of elements then the series has two middle values,
so median is calculated by taking out the average of both the value.
The given series is first arranged into ascending; $1, 2, 2, 3, 4, 5, 6, 7, 8, 9$
$\text{N} = 10$
$\text{median}= \frac{(10+1)}{2}$
$\text{th}$ term $= \frac{11}{2}$
$\text{th}$ term $= 5.5 $
$=\frac{( \text{value of 5th term }+ \text{value of $6^{th}$ term)}}{2}$
$= \frac{(4+5)}{2}$
$= \frac{9}{2}$
$= 4.5$
View full question & answer→MCQ 821 Mark
| Size |
$2$ |
$3$ |
$4$ |
$5$ |
$6$ |
$7$ |
$8$ |
| Frequency |
$10$ |
$12$ |
$25$ |
$20$ |
$25$ |
$15$ |
$11$ |
- A
$2$
- B
$8$
- ✓
Both $4$ and $6$
- D
Both $6$ and $8$
AnswerCorrect option: C. Both $4$ and $6$
Mode is that observation which have highest frequency.
Since, both $4$ and $6$ have highest frequency
i.e. $25$ and $25,$ they are the mode of the given distribution.
Hence, option $(C)$ is correct.
View full question & answer→MCQ 831 Mark
Choose the correct answer. Let ${x}_1, {x}_2, {x}_3, {x}_4, {x}_5$ be the observations with mean $m$ and standard deviation $s$. The standard deviation of the observations $k x_1, k x_2, k x_3, k x_4, k x_5$ is:
AnswerHere, $\text{m}=\frac{\sum\text{x}_\text{i}}{\text{N}},$
$\text{S}=\sqrt{\frac{\sum\text{x}_\text{i}^2}{5}-\Big(\frac{\sum\text{x}_\text{i}}{5}\Big)^2}$
$\therefore\ \text{SD}=\sqrt{\frac{\text{K}^2\sum\text{x}_\text{i}^2}{5}-\Big(\frac{\text{K}\sum\text{x}_\text{i}}{5}\Big)^2}$
$=\sqrt{\frac{\text{K}^2\sum\text{x}_\text{i}^2}{5}-\text{K}^2\Big(\frac{\sum\text{x}_\text{i}}{5}\Big)^2}$
$=\text{K}=\sqrt{\frac{\sum\text{x}_\text{i}^2}{5}-\Big(\frac{\sum\text{x}_\text{i}}{5}\Big)^2}$
$=\text{K}.\text{S}$
View full question & answer→MCQ 841 Mark
Mean of a set of $23$ values is $7$, if each value is multiplied by $23$ the new mean is:
Answer$\Rightarrow$ Sum of the observation $= 23 \times 7 = 161$
If each observation is multiplied by $23$ then the sum is also multiplied by $23.$
New sum $= 23 \times 161 = 3703$
$\Rightarrow$ New mean $ =\frac{3703}{23}=161$
View full question & answer→MCQ 851 Mark
A batsman scores runs in $10$ innings as $38, 70, 48, 34, 42, 55, 63, 46, 54$ and $44.$ The mean deviation about mean is:
- ✓
$8.6$
- B
$6.4$
- C
$10.6$
- D
$10.6$
Answer$N = 10$
$\overline{\text{X}}=\frac{38+70+48+34+42+55+63+46+54+44}{10}$
$=\frac{494}{10}$
$=49.4$
|
$x_i$
|
$\text{d}_\text{i}=\big|\text{x}_\text{i}-49.4\big|$
|
| $34$ |
$15.4$ |
| $38$ |
$11.4$ |
| $42$ |
$7.4$ |
| $44$ |
$5.4$ |
| $46$ |
$3.4$ |
| $48$ |
$1.4$ |
| $54$ |
$4.6$ |
| $55$ |
$5.6$ |
| $63$ |
$13.6$ |
| $70$ |
$20.6$ |
|
|
$\sum\limits^{\text{n}}_{\text{i}=}\text{d}_\text{i}=88.8$
|
Mean deviation from the mean $=\frac{88.8}{10}= 8.88$
Disclaimer: No option is matching the answer. View full question & answer→MCQ 861 Mark
Median divides the total frequency into $..........$ equal parts:
AnswerThe median of the data series is the middle term or the mean of the two middle terms.
Hence, it divides the data series or the frequency of terms into two equal halves.
View full question & answer→MCQ 871 Mark
The average of $\displaystyle 1\frac{1}{6},2\frac{1}{3},6\frac{2}{3}161,231,632$ and $\displaystyle 8\frac{5}{6}865$ is:
- A
$\displaystyle 6\frac{3}{4}$
- B
$\displaystyle 5\frac{3}{4}$
- ✓
$\displaystyle 4\frac{3}{4}$
- D
$\displaystyle 4\frac{3}{4}$
AnswerCorrect option: C. $\displaystyle 4\frac{3}{4}$
$=\displaystyle 1\frac{1}{6}+2\frac{1}{3}+6\frac{2}{3}+ 8\frac{5}{6}$
$=\frac{7}{6}+ \frac{7}{3 }+ \frac{20}{3 }+\frac{53}{6}$
$ = \frac{6+7+14+40+53}{6} $
$= \frac{114}{6}$
$=19$
$\therefore \text{Average}=\frac{19}{4}=4\dfrac{3}{4}$
View full question & answer→MCQ 881 Mark
The average age of a teacher and three students is $20$ years. If all students are of equal age and the difference between the age of the teacher and that of a student is $20$ years, then the age of the teacher is:
- A
$25$ years
- B
$30$ years
- ✓
$35$ years
- D
$36$ years
AnswerCorrect option: C. $35$ years
Let the age of each student be $x$ years
Then, the age of teacher will be $(x + 20)$ years
Mean age $=\frac{\left (\text{x}+20 \right )+3\text{x}}{4}$
$20=\frac{\text{4x}+20}{4}$
$\Rightarrow x = 15$
Hence, age of the teacher $= 35$ years
View full question & answer→MCQ 891 Mark
The average of four consecutive even numbers is one fourth of the sum of these numbers. What is the difference between the first and last number?
AnswerLet the numbers be $2x - 2, 2x, 2x + 2$ and $2x + 4,$
where $x$ is a natural number.
Then the difference between the first and last number $= 2x + 4 - (2x - 2) = 6.$
View full question & answer→MCQ 901 Mark
The most frequent value in a data set is?
AnswerMode is the highest occurring figure in a series.
It is the value in a series of observation that repeats maximum number of times and, which represents the whole series as most of the values, in the series revolves around this value.
Therefore, mode is the value that occurs the most frequent times in a series.
View full question & answer→MCQ 911 Mark
Consider the numbers $1, 2, 3, 4, 5, 6, 7, 8, 9, 10.$ If $1$ is added to each number, the variance of the numbers so obtained is:
- ✓
$8.25$
- B
$2.87$
- C
$3.87$
- D
$3.87$
AnswerCorrect option: A. $8.25$
The given numbers are $1, 2, 3, 4, 5, 6, 7, 8, 9, 10.$
If $1$ is added to each number, then the new numbers obtained are
$2, 3, 4, 5, 6, 7, 8, 9, 10, 11$
Now,
$\sum\text{x}_\text{i}= 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 = 65$
$\sum\text{x}_\text{i}^2= 4 + 9 + 16 + 25 + 36 + 49 + 64 + 81 + 100 + 121 = 505$
$\therefore$ Variance of the numbers so obtained
$=\frac{\sum\text{x}_\text{i}^2}{10}-\Big(\frac{\sum\text{x}_\text{i}}{10}\Big)^2$
$=\frac{505}{10}-\Big(\frac{65}{10}\Big)^2$
$=50.5-42.25$
$=8.25$
View full question & answer→MCQ 921 Mark
A child says that the median of $3, 14, 18, 20, 5$ is $18.$ What concept does the child missed about finding the median?
AnswerTo calculate the median of any data series.
The data series has to be arranged in the ascending order.
The child hasn't arranged the data series in ascending order.
View full question & answer→MCQ 931 Mark
The captain of a cricket team of $11$ members is $26$ years old and the wicket keeper is $3$ years older. If the ages of these two are excluded, the average age of the remaining players is one year less than the average age of the whole team. What is the average age of the team?
- ✓
$23$ years
- B
$24$ years
- C
$25$ years
- D
$25$ years
AnswerCorrect option: A. $23$ years
Let the average age of the whole team by $x$ years.
$= 11 x - (26 + 29) = 9(x - 1)$
$= 11x - 9x = 46$
$= 2x = 46$
$\Rightarrow 2x = 46$
$= x = 23$
$\Rightarrow x = 23.$
So, average age of the team is $23$ years.
View full question & answer→MCQ 941 Mark
Two high school classes took the same test. One class of $20$ students made an average grade of $80\%;$ the other class of $30$ students made an average grade of $70\%.$ The average grade for all students in both classes is:
- A
$75\%$
- ✓
$74\%$
- C
$77\%$
- D
$77\%$
AnswerCorrect option: B. $74\%$
$\text{Average}=\frac{20.80+30.70}{20+30}=74$
View full question & answer→MCQ 951 Mark
A school has $20$ teachers one of them retires at the age of $60$ years and a new teacher replaces him this change reduces the average age of the staff by $2$ years the age of new teacher is:
- A
$28$ years
- B
$25$ years
- ✓
$20$ years
- D
$21$ years
AnswerCorrect option: C. $20$ years
Let the average age of the staff $= x$
Age of the new teacher$= y$
According to the questionNew age of the staff reduced by
$2$ years $\Rightarrow \dfrac{20\text{x}-60+\text{y}}{20}$
$\text{ x}-2\Rightarrow 20\text{x}−60+\text{y}$
$\Rightarrow 20\text{x}-60+\text{y}$
$ =20(\text{x}-2)$
$\Rightarrow 20\text{x}−60+\text{y}$
$\Rightarrow 20\text{x}−60+\text{y}$
$\Rightarrow\text{y}=60-40=20$
$\Rightarrow \text{y}=60−40= 20$
Hence the age of the new teacher is $20$ years.
View full question & answer→MCQ 961 Mark
For a frequency distribution standard deviation is computed by applying the formula:
- ✓
$\sigma=\sqrt{\frac{\sum\text{fd}^2}{\sum\text{f}}-\Big(\frac{\sum\text{fd}}{\sum\text{f}}\Big)^2}$
- B
$\sigma=\sqrt{\Big(\frac{\sum\text{fd}}{\sum\text{f}}\Big)^2-\frac{\sum\text{fd}^2}{\sum\text{f}}}$
- C
$\sigma=\sqrt{\frac{\sum\text{fd}^2}{\sum\text{f}}-\frac{\sum\text{fd}}{\sum\text{f}}}$
- D
$\sigma=\sqrt{\frac{\sum\text{fd}^2}{\sum\text{f}}-\frac{\sum\text{fd}}{\sum\text{f}}}$
AnswerCorrect option: A. $\sigma=\sqrt{\frac{\sum\text{fd}^2}{\sum\text{f}}-\Big(\frac{\sum\text{fd}}{\sum\text{f}}\Big)^2}$
View full question & answer→MCQ 971 Mark
Choose the correct answer. Following are the marks obtained by $9$ students in a mathematics test: $50, 69, 20, 33, 53, 39, 40, 65, 59$ The mean deviation from the median is:
- A
$9$
- B
$10.5$
- ✓
$12.67$
- D
$10.67$
AnswerCorrect option: C. $12.67$
$\therefore\ \text{Median}=5^{\text{th}}\text{ term}$
$\text{M}_\text{e}=50$
| $x_i$ |
$d_i = |x_i - M_e|$ |
| $20$ |
$30$ |
| $33$ |
$17$ |
| $39$ |
$11$ |
| $40$ |
$10$ |
| $50$ |
$0$ |
| $53$ |
$3$ |
| $59$ |
$9$ |
| $65$ |
$15$ |
| $69$ |
$19$ |
| $N = 2$ |
$\sum\text{d}_\text{i}=114$ |
$\therefore\ \text{MD}=\frac{114}{9}=12.67$ View full question & answer→MCQ 981 Mark
AnswerMode is the value that occurs most often For example:
$13, 13, 12, 14, 13$ The Mode of the following is $13.$
View full question & answer→MCQ 991 Mark
The mean of $20$ observations is $15$ On checking it was found that the two observations were wrongly copied as $3$ and $6.$ The correct values are $8$ and $4$ , then correct mean will be given by:
- ✓
$15.15$
- B
$14.69$
- C
$14.74$
- D
$15.54$
AnswerCorrect option: A. $15.15$
Mean of $20$ observatios $= 15$
Sum of $20$ observations $= 15 \times 20 = 300$
Correct sum $= 300 + 8 + 4 - 3 - 6 = 300$
Correct mean $= \dfrac{303}{20} = 15.15$
View full question & answer→MCQ 1001 Mark
The mean of $6$ numbers is $42$ If one number is excluded, the mean of remaining numbers is $45.$ Find the excluded number:
Answermean of $6$ numbers $= 42$
Sum of $6$ numbers $= 42 \times 6 = 252$
After excluding one number,
mean of $5$ numbers $= 45$
Sum of $5$ numbers $= 45 \times 5 = 225$
Thus, the number excluded $= 252 - 225 = 27$
View full question & answer→MCQ 1011 Mark
For a frequency distribution mean deviation from mean is computed by:
- A
$\text{M.D.}=\frac{\sum\text{f}}{\sum\text{f}\ |\text{d}|}$
- B
$\text{M.D.}=\frac{\sum\text{d}}{\sum\text{f}}$
- C
$\text{M.D.}=\frac{\sum\text{f|d|}}{\sum\text{f}}$
- ✓
$\text{M.D.}=\frac{\sum\text{fd}}{\sum\text{f}}$
AnswerCorrect option: D. $\text{M.D.}=\frac{\sum\text{fd}}{\sum\text{f}}$
View full question & answer→MCQ 1021 Mark
The weights in kilogram of $9$ members in a school boxing team are $54, 59, x, 53, 73, 49, 50, 58, 45$ If the average is $56$ then $x$ is:
- A
$61\ Kg$
- B
$62\ Kg$
- C
$64\ Kg$
- ✓
$63\ Kg$
AnswerCorrect option: D. $63\ Kg$
$\displaystyle \frac{54+59+\text{x}+53+73+49+50+58+45}{9}=56$
On simplification $\text{x} = 63$
View full question & answer→MCQ 1031 Mark
Choose the correct answer. Consider the numbers $1, 2, 3, 4, 5, 6, 7, 8, 9, 10.$ If $1$ is added to each number, the variance of the numbers so obtained is:
- A
$6.5$
- B
$2.87$
- C
$3.87$
- ✓
$8.25$
AnswerCorrect option: D. $8.25$
Given numbers are $1, 2, 3,4, 5, 6, 7, 8, 9$ and $10$
If $1$ is added to each number, then observations will be $2, 3,4, 5, 6,7, 8, 9, 10$ and $11$
$\therefore\ \sum\text{x}_\text{i}=2+3+4+\ ....\ +11$
$=\frac{10}{2}\big[2\times2+9\times1\big]=5[4+9]=65$
and $\sum\text{x}^2_\text{i}=2^2+3^2+4^2+5^2+\ .....\ +11^2=(1^2+2^2+3^2+\ .....\ +11^2)-(1^2)$
$=\frac{11\times12\times23}{6}-1=505$
$\therefore\ \text{s}^2=\frac{\sum\text{x}^2_\text{i}}{\text{n}}-\Big(\frac{\sum\text{x}_\text{i}}{10}\Big)^2$
$=\frac{505}{10}-\Big(\frac{65}{10}\Big)^2$
$=50.5-(6.5)^2$
$=50.5-42.35$
$=8.25$
View full question & answer→MCQ 1041 Mark
Which average shows the most common variable in the data set?
AnswerMode is the highest occurring figure in a series.
It is the value in a series of observation that repeats maximum number of times and
which represents the whole series as most of the values in the series revolves around this value.
Therefore, the most common variable in the series of observations is known as mode.
View full question & answer→MCQ 1051 Mark
The mean of five numbers is $18.$ If one number is excluded, then their mean is $16,$ the excluded number is $............$
AnswerMean of $55$ numbers $= 18$
Sum of these $55$ numbers $= 18 \times 5 = 90$
Let number that has been excluded be $x$ New mean $=\dfrac{90-\text{x}}{4} = 16$
Solving this, we get $90 - x = 64$
$x = 26$
View full question & answer→MCQ 1061 Mark
In a factory, the average salary of the employees is $Rs. 70.$ If the average salary of $12$ officers is $Rs. 400$ and that of the remaining employees is $Rs. 60,$ then the number of employees are $...........$
Answer$\Rightarrow$ Let total number of employees be
$x \Rightarrow$ Average salary of total employee$= Rs. 70$
Average of $12$ employees $= Rs. 400$
$\Rightarrow$ Average of remaining employees $\text{Rs}.60$
$\therefore 70=\frac{400\times 12+(\text{x}-12)\times 60}{\text{x}}$
$\therefore 70\text{x}=4800+60\text{x}-720$
$\therefore 70\text{x}=4080+60\text{x}$
$\therefore 10\text{x}=4080\therefore\text{x}=408$
Total number of employees are $408.$
View full question & answer→MCQ 1071 Mark
Choose the correct answer. The standard deviation of the data $6, 5, 9, 13, 12, 8, 10$ is:
- ✓
$\sqrt{\frac{52}{7}}$
- B
$\frac{52}{7}$
- C
$\sqrt{7}$
- D
$\sqrt{6}$
AnswerCorrect option: A. $\sqrt{\frac{52}{7}}$
Given data are $6, 5, 9, 13, 12, 8$ and $10$
| $x_i$ |
${x_i}^2$ |
| $6$ |
$36$ |
| $5$ |
$25$ |
| $9$ |
$81$ |
| $13$ |
$169$ |
| $12$ |
$144$ |
| $8$ |
$64$ |
| $10$ |
$100$ |
| $\sum\text{x}_\text{i}=63$ |
$\sum\text{x}_\text{i}^2=619$ |
$\therefore\ \text{SD}=\sigma=\sqrt{\frac{\sum\text{x}_\text{i}^2}{\text{N}}-\Big(\frac{\sum\text{x}_\text{i}}{\text{N}}\Big)^2}$
$=\sqrt{\frac{619}{7}-\Big(\frac{63}{7}\Big)^2}$
$=\sqrt{\frac{4333-396}{49}}$
$=\sqrt{\frac{396}{49}}$
$=\sqrt{\frac{52}{7}}$ View full question & answer→MCQ 1081 Mark
A grocer has a sale of $Rs. 6435, Rs. 6927, Rs. 6855, Rs. 7230$ and $Rs. 6562$. for $5$ consecutive months. How much sale must he have in the sixth month so that he gets an average sale of $Rs. 6500?$
- ✓
$Rs. 4991$
- B
$Rs. 5991$
- C
$Rs. 6991$
- D
$Rs. 6001$
AnswerCorrect option: A. $Rs. 4991$
Total sale of $5$ months $= Rs. (6435 + 6927 + 7230 + 6562) = Rs. 34009.$
Required sale $= Rs. [(6500 \times 6) - 34009]$
$= Rs. (39000 - 34009)$
$= Rs. 4991.$
View full question & answer→MCQ 1091 Mark
If the mean of first $n$ natural numbers is equal to $ \dfrac{\text{n}+7}{3}$ then $nn$ is equal to:
Answer$1\ +\ 2+\ 3\ +....\text{n}=\frac{\text{n}\times(\text{n}\ +\ 1)}{2} $
$\text{Then u}=\frac{\text{n}\times(\text{n}+1)}{2\times \text{n}}$
$\therefore \text{u}=\frac{(\text{n}+1)}{2}$
But Given: $ \text{u}=\dfrac{(\text{n}+7)}{3}$
Thus, $ \text{u}=\frac{(\text{n}+1)}{2}=\frac{(\text{n}+7)}{3}$
Solving above we get,
$n=11$
View full question & answer→MCQ 1101 Mark
The median of the following data $46, 64, 87, 41, 58, 77, 35, 90, 55, 33, 92$ is:
AnswerArrange the given data in ascending order.
We have, $33,35,41,46,55,58,64,77,87,90$ and $92.$
The sixth entry is $58.$
Median is $58.$
View full question & answer→MCQ 1111 Mark
The mean of all possible factor of $10$ is:
AnswerFactors of $10$ are, $1, 2, 5, 10$
Hence required mean $ =\frac{1+2+5+10}{4}$
$ =\frac{18}{4}$
$=4.5$
View full question & answer→MCQ 1121 Mark
What is the modal value for the numbers $5, 8, 6, 4, 10, 15, 18, 10?$
AnswerMode is the highest occurring figure in a series.
It is the value in a series of observation that repeats maximum number of times and, which represents the whole series as most of the values in the series revolves around this value.
Since in the given series, $10$ is occurring the highest number of times.
Therefore, $10$ is the mode of the series of given observations.
View full question & answer→MCQ 1131 Mark
Find the mean of all the positive factors of $72:$
- ✓
$16.25$
- B
$17.25$
- C
$18.25$
- D
$19.25$
AnswerCorrect option: A. $16.25$
Factors of $72$ are: $1, 2, 3, 4, 6, 8, 9, 12, 18, 24, 36, 72$
$\text{mean}=\frac{\text{sum}}{\text{count}}$
$\text{Mean}=\frac{1+2+3+4+6+8+9+12+18+24+36+72}{12}$
$ \text{mean}=\frac{195}{12}$
$ \text{Mean}=16.25$
View full question & answer→MCQ 1141 Mark
In the first $10$ overs of a cricket game, the run rate was only $3.2$ What should be the run rate in the remaining $40$ overs to reach the target of $282$ runs?
AnswerCorrect option: A. $6.25$
For first $10$ overs, run rate $= 3.2$
$\Rightarrow$ Runs scored $= 3.2 \times 10 = 32$
Total runs to be scored $= 282$
Runs Left to be scored in $4040$ overs $= \frac{282-32}{40}$
$ =\dfrac{250}{40}$
Required Run$-$Rate in remaining overs $= 6.25$
View full question & answer→MCQ 1151 Mark
The mean of $30, 32, 24, 34, 26, 28, 30, 35, 33, 25$ is $29.7$ If true then enter $1$ and if false then enter $0:$
AnswerMean of the series: $30, 32, 24, 34, 26, 28, 30, 35, 33, 25$
$ \text{mean} = \frac{\text{Sum}}{\text{Number of observations}}$
$\text{mean} = \frac{30 + 32 + 24 + 34 + 26 + 28 + 30 + 35 + 33 + 25}{10}$
$ \text{mean} = \frac{297}{10} = 29.7$
View full question & answer→MCQ 1161 Mark
The average age of two brothers is $9$ years It is increased by $9$ years when their mothers age is also included then the age of mother is:
- A
$35$ years
- ✓
$36$ years
- C
$37$ years
- D
$38$ years
AnswerCorrect option: B. $36$ years
Average age of the two brother $= 9$ years
$\therefore$ Age of two brother $= 9 \times 2 = 18$ years
If their mother age is included then the average age is increased by $9$
$\therefore$ Average age of $3 = 9 + 9 = 18$ years
Now Total age of three $= 3 \times 18 = 54$ Years
$\therefore$ Mothers age $= 54 - 18 = 36$ years.
View full question & answer→MCQ 1171 Mark
| $x$ |
$10$ |
$15$ |
$20$ |
$25$ |
$35$ |
| $f$ |
$6$ |
$p$ |
$26$ |
$10$ |
$8$ |
Answer$∑ \text{fx} = 15\text{p} + 1110; ∑ \text{f} = 50 + \text{p}$
$\text{x} = 21$
$21=\frac{15_p\ +\ 1110}{50 \ +\ p}$
$\Rightarrow 21\text{p} − 15\text{p} = 1110 − 1050$
$\Rightarrow\text{p}=\frac{60}{6}=10$
View full question & answer→MCQ 1181 Mark
Choose the correct answer. Let $x_1, x_2, ... x_n$ be $n$ observations. Let $w_i= lx_i+ k$ for $i = 1, 2, ... n,$ where $l$ and $k$ are constants. If the mean of $x_i’s$ is $48$ and their standard deviation is $12,$ the mean of $w_i’s$ is $55$ and standard deviation of $w_i’s$ is $15,$ the values of $l$ and $k$ should be:
- ✓
$l = 1.25, k = -5$
- B
$l = -1.25, k = 5$
- C
$l = 2.5, k = 5$
- D
$l = 2.5, k = -5$
AnswerCorrect option: A. $l = 1.25, k = -5$
Given, $\text{w}_\text{i}=\text{lx}_\text{i}+\text{k},$
$\bar{\text{x}}_\text{i}=48,\text{ sx}_\text{i}=12,\text{ w}_\text{i}=55$ and $\text{sw}_\text{i}=15$
Then, $\bar{\text{w}}_\text{i}=\text{l}\bar{\text{x}}_\text{i}+\text{k}$
where $\big[\bar{\text{w}}_\text{i}$ is mean $w_\text{i}{'\text{s}}$ and $\bar{\text{x}}_\text{i}$ is mean of $\text{x}_\text{i}{'\text{s}}\big]$
$\Rightarrow55=\text{l}\times48+\text{k}\ ...(\text{i})$
Now, $\text{SD}$ of $\text{w}_\text{i}=\text{l}(\text{SD}$ of $\text{x}_\text{i})$
$\Rightarrow15=\text{l}\times12$
$\Rightarrow\text{l}=\frac{15}{12}=12.5$
From Eq. $(i)$ we get $k = 55 - 1.25 \times 48 = -5$
View full question & answer→MCQ 1191 Mark
Which one of the following statements is correct?
- A
The Standard deviation for a given distribution is the square of the variance.
- ✓
The standard deviation for a given distribution is the square root of the variance.
- C
The standard deviation for a given distribution is root equal to the variance.
- D
The standard deviation for a given distribution is equal to the variance.
AnswerCorrect option: B. The standard deviation for a given distribution is the square root of the variance.
View full question & answer→MCQ 1201 Mark
Choose the correct answer. When tested, the lives $($in hours$)$ of $5$ bulbs were noted as follows: $1357, 1090, 1666, 1494, 1623$ The mean deviations $($in hours$)$ from their mean is:
AnswerThe lines of $5$ bulbs are given by
$1357, 1090, 1666, 1494, 1623$
$\therefore\ \text{Mean}=\frac{1357+1090+1666+1494+1623}{5}$
$\Rightarrow\bar{\text{x}}=\frac{7230}{5}=1446$
| $x_i$ |
$\text{d}_\text{i}=|\text{x}_{\text{i}}-\bar{\text{x}}|$ |
| $1357$ |
$89$ |
| $1090$ |
$356$ |
| $1666$ |
$220$ |
| $1494$ |
$48$ |
| $1623$ |
$177$ |
| Total |
$\sum\text{d}_\text{i}=890$ |
$\therefore\ \text{MD}=\frac{\sum\text{d}_\text{i}}{\text{n}}=\frac{890}{5}=178$ View full question & answer→MCQ 1211 Mark
The mean of $20$ observations is $12.5$ By error, one observation was noted as $-15$ instead of $15.$ Then the correct mean is $.............$
AnswerMean of $20$ observations $= 12.5$
Sum of $20$ observations $= 12.5 \times 20 = 250$
Since $15$ was misread as,
$15$ New sum $= 250 - (-15) + 15 = 280$
The correct $\text{mean} = \dfrac{280}{20} = 14$
View full question & answer→MCQ 1221 Mark
The mean of the following data is : $45, 35, 20, 30, 15, 25, 40:$
AnswerMean is given by $ \text{mean}=\dfrac{\text{sum of the elements}}{\text{total number of elements}}$
$=\text{mean}=\frac{45+35+20+30+15+25+40}{7}$
$=\frac{210}{7}$
$ =30$
View full question & answer→MCQ 1231 Mark
The median $31, 16,19, 25, 14, 13,12, 4, 28, 45$ is.
AnswerCorrect option: C. $17.5$
Arranging the given data in ascending order $4, 12, 13, 16, 19, 28, 31,$
The middle terms are $16, 19$ Hence median
$=\frac{16 \ +\ 19}{2}$
$=17.5$
View full question & answer→MCQ 1241 Mark
Value of the middle$-$most observation $(s)$ is called:
AnswerTo find the Median, place the numbers in value order and find the middle number.
If there are two middle numbers,
take the mean of the two numbers and this,
will be the median of the data set.
The middle most observation of a data series is called the median of the series.
View full question & answer→MCQ 1251 Mark
If the median of $ \frac{\text{x}}{5}\text{x} \frac{\text{x}}{4} \frac{\text{x}}{2}\ \text{and}\ \frac{\text{x}}{3}$ $($where $x > 0)$ is $8$ then the value of $x$ would be:
AnswerArranging is ascending order the values are
$\frac{\text{x}}{5},\frac{\text{x}}{4},\frac{\text{x}}{3},\frac{\text{x}}{2},\text{x}$
Middle value $ =\frac{\text{x}}{3}$
$\Rightarrow\frac{\text{x}}{3}=\text{x}=24$
View full question & answer→MCQ 1261 Mark
Let $x_1, x_2,...., x_n$ be values taken by a variable $X$ and $y_1, y_2,...., y_n$ be the values taken by a variable $Y$ such that $y_i=a x_i+b, i= , i = 1, 2,..., n$. Then,
- ✓
$\ce{Var (Y) = a^2 Var (X)}$
- B
$\ce{Var (X) = a^2 Var (Y)}$
- C
$\text{Var (Y) = Var (X) + b}$
- D
$\text{Var (X) = Var (X) + b}$
AnswerCorrect option: A. $\ce{Var (Y) = a^2 Var (X)}$
$\text{Var}(\text{x})=\frac{\sum\limits^\text{n}_{\text{i}=1}\Big(\text{x}_\text{i}-\overline{\text{X}}\Big)^2}{\text{n}}$ where Mean $\Big(\overline{\text{X}}\Big)=\frac{\sum\limits^\text{n}_{\text{i}=1}\text{x}_\text{i}}{\text{n}}$
$\text{Var}(\text{Y})=\frac{\sum\limits^\text{n}_{\text{i}=1}\Big(\text{y}_\text{i}-\overline{\text{Y}}\Big)^2}{\text{n}}$ and $\overline{\text{Y}}=\frac{\sum\limits^\text{n}_{\text{i}=1}\text{y}_\text{i}}{\text{n}}$
We have,
$\text{y}_\text{i}=\text{ax}_\text{i}+\text{b}$
$\overline{\text{Y}}=\frac{\sum\limits^\text{n}_{\text{i}=1}\text{y}_\text{i}}{\text{n}}$
$\overline{\text{Y}}=\frac{\sum\limits^\text{n}_{\text{i}=1}\text{ax}_\text{i}+\text{b}}{\text{n}}$
$=\frac{\sum\limits^\text{n}_{\text{i}=1}\text{x}_\text{i}}{\text{n}}+\frac{\text{nd}}{\text{n}}$
$=\text{a}\overline{\text{X}}+\text{b}$
$\text{Var}(\text{Y})=\frac{\sum\limits^\text{n}_{\text{i}=1}\Big(\text{y}_\text{i}-\overline{\text{Y}}\Big)^2}{\text{n}}$
$=\frac{\sum\limits^\text{n}_{\text{i}=1}\Big\{\text{ax}_\text{i}+\text{b}-\big(\text{a}\overline{\text{X}}+\text{b}\big)\Big\}^2}{\text{n}}$
$=\frac{\sum\limits^\text{n}_{\text{i}=1}\big(\text{ax}_\text{i}-\text{a}\overline{\text{X}}\big)^2}{\text{n}}$
$=\text{a}^2\frac{\sum\limits^\text{n}_{\text{i}=1}\big(\text{x}_\text{i}-\overline{\text{X}}\big)^2}{\text{n}}$
$=\text{a}^2\text{Var}(\text{X})$
View full question & answer→MCQ 1271 Mark
The average of $15$ numbers is $18$ The average of first $8$ is $19$ and that last $8$ is $17$ then the $8^{th}$ number is:
AnswerAverage of $15$ numbers $15 \times 18= 270$
Average of first $8$ number is $19$
$\therefore$ Sum of first $8$ number $=19 \times 8 = 152 = 19 \times 8 = 152$
Average of first $8$ number $= 17$
$\therefore$ Sum of $8$ number $= 8 \times 17 = 136 = 8 \times 17 = 136$
$\therefore 8^{th}$ number is $= (152 + 136) - 270$
$\Rightarrow 288 - 270=18$
View full question & answer→MCQ 1281 Mark
Aman is his $12^{th}$ innings makes a score of $63$ runs and increases his average score to $2$. What is his average after the $12^{th}$ innings ?
AnswerLet the average score till $11^{th}$ innings be $x$ according to question
$\Rightarrow \frac{11\text{x}+63}{12}= x + 2$
$\Rightarrow 11x + 63 = 12x + 24$
$\Rightarrow x = 39$
$12^{th}$ inning average
$\Rightarrow 39 + 2 = 41$
View full question & answer→MCQ 1291 Mark
The mean of $x_1, x_2....x_{50}\ M,$ if every $x_i = 1,2...50$ is replaced by $\frac{\text{x}_i}{50}$ then the mean is:
AnswerCorrect option: D. $ \displaystyle \frac{50}{\text{M}}$
Given $ \text{mean}= \frac{\text{x}_{1}+\text{x}_{2}+\text{x}_{3}...........\text{x}_{50}}{50}$
$ \text{mean} =\frac{\frac{\text{x}_{1}}{50}+\frac{\text{x}_{2}}{50}+...........\frac{\text{x}_{50}}{50}}{50}$
$ =\frac{\text{x}_1+\text{x}_2+....\text{x}_{50}}{50\times50}$
$=\displaystyle \frac{\text{50}}{M}$
View full question & answer→MCQ 1301 Mark
The modal value is the value of the variate which divides the total frequency into two equal parts:
AnswerFalse. Modal value is the value which occurs maximum number of times in the data.
View full question & answer→MCQ 1311 Mark
Find the mean of $23, 28, 13, 16, 20:$
AnswerGiven observations $23, 28, 13, 16, 20$ No. of observations are $5$ Mean of
observations $\dfrac{23+28+13+16+20}{5}$
$ =\dfrac{100}{5}$
$ =20$
View full question & answer→MCQ 1321 Mark
The $..........$ of a set of data is the middlemost number in the set.
AnswerThe median of a set of data is the middlemost number in the set.
Example: $3, 4, 5, 1, 1, 8, 10$
So, first arrange the data in order.
So, $1, 1, 3, 4, 5, 8, 10$
The middle number is median $= 4$
View full question & answer→MCQ 1331 Mark
The combined mean of three groups is $12$ and the combined mean of first two groups is $3.$ If the first, second and third group have their mean as $2, 3$ and $5$ times respectively, then the mean of third group is:
AnswerLet in common no. in each group is $X$
Then Member of each group is $2X, 3X$ and $5X$
Total of three group $= (2X + 3X + 5X) 12 = 120x$
And total of Two group $= (2X + 3X)3 = 15X ($given mean of two group is $3)$
Then total of third group $= 120X - 15X = 115X$
Mean of third group $ =\frac{115\text{X}}{5\text{X}}=21$
View full question & answer→MCQ 1341 Mark
In the formula for mode of a grouped data, $\text{ mode} =\text{l}+\left \{\frac {\text{f}_1-\text{f}_0}{2\text{f}_2-\text{f}_0-\text{f}_2}\right \}\times\text{h}$ where symbols have their usual meaning $f_0$ represents:
- A
- B
Frequency of median class
- ✓
Frequency of the class preceding the modal class
- D
Frequency of the class preceding the Medium class
AnswerCorrect option: C. Frequency of the class preceding the modal class
In the formula for mode of a grouped data,
$\text{ mode} =\text{l}+\left \{\frac {\text{f}_1-\text{f}_0}{2\text{f}_2-\text{f}_0-\text{f}_2}\right \}\times\text{h}$ where symbols have their usual meaning
$f_0$ represents Frequency of the class preceding the modal classwhere
$f =$ Frequency,
$1 =$ Lowest value of the modal range,
$f_1=$ Frequency of modal class,
$f_2$ Frequency of class succeeding the modal class and
$f_0 =$ Frequency of class preceding the modal class.
Hence, option $C$ is correct.
View full question & answer→MCQ 1351 Mark
The average of $50$ numbers is $38.$ If two numbers namely $45$ and $55$ are discarded, the average of the remaining numbers is:
- ✓
$37.5$
- B
$38.5$
- C
$36.5$
- D
$35.5$
AnswerCorrect option: A. $37.5$
Average of remaining $48$ numbers
$=\frac{(50 \times38 )- 55 - 45}{48}$
$= 37.5$
View full question & answer→MCQ 1361 Mark
The $...........$ is the difference between the greatest and the least value of the variate:
AnswerRange as the name indicates gives us all the area available under light and hence statement is true.
View full question & answer→MCQ 1371 Mark
State true or false: The mode is the most frequently occurring observation:
AnswerThe observation occurring the most number of times or which has highest frequency is called the mode.
Thus, the given statement is true.
View full question & answer→MCQ 1381 Mark
The mean of the following natural numbers $1, 2, 3,...10$ is:
AnswerNumbers are $1, 2, 3,..10$
Sum of the numbers $= {\frac{\text{n}(\text{n}+1)}{2}}= \frac{10\times11}{2} = 55$
$\text{Mean} = \frac{55}{10} =5.5$
View full question & answer→MCQ 1391 Mark
The mean of $200$ items was $50.$ Later on, it was discovered that two items were misread as $92$ and $8$ instead of $192$ and $8.$ The correct mean is:
AnswerCorrect option: C. $50.9$
Mean of $200$ observations $= 50$
Sum of $200$ observations $= 50 \times 200 = 10000$
After replacing the misread observation $92$ to $192$ and $8$ to $88$
Sum of $200200$ observations $= 10000 - 92 + 192 - 8 + 88 = 10180$
$\text{New mean} = \frac{10180}{200} = 50.9$
View full question & answer→MCQ 1401 Mark
Choose the correct answer. The mean of $100$ observations is $50$ and their standard deviation is $5.$ The sum of all squares of all the observations is:
- A
$50000$
- B
$250000$
- ✓
$252500$
- D
$25000$
AnswerCorrect option: C. $252500$
Here, $\bar{\text{x}}=\frac{\sum\text{x}_\text{i}}{\text{n}}$
$\Rightarrow50=\frac{\sum\text{x}_\text{i}}{100}$
$\Rightarrow{\sum\text{x}_\text{i}}=5000$
$\therefore\ \text{SD}=\sqrt{\frac{\sum\text{x}_\text{i}^2}{\text{n}}-\Big(\frac{\sum\text{x}_\text{i}}{\text{n}}\Big)^2}$
$\Rightarrow\ 5=\sqrt{\frac{\sum\text{x}_\text{i}^2}{\text{n}}-\Big(\frac{5000}{100}\Big)^2}$
$\Rightarrow25=\frac{\sum\text{x}_\text{i}^2}{100}=2525$
$\therefore\ {\sum\text{x}_\text{i}^2}=2525\times100=252500$
View full question & answer→MCQ 1411 Mark
If two variates $X$ and $Y$ are connected by the relation $\text{Y}=\frac{\text{aX}+\text{b}}{\text{c}},$ where $a, b, c$ are constants such that $ac < 0,$ then
- A
$\sigma\text{Y}=\frac{\text{a}}{\text{c}}\sigma\text{X}$
- ✓
$\sigma\text{Y}=-\frac{\text{a}}{\text{c}}\sigma\text{X}$
- C
$\sigma\text{Y}=-\frac{\text{a}}{\text{c}}\sigma\text{X}-\text{b}$
- D
$\sigma\text{Y}=-\frac{\text{a}}{\text{c}}\sigma\text{X}+\text{b}$
AnswerCorrect option: B. $\sigma\text{Y}=-\frac{\text{a}}{\text{c}}\sigma\text{X}$
$\text{Y}=\frac{\text{aX}+\text{b}}{\text{c}}$
$\overline{\text{Y}}=\frac{\sum\limits_{\text{i}=1}^\text{n}\frac{\text{aX}+\text{b}}{\text{c}}}{\text{n}}$
$=\frac{\frac{\text{a}\sum\limits_{\text{i}=1}^\text{n}\text{X}+\text{nb}}{\text{c}}}{\text{n}}$
$=\frac{\frac{\text{a}}{\text{c}}\sum\limits_{\text{i}=1}^\text{n}\text{X}}{\text{n}}+\frac{\text{b}}{\text{c}}$
$=\frac{\text{a}\overline{\text{X}}}{\text{c}}+\frac{\text{b}}{\text{c}}$
We know:
$\text{Var}(\text{X})=\frac{\sum\limits_{\text{i}=1}^\text{n}\big(\text{x}_\text{i}-\overline{\text{X}}\big)^2}{\text{n}}$
$=\sigma^2$
$\text{Var}(\text{Y})=\frac{\sum\limits_{\text{i}=1}^\text{n}\big(\text{y}_\text{i}-\overline{\text{Y}}\big)^2}{\text{n}}$
$=\frac{\sum\limits_{\text{i}=1}^\text{n}\big(\frac{\text{aX}}{\text{c}}+\frac{\text{b}}{\text{c}}-\frac{\text{a}}{\text{c}}\overline{\text{X}}-\frac{\text{b}}{\text{c}}\big)^2}{\text{n}}$
$=\frac{\sum\limits_{\text{i}=1}^\text{n}\big(\frac{\text{aX}}{\text{c}}-\frac{\text{a}}{\text{c}}\overline{\text{X}}\big)^2}{\text{n}}$
$=\Big(\frac{\text{a}}{\text{c}}\Big)^2\frac{\sum\limits_{\text{i}=1}^{\text{n}}\big(\text{x}_\text{i}-\overline{\text{X}}\big)^2}{\text{n}}$
$=\Big(\frac{\text{a}}{\text{c}}\Big)^2\sigma^2$
$\text{SD}$ of ${Y}\big(\sigma_\text{y}\big)=\sqrt{\Big(\frac{\text{a}}{\text{c}}\Big)^2\sigma^2}$
$=\Big|\frac{\text{a}}{\text{c}}\Big|\sigma$
$\text{ac}<0$
$\Rightarrow\text{a}<0$ or $\text{c}<0$
$\therefore\Big|\frac{\text{a}}{\text{c}}\Big|=-\frac{\text{a}}{\text{c}}$
$\Rightarrow\sigma\text{Y}=-\frac{\text{a}}{\text{c}}\sigma\text{X}$
View full question & answer→MCQ 1421 Mark
If $16$ observations are arranged in ascending order, then median is:
AnswerCorrect option: C. $\frac{8^\text{th}\ \text{observation}\ +\ 9^\text{th}\ \text{observation}}{2}$
For even number of observations median is the mean of $ \frac{\text{n}}{2}^{th}$
observation and $ \Big(\frac{\text{n}}{2}+1\Big)^{th}$ observation:
So, median of $16$ observation$= \frac{8^\text{th}\ \text{observation}\ +\ 9^\text{th}\ \text{observation}}{2}$
View full question & answer→MCQ 1431 Mark
If for a sample of size $60,$ we have the following information $\sum\text{x}_\text{i}^2=18000$ and $\sum\text{x}_\text{i}=960$ then the variance is:
AnswerGiven $\sum\text{x}_\text{i}^2=18000,\ \sum\text{x}_\text{i}=960$ and $n = 60$
$\therefore$ Variance
$=\frac{\sum\text{x}_\text{i}^2}{\text{n}}-\bigg(\frac{\sum\text{x}_\text{i}}{\text{n}}\bigg)^2$
$=\frac{18000}{60}-\Big(\frac{960}{60}\Big)^2$
$=300-256$
$=44$
Hence, the correct answer is option $(d).$
View full question & answer→MCQ 1441 Mark
The mean of $10$ observation is $25.$ If one observation namely $25,$ is deleted, the new mean is:
AnswerMean of $1010$ observations $= 25$
Sum of $1010$ observations $= 25 \times 10 = 250$
After removing an observation with a value $= 25$
New sum $= 250 - 25 = 225$
New $ \text{mean} = \frac{225}{9} = 25$
View full question & answer→MCQ 1451 Mark
Emmy did a survey of how many games each of $2020$ friends owned, and got the following data: $5, 7, 12, 13, 4, 6, 8, 12, 9, 16, 13, 12, 5, 13, 7, 17, 3, 9, 12, 14.$ Find the mean:
- A
$8.55$
- B
$7.59$
- C
$5.49$
- ✓
$9.85$
AnswerCorrect option: D. $9.85$
View full question & answer→MCQ 1461 Mark
The variance is the $.............$ of the standard deviation:
View full question & answer→MCQ 1471 Mark
Seven of the eight numbers in a distribution are $11, 16,6, 10, 13, 11, 13.$
Given that the mean of the distribution is $12,$ if $12$ will be included then find the new mean of the distribution.
AnswerCorrect option: B. $11.5$
$\text{mean}=\frac{6+10+11+11+13+13+13+16+12}{8}=11.5$
View full question & answer→MCQ 1481 Mark
The average age of $15$ students of a class is $15$ years. Out of these, the average age of $5$ students is $14$ years and that of the other nine students is $16$ years. What is the age of the $15^{th}$ student?
- A
$17$ years
- B
$13$ years
- C
$19$ years
- ✓
$11$ years
AnswerCorrect option: D. $11$ years
Total age of $15$ students $= (15 \times 15)$ years $= 225$ years
Total age of $5$ students $= (5 \times 14)$ years $= 70$ years
Total age of other $9$ students $= (9 \times 16)$ years $= 144$ years
$\therefore$ Age of the $15^{th}$ student $= 225 - (70 + 144) = 225 - 214 = 11$ years.
View full question & answer→MCQ 1491 Mark
The mean deviation of the data $3, 10, 10, 4, 7, 10, 5$ from the mean is:
AnswerCorrect option: B. $2.57$
The given observations are $3, 10, 10, 4, 7, 10, 5.$
$\therefore\text{Mean},\ \overline{\text{x}}=\frac{3+10+10+4+7+10+5}{7}=\frac{49}{7}=7$
Now,
Mean deviation from mean, $MD$
$=\frac{\sum|\text{x}_\text{i}-7|}{7}$
$=\frac{|3-7|+|10-7|+|10-7|+|4-7|+|7-7|+|10-7|+|5-7|}{7}$
$=\frac{4+3+3+3+0+3+2}{7}$
$=\frac{18}{7}$
$=2.57$
Hence, the correct answer is $(b).$
View full question & answer→MCQ 1501 Mark
If different values of variable $x$ are $9.8, 5.4, 3.7, 1.7, 1.8, 2.6, 2.8, 8.6, 10.5$ and $11.1;$ find the mean:
- ✓
$5.8$
- B
$7.8$
- C
$9.8$
- D
$11.8$
AnswerValues of $x$ are: $9.8, 5.4, 3.7, 1.7, 1.8, 2.6, 2.8, 8.6, 10.5$ and $11.1$
$\text{Mean}=\frac{\ \text{Sum}}{\text{Count}}$
$\text{Mean}=\frac{9.8+5.4+3.7+1.7+1.8+2.6+2.8+8.6+10.5+11.1}{10}$
$ \text{Mean}=\frac{58}{10}$
$\text{Mean}=5.8$
View full question & answer→MCQ 1511 Mark
The average weight of a group of $20$ boys was calculated to be $89.4\ kg$ and it was later discovered that one weight was misread as $78\ kg$ instead of $87\ kg.$ The correct average weight is:
- A
$88.95\ kg$
- ✓
$89.85\ kg$
- C
$89.55\ kg$
- D
$87.55\ kg$
AnswerCorrect option: B. $89.85\ kg$
Difference in weight $= 87 - 78 = 9\ kg$
$\therefore$ Correct average weight $= 89.4 +\frac {9}{20}$
$= 89.4 + 0.45$
$= 89.85\ kg$
View full question & answer→MCQ 1521 Mark
Find the incorrect formula from the following:
- A
$ \text{mode}=\text{L}+\Big(\frac{\text{f}\text{m}-\text{f}1}{\text{f}\text{m}-\text{f}1-\text{f}2}\Big)\text{h}$
- B
$\text{Mean} = \frac{\sum {\text{ fixi }} }{\sum { \text{xi }} }$
- ✓
$\overline{\text{x}}= \frac{\sum {\text{ fixi }} }{\sum { \text{fi }} }$
- D
$\overline{\text{x}}= \frac{\sum {\text{ fi }} }{\sum { \text{fixi }} }$
AnswerCorrect option: C. $\overline{\text{x}}= \frac{\sum {\text{ fixi }} }{\sum { \text{fi }} }$
$\text{ A Median} = \text{L} + \begin{pmatrix} \frac{N}{2} - \text{c.f.} \end{pmatrix} \frac{\text{h}}{\text{f}} , $
$ \text{mode}=\text{L}+\Big(\frac{\text{f}\text{m}-\text{f}1}{\text{f}\text{m}-\text{f}1-\text{f}2}\Big)\text{h}$ Where,
$L$ is lower boundary of median class,
$N$ is sum of frequencies, $c.f.$ is cumulative frequency,
$h$ is width of classes, $f$ is frequency of mode class,
$\text{f}_\text{m}$ is frequency of modal class,
$\text{f}_\text{1}$ is frequency of pre$-$modal class,
$\text{f}_\text{2}$ is frequency of post$-$modal class Mean is
given $= \frac{\sum {\text{ fixi }} }{\sum { \text{fi }} }$
Thus, Mean given in option $C$ is incorrect.
View full question & answer→MCQ 1531 Mark
The algebraic sum of the deviations of a set of value of a data from their mean is:
MCQ 1541 Mark
The mean deviation from the median is:
- A
Equal to that measured from another value.
- B
Maximum if all observations are positive.
- C
Greater than that measured from any other value.
- ✓
Less than that measured from any other value.
AnswerCorrect option: D. Less than that measured from any other value.
In a frequency distribution, the sum of absolute values of deviations from the mean and mode is always more than the sum of the deviations from the median.
View full question & answer→MCQ 1551 Mark
Consider the first $10$ positive integers. If we multiply each number by $-1$ and then add $1$ to each number, the variance of the numbers so obtained is:
- ✓
$8.25$
- B
$6.5$
- C
$3.87$
- D
$4.87$
AnswerCorrect option: A. $8.25$
The first $10$ positive integers are $1, 2, 3, 4, 5, 6, 7, 8, 9, 10.$
Multiplying each number by $−1,$ we get
$-1, -2, -3, -4, -5, -6, -7, -8, -9, -10$
Adding $1$ to each of these numbers, we get
$0, -1, -2, -3, -4, -5, -6, -7, -8, -9$
Now,
$\sum\text{x}_\text{i}= 0 + (-1) + (-2) + (-3) + (-4) + (-5) + (-6) + (-7) + (-8) + (-9) = -45$
$\sum\text{x}_\text{i}^2= 0 + 1 + 4 + 9 + 16 + 25 + 36 + 49 + 64 + 81 = 285$
$\therefore$ Variance of the obtained numbers
$=\frac{\sum\text{x}_\text{i}^2}{10}-\Big(\frac{\sum\text{x}_\text{i}}{10}\Big)^2$
$=\frac{285}{10}-\Big(\frac{-45}{10}\Big)^2$
$=28.5-20.25$
$=8.25$
View full question & answer→MCQ 1561 Mark
Calculate the mode for $17, 12, 19, 11, 20, 11, 20, 19, 10, 25, 19:$
AnswerMode is the value which occurs most often in the data set of values.
Given data set is $17, 12, 19, 11, 20, 11, 20, 19, 10, 25, 19$
In the above data set, value $19$ has occurred many times i.e., $3$ times.
Therefore the mode of the given data set is $19$
View full question & answer→MCQ 1571 Mark
The average weight of $20$ students was calculated $70\ kg.$ It was later discovered that one weight was misread as $70$ instead of $90,$ the correct average in $kg$ is
AnswerWrong average $=70$
$\therefore$ Wrong sum of weight of $20$ students $= 20 \times 70$
$ \therefore$ Correct sum of weights of $20$ students
$= 1400 -$ wrong weight $+$ correct weight
$= 1400 - 70 + 90$
$= 1420$
$\therefore \text{Correct mean} = \frac{1420}{20}=71 \text{kg.}$
View full question & answer→MCQ 1581 Mark
The sum of $12$ observations is $600$ then their mean is $............$
Answer$\text{Mean} = \frac { \text{Sum of observations}}{\text{Total number of observations}} = \dfrac {600}{12} = 50$
View full question & answer→MCQ 1591 Mark
Mean of twenty observations is $15.$ It two observations $3$ and $14$ are replaced by $8$ and $9$ respectively, then the new mean will be:
AnswerTotal $= 20 \times 15 = 300$
When observations $3$ and $14$ are replaced by $8$ and $9$ respectively,
new sum will be $300 - 3 - 14 + 8 + 9 = 300$
New mean will be again $15$
View full question & answer→MCQ 1601 Mark
Let the observations at hand be arranged in increasing order. Which one of the following measures will not be affected when the smallest and the largest observations are removed?
AnswerIf largest and smallest are removed the centre will remain intact.
Hence median will not be disturbed.
View full question & answer→MCQ 1611 Mark
The value of $ \displaystyle\sum _{ \text{p,q,r} }^{ }{ {\text{ p}}^{ 2 } } - \displaystyle\sum _{ \text{p,q,r} }^{ }{ {\text{q}}^{2}}$ is:
- A
$ { \text{p} }^{ 2 }+{ \text{q} }^{ 2 }+{ \text{r} }^{ 2 }$
- ✓
$ 0$
- C
$ 2{ \text{p} }^{ 2 }+2{ \text{q} }^{ 2 }-2{ \text{r} }$
- D
$ 2{ \text{p} }^{ 2 }+2{ \text{q} }^{ 2 }+2{ \text{r} }$
AnswerBoth the summations runs over $ \text{p,q,r}$
$\therefore \displaystyle\sum _ { \text{p,q,r }}^{ }{ { \text{p} }^{ 2 } } - \displaystyle\sum _ { \text{p,q,r }}^{ }{ { \text{p} }^{ 2 } } - (\text{p}^2 +\text{q}^2+\text{r}^2) - (\text{p}^2 +\text{q}^2+\text{r}^2)= 0$
Hence, the answer is $0.$
View full question & answer→MCQ 1621 Mark
If $v$ is the variance and $\sigma$ is the standard deviation, then:
AnswerCorrect option: C. $\text{V}=\sigma^2$
The variance is the square of the standard deviation.
View full question & answer→MCQ 1631 Mark
The mean of three numbers is $10.$ The mean of other four numbers is $12.$ Find the mean of all the numbers:
- A
$13.5$
- ✓
$11.15$
- C
$14.15$
- D
$12.15$
AnswerCorrect option: B. $11.15$
Mean of $3$ nos $= 10$ Total of $3$ nos is $10 \times 3 = 30$ Mean of other $4$ nos is $12$
Total of $4$ nos is $12 \times 4 = 48$
total of $4 + 3 = 7$
numbers is $48 + 30 = 78$
Mean of $7$ numbers is $ \cfrac{78}{7} = 11.15.$
View full question & answer→MCQ 1641 Mark
Find the average of all the number between $6$ and $34$ which are divisible by $5.$
AnswerNumbers between $6$ and $34$ divisible by $5$ are $10, 15, 20, 25, 30.$
Required average $= \frac{10+15+20+25+30}{5}=\frac {100}{5}= 20$
View full question & answer→MCQ 1651 Mark
If the $S.D.$ of a set of observations is $8$ and if each observation is divided by $−2,$ the $S.D.$ of the new set of observations will be:
AnswerIf a set of observations, with $SD \sigma \sigma$ , are multiplied with a non$-$zero real number a, then $SD$ of the new observations will be $|\text{a}|\sigma.$
Dividing the set of observations by $-2$ is same as multiplying the observations by $\frac{1}{-2}.$
New $\text{S.D.}=\Big|-\frac{1}{2}\Big|\times8$
$=\frac{8}{2}$
$=4$
View full question & answer→MCQ 1661 Mark
The mean deviation for $n$ observations $x_1, x_2, ...,x_n$ from their mean $\overline{\text{X}}$ is given by:
- A
$\sum^\text{n}_{\text{i}=1}\Big(\text{x}_\text{i}-\overline{\text{X}}\Big)$
- ✓
$\frac{1}{\text{n}}\sum^\text{n}_{\text{i}=1}\Big(\text{x}_\text{i}-\overline{\text{X}}\Big)$
- C
$\sum^\text{n}_{\text{i}=1}\Big(\text{x}_\text{i}-\overline{\text{X}}\Big)^2$
- D
$\sum^\text{n}_{\text{i}=1}\Big(\text{x}_\text{i}-\overline{\text{X}}\Big)^2$
AnswerCorrect option: B. $\frac{1}{\text{n}}\sum^\text{n}_{\text{i}=1}\Big(\text{x}_\text{i}-\overline{\text{X}}\Big)$
The mean deviation for $n$ observations $x_1, x_2, ...,x_n$ from their mean $\overline{\text{X}}$ is $\frac{1}{\text{n}}\sum^\text{n}_{\text{i}=1}\Big|\text{x}_\text{i}-\overline{\text{X}}\Big|.$
Disclaimer: There is some printing error in option $(b)$ given in the question.
The answer would be option $(b)$ if it given as $\frac{1}{\text{n}}\sum^\text{n}_{\text{i}=1}\Big|\text{x}_\text{i}-\overline{\text{X}}\Big|.$
View full question & answer→