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→