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→