Statistics

$\displaystyle \textbf{Question 1. }\text{For a frequency distribution mean deviation from mean is computed by}$
$\displaystyle \text{(a) }M.D.=\frac{\sum f}{\sum f|d|}\qquad \text{(b) }M.D.=\frac{\sum d}{\sum f}$
$\displaystyle \text{(c) }M.D.=\frac{\sum fd}{\sum f}\qquad \text{(d) }M.D.=\frac{\sum f|d|}{\sum f}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(d) }\;M.D.=\frac{\sum f|d|}{\sum f}$
$\\$

$\displaystyle \textbf{Question 2. }\text{For a frequency distribution standard deviation is computed by applying the formula}$
$\displaystyle \text{(a) }\sigma=\sqrt{\frac{\sum fd^2}{\sum f}-\left(\frac{\sum fd}{\sum f}\right)^2}$
$\displaystyle \text{(b) }\sigma=\sqrt{\left(\frac{\sum fd}{\sum f}\right)^2-\frac{\sum fd^2}{\sum f}}$
$\displaystyle \text{(c) }\sigma=\sqrt{\frac{\sum fd^2}{\sum f}}-\frac{\sum fd}{\sum f}$
$\displaystyle \text{(d) }\sigma=\sqrt{\left(\frac{\sum fd}{\sum f}\right)^2}-\frac{\sum fd^2}{\sum f}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(a) }\;\sigma=\sqrt{\frac{\sum fd^2}{\sum f}-\left(\frac{\sum fd}{\sum f}\right)^2}$
$\\$

$\displaystyle \textbf{Question 3. }\text{If }v\text{ is the variance and }\sigma\text{ is the standard deviation, then}$
$\displaystyle \text{(a) }v=\frac{1}{\sigma^2}\qquad \text{(b) }v=\frac{1}{\sigma}$
$\displaystyle \text{(c) }v=\sigma^2\qquad \text{(d) }v^2=\sigma$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(c) }\;v=\sigma^2$
$\\$

$\displaystyle \textbf{Question 4. }\text{The mean deviation from the median is}$
$\displaystyle \text{(a) equal to that measured from another value}$
$\displaystyle \text{(b) maximum if all observations are positive}$
$\displaystyle \text{(c) greater than that measured from any other value}$
$\displaystyle \text{(d) less than that measured from any other value}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{(d) less than that measured from any other value}$
$\\$

$\displaystyle \textbf{Question 5. }\text{If }n=10,\overline{X}=12\text{ and }\sum x_i^2=1530,\text{ then the coefficient of variation is}$
$\displaystyle \text{(a) }36\%\qquad \text{(b) }41\%\qquad \text{(c) }25\%\qquad \text{(d) none of these}$
$\displaystyle \text{Answer:}$
$\displaystyle \sigma^2=\frac{\sum x_i^2}{n}-\overline{X}^{\,2}$
$\displaystyle =\frac{1530}{10}-12^2$
$\displaystyle =153-144=9$
$\displaystyle \sigma=3$
$\displaystyle \text{Coefficient of variation}=\frac{\sigma}{\overline{X}}\times100$
$\displaystyle =\frac{3}{12}\times100$
$\displaystyle =25\%$
$\displaystyle \therefore \text{(c) }25\%$
$\\$

$\displaystyle \textbf{Question 6. }\text{The standard deviation of the data:}$
$\displaystyle \begin{array}{c|ccccc}x & 1 & a & a^2 & \cdots & a^n\\ \hline f & {^nC_0} & {^nC_1} & {^nC_2} & \cdots & {^nC_n}\end{array}$
$\displaystyle \text{is}$
$\displaystyle \text{(a) }\left(\frac{1+a^2}{2}\right)^n-\left(\frac{1+a}{2}\right)^n \qquad \text{(b) }\left(\frac{1+a^2}{2}\right)^{2n}-\left(\frac{1+a}{2}\right)^n$
$\displaystyle \text{(c) }\left(\frac{1+a}{2}\right)^{2n}-\left(\frac{1+a^2}{2}\right)^n \qquad \text{(d) none of these}$
$\displaystyle \text{Answer:}$
$\displaystyle \sum f={^nC_0}+{^nC_1}+\cdots+{^nC_n}=2^n$
$\displaystyle \sum fx={^nC_0}+{^nC_1}a+{^nC_2}a^2+\cdots+{^nC_n}a^n=(1+a)^n$
$\displaystyle \sum fx^2={^nC_0}+{^nC_1}a^2+{^nC_2}a^4+\cdots+{^nC_n}a^{2n}=(1+a^2)^n$
$\displaystyle \sigma^2=\frac{\sum fx^2}{\sum f}-\left(\frac{\sum fx}{\sum f}\right)^2$
$\displaystyle =\frac{(1+a^2)^n}{2^n}-\left(\frac{(1+a)^n}{2^n}\right)^2$
$\displaystyle =\left(\frac{1+a^2}{2}\right)^n-\left(\frac{1+a}{2}\right)^{2n}$
$\displaystyle \therefore \sigma=\sqrt{\left(\frac{1+a^2}{2}\right)^n-\left(\frac{1+a}{2}\right)^{2n}}$
$\displaystyle \therefore \text{(d) none of these}$
$\\$

$\displaystyle \textbf{Question 7. }\text{The mean deviation of the series }a,a+d,a+2d,\ldots,a+2nd\text{ from its mean is}$
$\displaystyle \text{(a) }\frac{(n+1)d}{2n+1}\qquad \text{(b) }\frac{nd}{2n+1}\qquad \text{(c) }\frac{n(n+1)d}{2n+1}\qquad \text{(d) }\frac{(2n+1)d}{n(n+1)}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Number of terms}=2n+1$
$\displaystyle \text{Mean}=a+nd$
$\displaystyle \text{Sum of absolute deviations}=2[d+2d+3d+\cdots+nd]$
$\displaystyle =2d\cdot\frac{n(n+1)}{2}$
$\displaystyle =n(n+1)d$
$\displaystyle \text{Mean deviation}=\frac{n(n+1)d}{2n+1}$
$\displaystyle \therefore \text{(c) }\frac{n(n+1)d}{2n+1}$
$\\$

$\displaystyle \textbf{Question 8. }\text{A batsman scores runs in }10\text{ innings as }38,70,48,34,42,55,63,46,54\text{ and }44.$
$\displaystyle \text{The mean deviation about mean is}$
$\displaystyle \text{(a) }8.6\qquad \text{(b) }6.4\qquad \text{(c) }10.6\qquad \text{(d) }7.6$
$\displaystyle \text{Answer:}$
$\displaystyle \overline{x}=\frac{38+70+48+34+42+55+63+46+54+44}{10}$
$\displaystyle =\frac{494}{10}=49.4$
$\displaystyle \sum |x-\overline{x}|=11.4+20.6+1.4+15.4+7.4+5.6+13.6+3.4+4.6+5.4$
$\displaystyle =88.8$
$\displaystyle \text{Mean deviation about mean}=\frac{88.8}{10}$
$\displaystyle =8.88$
$\displaystyle \therefore \text{(a) Mean deviation about mean}=8.88$
$\\$

$\displaystyle \textbf{Question 9. }\text{The mean deviation of the numbers }3,4,5,6,7\text{ from the mean is}$
$\displaystyle \text{(a) }25\qquad \text{(b) }5\qquad \text{(c) }1.2\qquad \text{(d) }0$
$\displaystyle \text{Answer:}$
$\displaystyle \overline{x}=\frac{3+4+5+6+7}{5}=5$
$\displaystyle \sum |x-\overline{x}|=|3-5|+|4-5|+|5-5|+|6-5|+|7-5|$
$\displaystyle =2+1+0+1+2=6$
$\displaystyle \text{Mean deviation}=\frac{6}{5}=1.2$
$\displaystyle \therefore \text{(c) }1.2$
$\\$

$\displaystyle \textbf{Question 10. }\text{The sum of the squares deviations for }10\text{ observations taken from their mean }50\text{ is }250.\text{ The}$
$\displaystyle \text{coefficient of variation is}$
$\displaystyle \text{(a) }10\%\qquad \text{(b) }40\%\qquad \text{(c) }50\%\qquad \text{(d) none of these}$
$\displaystyle \text{Answer:}$
$\displaystyle \sigma=\sqrt{\frac{\sum (x-\overline{x})^2}{n}}$
$\displaystyle =\sqrt{\frac{250}{10}}$
$\displaystyle =\sqrt{25}=5$
$\displaystyle \text{Coefficient of variation}=\frac{\sigma}{\overline{x}}\times100$
$\displaystyle =\frac{5}{50}\times100$
$\displaystyle =10\%$
$\displaystyle \therefore \text{(a) }10\%$
$\\$

$\displaystyle \textbf{Question 11. }\text{Let }x_1,x_2,\ldots,x_n\text{ be values taken by a variable }X\text{ and }y_1,y_2,\ldots,y_n\text{ be the values taken by a}$
$\displaystyle \text{variable }Y\text{ such that }y_i=ax_i+b,\ i=1,2,\ldots,n.\text{ Then,}$
$\displaystyle \text{(a) Var}(Y)=a^2\text{ Var}(X)\qquad \text{(b) Var}(X)=a^2\text{ Var}(Y)$
$\displaystyle \text{(c) Var}(X)=\text{Var}(X)+b\qquad \text{(d) none of these}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Var}(aX+b)=a^2\text{Var}(X)$
$\displaystyle \therefore \text{Var}(Y)=a^2\text{Var}(X)$
$\displaystyle \therefore \text{(a) Var}(Y)=a^2\text{ Var}(X)$
$\\$

$\displaystyle \textbf{Question 12. }\text{If the standard deviation of a variable }X\text{ is }\sigma,$
$\displaystyle \text{ then the standard deviation of variable} \frac{aX+b}{c}\text{ is}$
$\displaystyle \text{(a) }a\sigma\qquad \text{(b) }\frac{a}{c}\sigma\qquad \text{(c) }\left|\frac{a}{c}\right|\sigma\qquad \text{(d) }\frac{a\sigma+b}{c}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{S.D.}(aX+b)=|a|\sigma$
$\displaystyle \text{S.D.}\left(\frac{aX+b}{c}\right)=\left|\frac{a}{c}\right|\sigma$
$\displaystyle \therefore \text{(c) }\left|\frac{a}{c}\right|\sigma$
$\\$

$\displaystyle \textbf{Question 13. }\text{If the S.D. of a set of observations is }8\text{ and if each observation is}$
$\displaystyle \text{divided by }-2,\text{ the S.D. of} \text{the new set of observations will be}$
$\displaystyle \text{(a) }-4\qquad \text{(b) }-8\qquad \text{(c) }8\qquad \text{(d) }4$
$\displaystyle \text{Answer:}$
$\displaystyle \text{New S.D.}=\left|\frac{1}{-2}\right|\times8$
$\displaystyle =\frac{1}{2}\times8$
$\displaystyle =4$
$\displaystyle \therefore \text{(d) }4$
$\\$

$\displaystyle \textbf{Question 14. }\text{If two variates }X\text{ and }Y\text{ are connected by the relation }Y=\frac{aX+b}{c},\text{ where }$
$\displaystyle a,b,c\text{ are constants } \text{such that }ac<0,\text{ then}$
$\displaystyle \text{(a) }\sigma_Y=\frac{a}{c}\sigma_X\qquad \text{(b) }\sigma_Y=-\frac{a}{c}\sigma_X\qquad \text{(c) }\sigma_Y=\frac{a}{c}\sigma_X+b\qquad \text{(d) none of these}$
$\displaystyle \text{Answer:}$
$\displaystyle Y=\frac{a}{c}X+\frac{b}{c}$
$\displaystyle \sigma_Y=\left|\frac{a}{c}\right|\sigma_X$
$\displaystyle \text{Since }ac<0,\text{ }\frac{a}{c}<0$
$\displaystyle \therefore \left|\frac{a}{c}\right|=-\frac{a}{c}$
$\displaystyle \sigma_Y=-\frac{a}{c}\sigma_X$
$\displaystyle \therefore \text{(b) }\sigma_Y=-\frac{a}{c}\sigma_X$
$\\$

$\displaystyle \textbf{Question 15. }\text{If for a sample of size }60,\text{ we have the following information }$
$\displaystyle \sum x_i^2=18000\text{ and }\sum x_i=960, \text{then the variance is}$
$\displaystyle \text{(a) }6.63\qquad \text{(b) }16\qquad \text{(c) }22\qquad \text{(d) }44$
$\displaystyle \text{Answer:}$
$\displaystyle \sigma^2=\frac{\sum x_i^2}{n}-\left(\frac{\sum x_i}{n}\right)^2$
$\displaystyle =\frac{18000}{60}-\left(\frac{960}{60}\right)^2$
$\displaystyle =300-16^2$
$\displaystyle =300-256=44$
$\displaystyle \therefore \text{(d) }44$
$\\$

$\displaystyle \textbf{Question 16. }\text{Let }a,b,c,d,e\text{ be the observations with mean }m\text{ and standard deviation}$
$\displaystyle s.\text{ The standard } \text{deviation of the observations }a+k,b+k,c+k,d+k,e+k\text{ is}$
$\displaystyle \text{(a) }s\qquad \text{(b) }ks\qquad \text{(c) }s+k\qquad \text{(d) }\frac{s}{k}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Adding a constant }k\text{ to each observation does not change standard deviation.}$
$\displaystyle \therefore \text{New standard deviation}=s$
$\displaystyle \therefore \text{(a) }s$
$\\$

$\displaystyle \textbf{Question 17. }\text{The standard deviation of first }10\text{ natural numbers is}$
$\displaystyle \text{(a) }5.5\qquad \text{(b) }3.87\qquad \text{(c) }2.97\qquad \text{(d) }2.87$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Variance of first }n\text{ natural numbers}=\frac{n^2-1}{12}$
$\displaystyle \sigma^2=\frac{10^2-1}{12}$
$\displaystyle =\frac{99}{12}=8.25$
$\displaystyle \sigma=\sqrt{8.25}$
$\displaystyle =2.87$
$\displaystyle \therefore \text{(d) }2.87$
$\\$

$\displaystyle \textbf{Question 18. }\text{Consider the first }10\text{ positive integers. If we multiply each number by }$
$\displaystyle -1\text{ and then add }1 \text{ to each number, the variance of the numbers so obtained is}$
$\displaystyle \text{(a) }8.25 \qquad \text{(b) }6.5 \qquad \text{(c) }3.87 \qquad \text{(d) }2.87$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The first }10\text{ positive integers are }1,2,3,\ldots,10.$
$\displaystyle \text{Variance of the first }n\text{ positive integers is } \sigma^{2}=\frac{n^{2}-1}{12}.$
$\displaystyle \text{For }n=10,$
$\displaystyle \sigma^{2}=\frac{10^{2}-1}{12}=\frac{99}{12}=8.25.$
$\displaystyle \text{The given transformation is }y=-x+1.$
$\displaystyle \text{For a linear transformation }y=ax+b,\text{ variance becomes }a^{2}\text{ times the original variance.}$
$\displaystyle \text{Here }a=-1.$
$\displaystyle \text{Therefore,}$
$\displaystyle \text{Var}(y)=(-1)^{2}\times 8.25=8.25.$
$\displaystyle \therefore \text{The variance of the new numbers is }8.25.$
$\displaystyle \text{Hence, option (a) is correct.}$
$\\$

$\displaystyle \textbf{Question 19. }\text{Consider the numbers }1,2,3,4,5,6,7,8,9,10.\text{ If }1\text{ is}$
$\displaystyle \text{added to each number, the variance of the numbers so obtained is}$
$\displaystyle \text{(a) }6.5 \qquad \text{(b) }2.87 \qquad \text{(c) }3.87 \qquad \text{(d) }8.25$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The given numbers are }1,2,3,\ldots,10.$
$\displaystyle \text{Variance of the first }n\text{ positive integers is } \sigma^{2}=\frac{n^{2}-1}{12}.$
$\displaystyle \text{For }n=10,$
$\displaystyle \sigma^{2}=\frac{10^{2}-1}{12}=\frac{99}{12}=8.25.$
$\displaystyle \text{When a constant is added to every observation, the variance remains unchanged.}$
$\displaystyle \text{Hence, after adding }1\text{ to each number, the variance is still }8.25.$
$\displaystyle \therefore \text{The variance of the new numbers is }8.25.$
$\displaystyle \text{Hence, option (d) is correct.}$
$\\$

$\displaystyle \textbf{Question 20. }\text{The mean of }100\text{ observations is }50\text{ and their standard deviation is }5.$
$\displaystyle \text{The sum of all squares of all the observations is}$
$\displaystyle \text{(a) }50000\qquad \text{(b) }250000\qquad \text{(c) }252500\qquad \text{(d) }255000$
$\displaystyle \text{Answer:}$
$\displaystyle \sigma^2=\frac{\sum x^2}{n}-\overline{x}^{\,2}$
$\displaystyle 5^2=\frac{\sum x^2}{100}-50^2$
$\displaystyle 25=\frac{\sum x^2}{100}-2500$
$\displaystyle \frac{\sum x^2}{100}=2525$
$\displaystyle \sum x^2=252500$
$\displaystyle \therefore \text{(c) }252500$
$\\$

$\displaystyle \textbf{Question 21. }\text{Let }x_1,x_2,\ldots,x_n\text{ be }n\text{ observations. Let } \\ y_i=ax_i+b\text{ for }i=1,2,\ldots,n,$
$\displaystyle \text{ where }a\text{ and }b\text{ are } \text{constants. If the mean of }x_i\text{'s is }48\text{ and their standard deviation is }12,$
$\displaystyle \text{ the mean of }y_i\text{'s is }55\text{ and } \text{standard deviation of }y_i\text{'s is }15,\text{ the values of }a\text{ and }b\text{ are}$
$\displaystyle \text{(a) }a=1.25,\ b=-5\qquad \text{(b) }a=-1.25,\ b=5\qquad \\ \text{(c) }a=2.5,\ b=-5\qquad \text{(d) }a=2.5,\ b=5$
$\displaystyle \text{Answer:}$
$\displaystyle \overline{Y}=a\overline{X}+b$
$\displaystyle 55=48a+b \qquad (i)$
$\displaystyle \sigma_Y=|a|\sigma_X$
$\displaystyle 15=|a|\times12$
$\displaystyle |a|=\frac{15}{12}=1.25$
$\displaystyle \text{Taking }a=1.25,\text{ from }(i),$
$\displaystyle 55=48(1.25)+b$
$\displaystyle 55=60+b$
$\displaystyle b=-5$
$\displaystyle \therefore \text{(a) }a=1.25,\ b=-5$
$\\$

$\displaystyle \textbf{Question 22. }\text{The mean deviation of the data }3,10,10,4,7,10,5\text{ from the mean is}$
$\displaystyle \text{(a) }2\qquad \text{(b) }2.57\qquad \text{(c) }3\qquad \text{(d) }3.57$
$\displaystyle \text{Answer:}$
$\displaystyle \overline{x}=\frac{3+10+10+4+7+10+5}{7}$
$\displaystyle =\frac{49}{7}=7$
$\displaystyle \sum |x-\overline{x}|=|3-7|+|10-7|+|10-7|+|4-7|+|7-7|+|10-7|+|5-7|$
$\displaystyle =4+3+3+3+0+3+2=18$
$\displaystyle \text{Mean deviation}=\frac{18}{7}=2.57$
$\displaystyle \therefore \text{(b) }2.57$
$\\$

$\displaystyle \textbf{Question 23. }\text{The mean deviation for }n\text{ observations }x_1,x_2,\ldots,x_n \\ \text{from their mean }\overline{X}\text{ is given by}$
$\displaystyle \text{(a) }\sum_{i=1}^{n}(x_i-\overline{X})\qquad \text{(b) }\frac{1}{n}\sum_{i=1}^{n}(x_i-\overline{X})$
$\displaystyle \text{(c) }\sum_{i=1}^{n}(x_i-\overline{X})^2\qquad \text{(d) }\frac{1}{n}\sum_{i=1}^{n}(x_i-\overline{X})^2$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The mean deviation about the mean }\overline{X}\text{ is defined as the average of the absolute}$
$\displaystyle \text{deviations of the observations from their mean.}$
$\displaystyle \text{Thus,}$
$\displaystyle \mathrm{M.D.}=\frac{1}{n}\sum_{i=1}^{n}|x_{i}-\overline{X}|.$
$\displaystyle \text{The absolute value bars are not shown in any of the given options.}$
$\displaystyle \text{Among the given choices, option (b) represents the intended form of mean deviation,}$
$\displaystyle \text{whereas options (c) and (d) involve squared deviations and are related to variance.}$
$\displaystyle \therefore \text{The correct answer is option (b).}$
$\\$

$\displaystyle \textbf{Question 24. }\text{Let }x_1,x_2,\ldots,x_n\text{ be }n\text{ observations and }\overline{X}$
$\displaystyle \text{be their arithmetic mean. The standard deviation is given by}$
$\displaystyle \text{(a) }\sum_{i=1}^{n}(x_i-\overline{X})^2\qquad \text{(b) }\frac{1}{n}\sum_{i=1}^{n}(x_i-\overline{X})^2$
$\displaystyle \text{(c) }\sqrt{\frac{1}{n}\sum_{i=1}^{n}(x_i-\overline{X})^2}\qquad \text{(d) }\sqrt{\frac{1}{n}\sum_{i=1}^{n}x_i^2-\overline{X}^{\,2}}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Standard deviation}=\sqrt{\frac{1}{n}\sum_{i=1}^{n}(x_i-\overline{X})^2}$
$\displaystyle \therefore \text{(c) }\sqrt{\frac{1}{n}\sum_{i=1}^{n}(x_i-\overline{X})^2}$
$\\$

$\displaystyle \textbf{Question 25. }\text{The standard deviation of the observations }6,5,9,13,12,8,10\text{ is}$
$\displaystyle \text{(a) }6\qquad \text{(b) }\sqrt{6}\qquad \text{(c) }\frac{52}{7}\qquad \text{(d) }\sqrt{\frac{52}{7}}$
$\displaystyle \text{Answer:}$
$\displaystyle \overline{x}=\frac{6+5+9+13+12+8+10}{7}$
$\displaystyle =\frac{63}{7}=9$
$\displaystyle \sum (x-\overline{x})^2=(6-9)^2+(5-9)^2+(9-9)^2+(13-9)^2+(12-9)^2+(8-9)^2+(10-9)^2$
$\displaystyle =9+16+0+16+9+1+1=52$
$\displaystyle \sigma=\sqrt{\frac{\sum (x-\overline{x})^2}{n}}$
$\displaystyle =\sqrt{\frac{52}{7}}$
$\displaystyle \therefore \text{(d) }\sqrt{\frac{52}{7}}$
$\\$