\displaystyle \text{MULTIPLE CHOICE QUESTIONS (MCQs)}

\displaystyle \textbf{Question 1: }\text{Standard deviation of first 10 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{For the first }n\text{ natural numbers, }\sigma=\sqrt{\frac{n^2-1}{12}}.
\displaystyle \therefore \sigma=\sqrt{\frac{10^2-1}{12}}=\sqrt{\frac{99}{12}}.
\displaystyle =\sqrt{8.25}\approx2.87.
\displaystyle \therefore \text{The correct option is (d).}
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{If }\bar{x}\text{ is the mean of }n\text{ observations }x_1,x_2,x_3,\ldots,x_n,\text{ then}
\displaystyle \frac{1}{n}\sum_{i=1}^{n}|x_i-\bar{x}|=\text{ ?}
\displaystyle \text{(a) M.D. about mean}\qquad\text{(b) S.D.}\qquad\text{(c) }0\qquad\text{(d) None of these}
\displaystyle \text{Answer:}
\displaystyle \text{M.D. about mean}=\frac{1}{n}\sum_{i=1}^{n}|x_i-\bar{x}|.
\displaystyle \therefore \text{The given expression represents the mean deviation about mean.}
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{S.D. of 15 items is }6\text{ and if each item is decreased by }1,\text{ then S.D. will be:}
\displaystyle \text{(a) }5\qquad\text{(b) }7\qquad\text{(c) }\frac{1}{15}\qquad\text{(d) }6
\displaystyle \text{Answer:}
\displaystyle \text{S.D. is not affected by adding or subtracting the same constant from each observation.}
\displaystyle \therefore \text{New S.D.}=6.
\displaystyle \therefore \text{The correct option is (d).}
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{The range of the data }35,50,48,62,27,39,43,72,56,68\text{ is:}
\displaystyle \text{(a) }25\qquad\text{(b) }45\qquad\text{(c) }35\qquad\text{(d) }15
\displaystyle \text{Answer:}
\displaystyle \text{Range}=\text{Largest value}-\text{Smallest value}.
\displaystyle =72-27=45.
\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{A set of }n\text{ values }x_1,x_2,x_3,\ldots,x_n\text{ has S.D. }=\sigma.\text{ Then the S.D. of the}
\displaystyle \text{values }x_1+k,x_2+k,x_3+k,\ldots,x_n+k\text{ will be:}
\displaystyle \text{(a) }\sigma\qquad\text{(b) }\sigma+k\qquad\text{(c) }\sigma-k\qquad\text{(d) }k\sigma
\displaystyle \text{Answer:}
\displaystyle \text{S.D. is not affected by adding the same constant to every observation.}
\displaystyle \therefore \text{New S.D.}=\sigma.
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{If }x_i=ay_i+b\ (a\ne0)\text{ for each }i=1,2,3,\ldots,n,\text{ then:}
\displaystyle \text{(a) }\bar{y}=\bar{x}\qquad\text{(b) }\sigma_x=\sigma_y
\displaystyle \text{(c) }\sigma_x=|a|\sigma_y\qquad\text{(d) }\sigma_x=a\sigma_y
\displaystyle \text{Answer:}
\displaystyle x_i=ay_i+b.
\displaystyle \therefore \sigma_x=|a|\sigma_y.
\displaystyle \therefore \text{The correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{If the variance of the numbers }2,4,5,6,8,17\text{ is }23.33,\text{ then the}
\displaystyle \text{variance of }4,8,10,12,16,34\text{ will be:}
\displaystyle \text{(a) }23.33\qquad\text{(b) }46.66\qquad\text{(c) }93.32\qquad\text{(d) None of these}
\displaystyle \text{Answer:}
\displaystyle \text{Each observation in the second set is twice the corresponding observation in the first set.}
\displaystyle \text{If each observation is multiplied by }k,\text{ the variance is multiplied by }k^2.
\displaystyle \therefore \text{New variance}=2^2(23.33)=4(23.33)=93.32.
\displaystyle \therefore \text{The correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{When tested, the lives (in hours) of 5 bulbs were noted as follows:}
\displaystyle 1357,\ 1090,\ 1666,\ 1494,\ 1623.
\displaystyle \text{The mean deviation (in hours) from the mean is:}
\displaystyle \text{(a) }179\qquad\text{(b) }178\qquad\text{(c) }220\qquad\text{(d) }356
\displaystyle \text{Answer:}
\displaystyle \bar{x}=\frac{1357+1090+1666+1494+1623}{5}=\frac{7230}{5}=1446.
\displaystyle \sum|x_i-\bar{x}|=89+356+220+48+177=890.
\displaystyle \text{M.D. about mean}=\frac{\sum|x_i-\bar{x}|}{n}=\frac{890}{5}=178.
\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{Which of the following is one of the measures of dispersion?}
\displaystyle \text{(a) Mean}\qquad\text{(b) Range}\qquad\text{(c) Median}\qquad\text{(d) Mode}
\displaystyle \text{Answer:}
\displaystyle \text{Range is a measure of dispersion.}
\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{The mean deviation about the mean for the data }6,7,10,12,13,4,8,12\text{ is:}
\displaystyle \text{(a) }1.75\qquad\text{(b) }2.5\qquad\text{(c) }2.05\qquad\text{(d) }2.75
\displaystyle \text{Answer:}
\displaystyle \bar{x}=\frac{6+7+10+12+13+4+8+12}{8}=\frac{72}{8}=9.
\displaystyle \sum|x_i-\bar{x}|=3+2+1+3+4+5+1+3=22.
\displaystyle \text{M.D. about mean}=\frac{22}{8}=2.75.
\displaystyle \therefore \text{The correct option is (d).}
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{Following are the marks obtained by 9 students in a Mathematics test:}
\displaystyle 50,\ 69,\ 20,\ 33,\ 53,\ 39,\ 40,\ 65,\ 59.
\displaystyle \text{Mean deviation from the median is:}
\displaystyle \text{(a) }9\qquad\text{(b) }10.5\qquad\text{(c) }12.67\qquad\text{(d) }14.76
\displaystyle \text{Answer:}
\displaystyle \text{Arranging the observations: }20,33,39,40,50,53,59,65,69.
\displaystyle \text{Median}=50.
\displaystyle \sum|x_i-50|=30+17+11+10+0+3+9+15+19=114.
\displaystyle \text{M.D. about median}=\frac{114}{9}=12.67.
\displaystyle \therefore \text{The correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 12: }\text{The following information relates to a sample of size }60:
\displaystyle \sum x^2=18000,\qquad\sum x=960.
\displaystyle \text{The variance of the data is:}
\displaystyle \text{(a) }6.63\qquad\text{(b) }16\qquad\text{(c) }22\qquad\text{(d) }44
\displaystyle \text{Answer:}
\displaystyle \bar{x}=\frac{\sum x}{n}=\frac{960}{60}=16.
\displaystyle \sigma^2=\frac{\sum x^2}{n}-\bar{x}^2.
\displaystyle =\frac{18000}{60}-16^2.
\displaystyle =300-256=44.
\displaystyle \therefore \text{The correct option is (d).}
\displaystyle \\

\displaystyle \textbf{Question 13: }\text{The mean deviation about the mean for the following data is:}
\displaystyle \begin{array}{c|ccccc}x_i&10&30&50&70&90\\ \hline f_i&4&24&28&16&8\end{array}
\displaystyle \text{(a) }16\qquad\text{(b) }17\qquad\text{(c) }18\qquad\text{(d) }19
\displaystyle \text{Answer:}
\displaystyle \sum f_i=4+24+28+16+8=80.
\displaystyle \sum f_ix_i=40+720+1400+1120+720=4000.
\displaystyle \therefore \bar{x}=\frac{\sum f_ix_i}{\sum f_i}=\frac{4000}{80}=50.
\displaystyle \sum f_i|x_i-\bar{x}|=4(40)+24(20)+28(0)+16(20)+8(40).
\displaystyle =160+480+0+320+320=1280.
\displaystyle \text{M.D. about mean}=\frac{\sum f_i|x_i-\bar{x}|}{\sum f_i}=\frac{1280}{80}=16.
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 14: }\text{The possible value of }x\text{ if standard deviation of the numbers }2,3,2x\text{ and }11
\displaystyle \text{is }3.5\text{ is:}
\displaystyle \text{(a) }2\qquad\text{(b) }3\qquad\text{(c) }4\qquad\text{(d) }5
\displaystyle \text{Answer:}
\displaystyle \text{For }x=3,\text{ the observations are }2,3,6,11.
\displaystyle \bar{x}=\frac{2+3+6+11}{4}=\frac{22}{4}=5.5.
\displaystyle \sigma^2=\frac{(2-5.5)^2+(3-5.5)^2+(6-5.5)^2+(11-5.5)^2}{4}.
\displaystyle =\frac{12.25+6.25+0.25+30.25}{4}=\frac{49}{4}.
\displaystyle \therefore \sigma=\sqrt{\frac{49}{4}}=\frac{7}{2}=3.5.
\displaystyle \therefore x=3.
\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 15: }\text{The mean deviation from the median for the following series is:}
\displaystyle 1,\ 2,\ 3,\ 4,\ 4,\ 5,\ 6,\ 7
\displaystyle \text{(a) }3.4\qquad\text{(b) }4.3\qquad\text{(c) }4.4\qquad\text{(d) }3.3
\displaystyle \text{Answer:}
\displaystyle \text{Median}=\frac{4+4}{2}=4.
\displaystyle \sum|x_i-4|=3+2+1+0+0+1+2+3=12.
\displaystyle \text{M.D. about median}=\frac{12}{8}=1.5.
\displaystyle \therefore \text{none of the given options is correct.}
\displaystyle \\

\displaystyle \textbf{Question 16: }\text{The variance of the data is }121.\text{ Then the standard deviation of the data is:}
\displaystyle \text{(a) }11\qquad\text{(b) }12\qquad\text{(c) }13\qquad\text{(d) }1331
\displaystyle \text{Answer:}
\displaystyle \sigma^2=121.
\displaystyle \therefore \sigma=\sqrt{121}=11.
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 17: }\text{The mean deviation about the mean of first }n\text{ natural numbers, when}
\displaystyle n\text{ is an even number, is:}
\displaystyle \text{(a) }\frac{n^2}{4}\qquad\text{(b) }\frac{n}{4}\qquad\text{(c) }\frac{n^2}{2}\qquad\text{(d) }\frac{n^2-1}{4}
\displaystyle \text{Answer:}
\displaystyle \text{Let }n=2m.
\displaystyle \text{Mean}=\frac{n+1}{2}=m+\frac{1}{2}.
\displaystyle \sum|x_i-\bar{x}|=2\left(\frac{1}{2}+\frac{3}{2}+\cdots+\frac{2m-1}{2}\right).
\displaystyle =2\left(\frac{1+3+\cdots+(2m-1)}{2}\right).
\displaystyle =m^2.
\displaystyle \text{M.D. about mean}=\frac{m^2}{2m}=\frac{m}{2}=\frac{n}{4}.
\displaystyle \therefore \text{The correct option is (b).}
\displaystyle \\

\displaystyle \textbf{Question 18: }\text{The mean and standard deviation of 100 items are }50\text{ and }4
\displaystyle \text{respectively. The sum of squares of the items is:}
\displaystyle \text{(a) }251600\qquad\text{(b) }215600\qquad\text{(c) }216500\qquad\text{(d) }215060
\displaystyle \text{Answer:}
\displaystyle \sigma^2=\frac{\sum x^2}{n}-\bar{x}^{\,2}.
\displaystyle 4^2=\frac{\sum x^2}{100}-50^2.
\displaystyle 16=\frac{\sum x^2}{100}-2500.
\displaystyle \frac{\sum x^2}{100}=2516.
\displaystyle \therefore \sum x^2=251600.
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 19: }\text{The standard deviation of the data }6,5,9,13,12,8,10\text{ is:}
\displaystyle \text{(a) }\sqrt{\frac{52}{7}}\qquad\text{(b) }\frac{52}{7}\qquad\text{(c) }6\qquad\text{(d) }5
\displaystyle \text{Answer:}
\displaystyle \bar{x}=\frac{6+5+9+13+12+8+10}{7}=\frac{63}{7}=9.
\displaystyle \sum(x_i-\bar{x})^2=9+16+0+16+9+1+1=52.
\displaystyle \sigma^2=\frac{\sum(x_i-\bar{x})^2}{n}=\frac{52}{7}.
\displaystyle \therefore \sigma=\sqrt{\frac{52}{7}}.
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 20: }\text{The standard deviation of some temperature data in degree Celsius}
\displaystyle \text{is }5.\text{ If the data were converted into Fahrenheit, the variance would be:}
\displaystyle \text{(a) }81\qquad\text{(b) }57\qquad\text{(c) }36\qquad\text{(d) }25
\displaystyle \text{Answer:}
\displaystyle F=\frac{9}{5}C+32.
\displaystyle \text{Therefore, S.D. in Fahrenheit}=\frac{9}{5}\times5=9.
\displaystyle \text{Variance}=(\text{S.D.})^2=9^2=81.
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \text{CASE BASED/SOURCE BASED / PASSAGE BASED}

\displaystyle \textbf{Question 21: }\text{The scores of 10 students in a test with maximum marks 50 were as follows:}
\displaystyle 28,\ 36,\ 34,\ 28,\ 48,\ 22,\ 35,\ 27,\ 19,\ 41.
\displaystyle \text{(i) Find the variance of marks.}
\displaystyle \text{(ii) If 2 grace marks are awarded to each student, find the new variance.}
\displaystyle \text{Answer:}
\displaystyle \text{(i) }\bar{x}=\frac{28+36+34+28+48+22+35+27+19+41}{10}.
\displaystyle =\frac{318}{10}=31.8.
\displaystyle \sum x_i^2=28^2+36^2+34^2+28^2+48^2+22^2+35^2+27^2+19^2+41^2.
\displaystyle =10804.
\displaystyle \sigma^2=\frac{\sum x_i^2}{n}-\bar{x}^{\,2}.
\displaystyle =\frac{10804}{10}-(31.8)^2.
\displaystyle =1080.4-1011.24=69.16.
\displaystyle \therefore \text{The variance of marks is }69.16.

\displaystyle \text{(ii) Adding the same constant to every observation does not change the variance.}
\displaystyle \therefore \text{New variance}=69.16.
\displaystyle \\

\displaystyle \textbf{Question 22: }\text{The mean and standard deviation of some data for the time taken to}
\displaystyle \text{complete a test are calculated with the following results:}
\displaystyle \text{Number of observations}=25,\quad\text{mean}=18.2\text{ seconds},\quad\text{S.D.}=3.25\text{ seconds}.
\displaystyle \text{Further, another set of 15 observations }x_1,x_2,x_3,\ldots,x_{15}\text{ is available and}
\displaystyle \sum_{i=1}^{15}x_i=279,\qquad\sum_{i=1}^{15}x_i^2=5524.
\displaystyle \text{(i) Find the mean of all 40 observations.}
\displaystyle \text{(ii) Find the standard deviation of all 40 observations.}
\displaystyle \text{Answer:}
\displaystyle \text{For the first set, }n_1=25,\quad\bar{x}_1=18.2,\quad\sigma_1=3.25.
\displaystyle \sum x_1=n_1\bar{x}_1=25(18.2)=455.
\displaystyle \text{For the second set, }n_2=15,\quad\sum x_2=279.
\displaystyle \text{(i) Combined mean}=\frac{\sum x_1+\sum x_2}{n_1+n_2}.
\displaystyle =\frac{455+279}{25+15}=\frac{734}{40}=18.35.
\displaystyle \therefore \text{The mean of all 40 observations is }18.35\text{ seconds.}

\displaystyle \text{(ii) We use }\sigma^2=\frac{\sum x^2}{n}-\bar{x}^{\,2}.
\displaystyle \therefore \sum x_1^2=n_1(\sigma_1^2+\bar{x}_1^{\,2}).
\displaystyle =25\left[(3.25)^2+(18.2)^2\right].
\displaystyle =25(10.5625+331.24)=8545.0625.
\displaystyle \therefore \sum x^2=8545.0625+5524=14069.0625.
\displaystyle \text{Combined variance}=\frac{14069.0625}{40}-(18.35)^2.
\displaystyle =351.7265625-336.7225=15.0040625.
\displaystyle \therefore \sigma=\sqrt{15.0040625}\approx3.87.
\displaystyle \therefore \text{The standard deviation of all 40 observations is approximately }3.87\text{ seconds.}
\displaystyle \\

\displaystyle \textbf{Question 23: }\text{Let }x_1,x_2,x_3,\ldots,x_n\text{ be }n\text{ observations. If each observation is}
\displaystyle \text{increased, decreased, multiplied or divided by a non-zero constant }a,\text{ then the mean is also}
\displaystyle \text{increased, decreased, multiplied or divided by the same non-zero constant }a.
\displaystyle \text{In case of variance, if each observation is increased or decreased by the same constant }a,
\displaystyle \text{then the variance remains unchanged. But on multiplying or dividing each observation by the same}
\displaystyle \text{non-zero constant }a,\text{ the variance }\sigma^2\text{ becomes }a^2\sigma^2\text{ or }\frac{\sigma^2}{a^2}\text{ respectively.}
\displaystyle \text{Thus, variance is independent of change of origin but not of change of scale.}
\displaystyle \text{Based on the above information, answer the following questions:}
\displaystyle \text{(i) The mean of 10 observations is }18.\text{ If each observation is increased by }2,\text{ find the new mean.}
\displaystyle \text{(ii) The mean of 7 observations is }25.\text{ If }3\text{ is subtracted from each observation, find the new mean.}
\displaystyle \text{Answer:}
\displaystyle \text{(i) Original mean}=18.
\displaystyle \text{Since each observation is increased by }2,
\displaystyle \text{New mean}=18+2=20.
\displaystyle \therefore \text{The new mean is }20.

\displaystyle \text{(ii) Original mean}=25.
\displaystyle \text{Since }3\text{ is subtracted from each observation,}
\displaystyle \text{New mean}=25-3=22.
\displaystyle \therefore \text{The new mean is }22.
\displaystyle \\

\displaystyle \textbf{Question 24: }\text{Consider the following data:}
\displaystyle \begin{array}{|c|c|} \hline \text{Class}&\text{Frequency}\\ \hline 0-10&5\\10-20&7\\20-30&15\\30-40&16\\40-50&4\\50-60&2 \\ \hline \end{array}
\displaystyle \text{Based on the above information, answer the following questions:}
\displaystyle \text{(i) What is the median of the given data?}
\displaystyle \text{(ii) Find the mean deviation about median.}
\displaystyle \text{Answer:}
\displaystyle \text{(i) }N=5+7+15+16+4+2=49.
\displaystyle \frac{N}{2}=\frac{49}{2}=24.5.
\displaystyle \begin{array}{|c|c|c|} \hline \text{Class}&f&\text{Cumulative frequency}\\ \hline 0-10&5&5\\10-20&7&12\\20-30&15&27\\30-40&16&43\\40-50&4&47\\50-60&2&49 \\ \hline \end{array}
\displaystyle \therefore \text{Median class}=20-30.
\displaystyle \text{Median}=l+\left(\frac{\frac{N}{2}-c}{f}\right)h.
\displaystyle =20+\left(\frac{24.5-12}{15}\right)10.
\displaystyle =20+\frac{25}{3}=28.33.
\displaystyle \therefore \text{The median is approximately }28.33.

\displaystyle \text{(ii) The class marks are }5,15,25,35,45,55.
\displaystyle \begin{array}{|c|c|c|c|} \hline x_i&f_i&|x_i-M|&f_i|x_i-M|\\ \hline 5&5&23.33&116.67\\15&7&13.33&93.33\\25&15&3.33&50.00\\35&16&6.67&106.67\\45&4&16.67&66.67\\55&2&26.67&53.33 \\ \hline \end{array}
\displaystyle \sum f_i|x_i-M|=486.67.
\displaystyle \text{M.D. about median}=\frac{\sum f_i|x_i-M|}{\sum f_i}.
\displaystyle =\frac{486.67}{49}\approx9.93.
\displaystyle \therefore \text{The mean deviation about median is approximately }9.93.
\displaystyle \\

\displaystyle \textbf{Question 25: }\text{You are given some observations as }34,66,30,38,44,50,40,60,42,51.
\displaystyle \text{Based on these observations, answer the following questions:}
\displaystyle \text{(i) Find the mean deviation about the mean.}
\displaystyle \text{(ii) Find the mean deviation about the median.}
\displaystyle \text{Answer:}
\displaystyle \text{(i) }\bar{x}=\frac{34+66+30+38+44+50+40+60+42+51}{10}.
\displaystyle =\frac{455}{10}=45.5.
\displaystyle \sum|x_i-\bar{x}|=11.5+20.5+15.5+7.5+1.5+4.5+5.5+14.5+3.5+5.5.
\displaystyle =90.
\displaystyle \text{M.D. about mean}=\frac{90}{10}=9.
\displaystyle \therefore \text{The mean deviation about the mean is }9.

\displaystyle \text{(ii) Arranging the observations: }30,34,38,40,42,44,50,51,60,66.
\displaystyle \text{Median}=\frac{42+44}{2}=43.
\displaystyle \sum|x_i-43|=13+9+5+3+1+1+7+8+17+23.
\displaystyle =87.
\displaystyle \text{M.D. about median}=\frac{87}{10}=8.7.
\displaystyle \therefore \text{The mean deviation about the median is }8.7.
\displaystyle \\

\displaystyle \text{ASSERTION REASONING}

\displaystyle \textbf{Question 26: }\text{Assertion (A): The mean deviation about the mean for the data}
\displaystyle 4,\ 7,\ 8,\ 9,\ 10,\ 12,\ 13,\ 17\text{ is }3.
\displaystyle \text{Reason (R): The mean deviation about the mean for the data}
\displaystyle 38,\ 70,\ 48,\ 40,\ 42,\ 55,\ 63,\ 46,\ 54,\ 44\text{ is }8.5.
\displaystyle \text{(a) A is true, R is true; R is a correct explanation of A.}
\displaystyle \text{(b) A is true, R is true; R is not a correct explanation of A.}
\displaystyle \text{(c) A is true; R is false.}
\displaystyle \text{(d) A is false; R is true.}
\displaystyle \text{Answer:}
\displaystyle \text{For Assertion (A), }\bar{x}=\frac{4+7+8+9+10+12+13+17}{8}=10.
\displaystyle \sum|x_i-\bar{x}|=6+3+2+1+0+2+3+7=24.
\displaystyle \therefore \text{M.D. about mean}=\frac{24}{8}=3.
\displaystyle \therefore \text{Assertion (A) is true.}
\displaystyle \text{For Reason (R), }\bar{x}=\frac{38+70+48+40+42+55+63+46+54+44}{10}=50.
\displaystyle \sum|x_i-\bar{x}|=12+20+2+10+8+5+13+4+4+6=84.
\displaystyle \therefore \text{M.D. about mean}=\frac{84}{10}=8.4\ne8.5.
\displaystyle \therefore \text{Reason (R) is false.}
\displaystyle \therefore \text{The correct option is (c).}
\displaystyle \\

\displaystyle \textbf{Question 27: }\text{Assertion (A): The mean deviation about the mean as a measure}
\displaystyle \text{of dispersion has certain limitations.}
\displaystyle \text{Reason (R): The sum of absolute deviations from the mean is more than or equal to the}
\displaystyle \text{sum of absolute deviations from the median. Therefore, mean deviation about the mean has}
\displaystyle \text{certain limitations, particularly when the degree of variability is high.}
\displaystyle \text{(a) A is true, R is true; R is a correct explanation of A.}
\displaystyle \text{(b) A is true, R is true; R is not a correct explanation of A.}
\displaystyle \text{(c) A is true; R is false.}
\displaystyle \text{(d) A is false; R is true.}
\displaystyle \text{Answer:}
\displaystyle \text{Mean deviation has certain limitations as a measure of dispersion.}
\displaystyle \therefore \text{Assertion (A) is true.}
\displaystyle \text{The sum of absolute deviations is minimum when deviations are taken from the median.}
\displaystyle \therefore \sum|x_i-M|\le\sum|x_i-\bar{x}|.
\displaystyle \text{Hence, Reason (R) is true and explains the Assertion.}
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 28: }\text{Assertion (A): In order to find the dispersion of values of }x\text{ from}
\displaystyle \text{mean }\bar{x},\text{ we take absolute measure of dispersion.}
\displaystyle \text{Reason (R): Sum of the deviations from mean }(\bar{x})\text{ is zero.}
\displaystyle \text{(a) A is true, R is true; R is a correct explanation of A.}
\displaystyle \text{(b) A is true, R is true; R is not a correct explanation of A.}
\displaystyle \text{(c) A is true; R is false.}
\displaystyle \text{(d) A is false; R is true.}
\displaystyle \text{Answer:}
\displaystyle \sum_{i=1}^{n}(x_i-\bar{x})=0.
\displaystyle \text{Thus, positive and negative deviations from the mean cancel each other.}
\displaystyle \text{Hence, absolute deviations are taken to obtain a measure of dispersion.}
\displaystyle \therefore \text{Both A and R are true and R is the correct explanation of A.}
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \textbf{Question 29: }\text{Assertion (A): The variance of first }n\text{ even natural numbers is }\frac{n^2-1}{4}.
\displaystyle \text{Reason (R): The sum of first }n\text{ natural numbers is }\frac{n(n+1)}{2}\text{ and the}
\displaystyle \text{sum of squares of first }n\text{ natural numbers is }\frac{n(n+1)(2n+1)}{6}.
\displaystyle \text{(a) A is true, R is true; R is a correct explanation of A.}
\displaystyle \text{(b) A is true, R is true; R is not a correct explanation of A.}
\displaystyle \text{(c) A is true; R is false.}
\displaystyle \text{(d) A is false; R is true.}
\displaystyle \text{Answer:}
\displaystyle \text{The first }n\text{ even natural numbers are }2,4,6,\ldots,2n.
\displaystyle \bar{x}=\frac{2(1+2+\cdots+n)}{n}=n+1.
\displaystyle \frac{\sum x_i^2}{n}=\frac{4}{n}\frac{n(n+1)(2n+1)}{6}.
\displaystyle =\frac{2(n+1)(2n+1)}{3}.
\displaystyle \sigma^2=\frac{\sum x_i^2}{n}-\bar{x}^{\,2}.
\displaystyle =\frac{2(n+1)(2n+1)}{3}-(n+1)^2.
\displaystyle =\frac{n^2-1}{3}.
\displaystyle \therefore \text{Assertion (A) is false.}
\displaystyle \text{Both formulas stated in Reason (R) are correct.}
\displaystyle \therefore \text{Reason (R) is true.}
\displaystyle \therefore \text{The correct option is (d).}
\displaystyle \\

\displaystyle \textbf{Question 30: }\text{Assertion (A): If each of the observations }x_1,x_2,\ldots,x_n\text{ is}
\displaystyle \text{increased by }a,\text{ where }a\text{ is a negative or positive number, then the variance remains unchanged.}
\displaystyle \text{Reason (R): Adding or subtracting a positive or negative number to (or from) each}
\displaystyle \text{observation of a group does not affect the variance.}
\displaystyle \text{(a) A is true, R is true; R is a correct explanation of A.}
\displaystyle \text{(b) A is true, R is true; R is not a correct explanation of A.}
\displaystyle \text{(c) A is true; R is false.}
\displaystyle \text{(d) A is false; R is true.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }y_i=x_i+a.\text{ Then }\bar{y}=\bar{x}+a.
\displaystyle y_i-\bar{y}=(x_i+a)-(\bar{x}+a)=x_i-\bar{x}.
\displaystyle \therefore (y_i-\bar{y})^2=(x_i-\bar{x})^2.
\displaystyle \therefore \sigma_y^2=\sigma_x^2.
\displaystyle \text{Thus, adding or subtracting the same constant does not affect the variance.}
\displaystyle \therefore \text{Both A and R are true and R is the correct explanation of A.}
\displaystyle \therefore \text{The correct option is (a).}
\displaystyle \\

\displaystyle \text{VERY SHORT ANSWER QUESTIONS}

\displaystyle \textbf{Question 31: }\text{If for a distribution }\sum(x-5)=3,\ \sum(x-5)^2=43\text{ and the total}
\displaystyle \text{number of items is }18,\text{ find the S.D.}
\displaystyle \text{Answer:}
\displaystyle \sigma^2=\frac{\sum(x-5)^2}{n}-\left(\frac{\sum(x-5)}{n}\right)^2.
\displaystyle =\frac{43}{18}-\left(\frac{3}{18}\right)^2=\frac{43}{18}-\frac{1}{36}.
\displaystyle =\frac{85}{36}.
\displaystyle \therefore \sigma=\sqrt{\frac{85}{36}}=\frac{\sqrt{85}}{6}\approx1.54.
\displaystyle \\

\displaystyle \textbf{Question 32: }\text{Find the variance of first }n\text{ natural numbers.}
\displaystyle \text{Answer:}
\displaystyle \text{Variance}=\frac{n^2-1}{12}.
\displaystyle \\

\displaystyle \textbf{Question 33: }\text{The mean and standard deviation of 6 observations are }8\text{ and }4
\displaystyle \text{respectively. If each observation is multiplied by }3,\text{ find the new S.D.}
\displaystyle \text{Answer:}
\displaystyle \text{New S.D.}=|3|\times4=12.
\displaystyle \therefore \text{The new S.D. is }12.
\displaystyle \\

\displaystyle \textbf{Question 34: }\text{The mean and standard deviation of 100 observations are }40\text{ and }5.1
\displaystyle \text{respectively. If the observation }50\text{ is replaced by }40,\text{ find the new S.D.}
\displaystyle \text{Answer:}
\displaystyle \sum x=100(40)=4000.
\displaystyle \sigma^2=\frac{\sum x^2}{n}-\bar{x}^{\,2}.
\displaystyle (5.1)^2=\frac{\sum x^2}{100}-40^2.
\displaystyle \therefore \sum x^2=100\left[(5.1)^2+40^2\right]=162601.
\displaystyle \text{New }\sum x=4000-50+40=3990.
\displaystyle \therefore \text{New mean}=\frac{3990}{100}=39.9.
\displaystyle \text{New }\sum x^2=162601-50^2+40^2=161701.
\displaystyle \text{New variance}=\frac{161701}{100}-(39.9)^2.
\displaystyle =1617.01-1592.01=25.
\displaystyle \therefore \text{New S.D.}=\sqrt{25}=5.
\displaystyle \\

\displaystyle \textbf{Question 35: }\text{Let }x_1,x_2,x_3,x_4,x_5\text{ be the observations with mean }m\text{ and standard}
\displaystyle \text{deviation }s.\text{ Find the standard deviation of observations }Kx_1,Kx_2,Kx_3,Kx_4,Kx_5.
\displaystyle \text{Answer:}
\displaystyle \text{If every observation is multiplied by }K,\text{ the S.D. is multiplied by }|K|.
\displaystyle \therefore \text{New S.D.}=|K|s.
\displaystyle \text{If }K>0,\text{ then the new S.D.}=Ks.
\displaystyle \\

\displaystyle \textbf{Question 36: }\text{Find the mean deviation from the mean for the set of observations }1,0,5.
\displaystyle \text{Answer:}
\displaystyle \bar{x}=\frac{1+0+5}{3}=2.
\displaystyle \text{M.D. about mean}=\frac{|1-2|+|0-2|+|5-2|}{3}.
\displaystyle =\frac{1+2+3}{3}=\frac{6}{3}=2.
\displaystyle \therefore \text{The mean deviation from the mean is }2.
\displaystyle \\

\displaystyle \textbf{Question 37: }\text{Find the variance of the following data:}
\displaystyle 6,\ 8,\ 10,\ 12,\ 14,\ 16,\ 18,\ 20,\ 22,\ 24.
\displaystyle \text{Answer:}
\displaystyle \bar{x}=\frac{6+8+10+12+14+16+18+20+22+24}{10}=15.
\displaystyle \sum(x_i-\bar{x})^2=81+49+25+9+1+1+9+25+49+81=330.
\displaystyle \sigma^2=\frac{\sum(x_i-\bar{x})^2}{n}=\frac{330}{10}=33.
\displaystyle \therefore \text{The variance is }33.
\displaystyle \\

\displaystyle \textbf{Question 38: }\text{Find the mean deviation about the mean for the data:}
\displaystyle 38,\ 70,\ 48,\ 40,\ 42,\ 55,\ 63,\ 46,\ 54,\ 44.
\displaystyle \text{Answer:}
\displaystyle \bar{x}=\frac{38+70+48+40+42+55+63+46+54+44}{10}=\frac{500}{10}=50.
\displaystyle \sum|x_i-\bar{x}|=12+20+2+10+8+5+13+4+4+6=84.
\displaystyle \text{M.D. about mean}=\frac{84}{10}=8.4.
\displaystyle \therefore \text{The mean deviation about the mean is }8.4.
\displaystyle \\

\displaystyle \textbf{Question 39: }\text{Find the mean deviation about the median of the following data:}
\displaystyle 3,\ 6,\ 11,\ 12,\ 18.
\displaystyle \text{Answer:}
\displaystyle \text{Median}=11.
\displaystyle \sum|x_i-11|=|3-11|+|6-11|+|11-11|+|12-11|+|18-11|.
\displaystyle =8+5+0+1+7=21.
\displaystyle \text{M.D. about median}=\frac{21}{5}=4.2.
\displaystyle \therefore \text{The mean deviation about the median is }4.2.
\displaystyle \\

\displaystyle \textbf{Question 40: }\text{Find the variance of first five natural numbers.}
\displaystyle \text{Answer:}
\displaystyle \bar{x}=\frac{1+2+3+4+5}{5}=3.
\displaystyle \sigma^2=\frac{(1-3)^2+(2-3)^2+(3-3)^2+(4-3)^2+(5-3)^2}{5}.
\displaystyle =\frac{4+1+0+1+4}{5}=\frac{10}{5}=2.
\displaystyle \therefore \text{The variance is }2.
\displaystyle \\

\displaystyle \textbf{Question 41: }\text{The variance of 10 observations is }4.\text{ If each observation is}
\displaystyle \text{multiplied by }3,\text{ find the variance of the new data.}
\displaystyle \text{Answer:}
\displaystyle \text{If each observation is multiplied by }k,\text{ the variance is multiplied by }k^2.
\displaystyle \therefore \text{New variance}=3^2\times4=9\times4=36.
\displaystyle \therefore \text{The variance of the new data is }36.
\displaystyle \\

\displaystyle \textbf{Question 42: }\text{Find the standard deviation of }2,4,5,6,8,17.
\displaystyle \text{Answer:}
\displaystyle \bar{x}=\frac{2+4+5+6+8+17}{6}=\frac{42}{6}=7.
\displaystyle \sum(x_i-\bar{x})^2=25+9+4+1+1+100=140.
\displaystyle \sigma^2=\frac{\sum(x_i-\bar{x})^2}{n}=\frac{140}{6}=\frac{70}{3}.
\displaystyle \therefore \sigma=\sqrt{\frac{70}{3}}\approx4.83.
\displaystyle \therefore \text{The standard deviation is }\sqrt{\frac{70}{3}}.
\displaystyle \\

\displaystyle \textbf{Question 43: }\text{Find the mean deviation of the data }2,9,9,3,6,9,4\text{ from the mean.}
\displaystyle \text{Answer:}
\displaystyle \bar{x}=\frac{2+9+9+3+6+9+4}{7}=\frac{42}{7}=6.
\displaystyle \sum|x_i-\bar{x}|=4+3+3+3+0+3+2=18.
\displaystyle \text{M.D. about mean}=\frac{18}{7}\approx2.57.
\displaystyle \therefore \text{The mean deviation from the mean is }\frac{18}{7}\approx2.57.
\displaystyle \\

\displaystyle \textbf{Question 44: }\text{Variance of the data }2,4,5,6,8,17\text{ is }23.33.\text{ What will be the}
\displaystyle \text{variance of }6,12,15,18,24,51\text{?}
\displaystyle \text{Answer:}
\displaystyle 6=3(2),\ 12=3(4),\ 15=3(5),\ 18=3(6),\ 24=3(8),\ 51=3(17).
\displaystyle \text{Thus, each observation is multiplied by }3.
\displaystyle \text{If each observation is multiplied by }k,\text{ the variance is multiplied by }k^2.
\displaystyle \therefore \text{New variance}=3^2(23.33)=9(23.33)=209.97.
\displaystyle \therefore \text{The variance of the new data is }209.97.
\displaystyle \\

\displaystyle \textbf{Question 45: }\text{The variance of 20 observations is }5.\text{ If each observation is}
\displaystyle \text{multiplied by }2,\text{ find the new variance of the resulting observations.}
\displaystyle \text{Answer:}
\displaystyle \text{If each observation is multiplied by }k,\text{ the variance is multiplied by }k^2.
\displaystyle \therefore \text{New variance}=2^2(5)=4(5)=20.
\displaystyle \therefore \text{The new variance is }20.
\displaystyle \\

\displaystyle \textbf{Question 46: }\text{Let }a,b,c,d,e\text{ be the observations with mean }m\text{ and standard}
\displaystyle \text{deviation }s.\text{ Find the standard deviation of the observations }a+k,b+k,c+k,d+k,e+k.
\displaystyle \text{Answer:}
\displaystyle \text{Adding the same constant }k\text{ to every observation does not change the standard deviation.}
\displaystyle \therefore \text{The standard deviation of the new observations is }s.
\displaystyle \\

\displaystyle \textbf{Question 47: }\text{Consider the first 10 positive integers. If we multiply each number}
\displaystyle \text{by }-1\text{ and then add }1\text{ to each number, find the variance of the numbers so obtained.}
\displaystyle \text{Answer:}
\displaystyle \text{Variance of first }n\text{ natural numbers}=\frac{n^2-1}{12}.
\displaystyle \therefore \text{Variance of first 10 positive integers}=\frac{10^2-1}{12}.
\displaystyle =\frac{99}{12}=\frac{33}{4}=8.25.
\displaystyle \text{Multiplying each observation by }-1\text{ multiplies the variance by }(-1)^2=1.
\displaystyle \text{Adding }1\text{ to each observation does not affect the variance.}
\displaystyle \therefore \text{The required variance is }\frac{33}{4}=8.25.
\displaystyle \\

\displaystyle \textbf{Question 48: }\text{The standard deviation of some temperature data in }^\circ\text{C is }10.\text{ If the}
\displaystyle \text{data were converted into }^\circ\text{F, find the variance.}
\displaystyle \text{Answer:}
\displaystyle F=\frac{9}{5}C+32.
\displaystyle \therefore \text{S.D. in }^\circ\text{F}=\frac{9}{5}\times10=18.
\displaystyle \therefore \text{Variance}=18^2=324.
\displaystyle \\

\displaystyle \textbf{Question 49: }\text{If the variance of a data is }144,\text{ then the standard deviation of the}
\displaystyle \text{data is }\underline{\hspace{1.5cm}}.
\displaystyle \text{Answer:}
\displaystyle \text{S.D.}=\sqrt{\text{Variance}}=\sqrt{144}=12.
\displaystyle \therefore \text{The standard deviation is }12.
\displaystyle \\

\displaystyle \textbf{Question 50: }\text{Find the variance and standard deviation for the following data:}
\displaystyle 57,\ 64,\ 43,\ 67,\ 49,\ 59,\ 44,\ 47,\ 61,\ 59.
\displaystyle \text{Answer:}
\displaystyle \bar{x}=\frac{57+64+43+67+49+59+44+47+61+59}{10}=\frac{550}{10}=55.
\displaystyle \sum(x_i-\bar{x})^2=4+81+144+144+36+16+121+64+36+16=662.
\displaystyle \text{Variance}=\frac{\sum(x_i-\bar{x})^2}{n}=\frac{662}{10}=66.2.
\displaystyle \text{S.D.}=\sqrt{66.2}\approx8.14.
\displaystyle \therefore \text{Variance}=66.2\text{ and standard deviation}\approx8.14.
\displaystyle \\

\displaystyle \text{SHORT ANSWER QUESTIONS}

\displaystyle \textbf{Question 51: }\text{Find the mean and variance for the data }6,7,10,12,13,4,8,12.
\displaystyle \text{Answer:}
\displaystyle \bar{x}=\frac{6+7+10+12+13+4+8+12}{8}=\frac{72}{8}=9.
\displaystyle \sum x^2=36+49+100+144+169+16+64+144=722.
\displaystyle \sigma^2=\frac{\sum x^2}{n}-\bar{x}^{\,2}.
\displaystyle =\frac{722}{8}-9^2=90.25-81=9.25.
\displaystyle \therefore \text{Mean}=9\text{ and variance}=9.25.
\displaystyle \\

\displaystyle \textbf{Question 52: }\text{If mean and standard deviation of 100 items are }50\text{ and }4
\displaystyle \text{respectively, find the sum of all the items and the sum of the squares of the items.}
\displaystyle \text{Answer:}
\displaystyle \bar{x}=\frac{\sum x}{n}.
\displaystyle 50=\frac{\sum x}{100}.
\displaystyle \therefore \sum x=5000.
\displaystyle \sigma^2=\frac{\sum x^2}{n}-\bar{x}^{\,2}.
\displaystyle 4^2=\frac{\sum x^2}{100}-50^2.
\displaystyle 16=\frac{\sum x^2}{100}-2500.
\displaystyle \therefore \frac{\sum x^2}{100}=2516.
\displaystyle \therefore \sum x^2=251600.
\displaystyle \therefore \text{Sum of all items}=5000\text{ and sum of their squares}=251600.
\displaystyle \\

\displaystyle \textbf{Question 53: }\text{The standard deviation for the data }6,7,10,12,13,4,8,12\text{ is }\underline{\hspace{1.5cm}}.
\displaystyle \text{Answer:}
\displaystyle \bar{x}=\frac{6+7+10+12+13+4+8+12}{8}=9.
\displaystyle \sum x^2=722.
\displaystyle \sigma^2=\frac{722}{8}-9^2=90.25-81=9.25.
\displaystyle \therefore \sigma=\sqrt{9.25}\approx3.04.
\displaystyle \therefore \text{The standard deviation is approximately }3.04.
\displaystyle \\

\displaystyle \textbf{Question 54: }\text{Following are the marks obtained by 9 students in a Mathematics test:}
\displaystyle 50,\ 69,\ 20,\ 33,\ 53,\ 39,\ 40,\ 65,\ 59.
\displaystyle \text{Find the mean deviation from the median.}
\displaystyle \text{Answer:}
\displaystyle \text{Arranging the observations: }20,33,39,40,50,53,59,65,69.
\displaystyle \text{Median}=50.
\displaystyle \sum|x_i-50|=30+17+11+10+0+3+9+15+19=114.
\displaystyle \text{M.D. about median}=\frac{114}{9}\approx12.67.
\displaystyle \therefore \text{The mean deviation from the median is approximately }12.67.
\displaystyle \\

\displaystyle \textbf{Question 55: }\text{Find the mean deviation about the mean for the following data:}
\displaystyle 12,3,18,17,4,9,17,19,20,15,8,17,2,3,16,11,3,1,0,5.
\displaystyle \text{Answer:}
\displaystyle \bar{x}=\frac{\sum x_i}{n}=\frac{200}{20}=10.
\displaystyle \sum|x_i-\bar{x}|=124.
\displaystyle \text{M.D. about mean}=\frac{\sum|x_i-\bar{x}|}{n}.
\displaystyle =\frac{124}{20}=6.2.
\displaystyle \therefore \text{The mean deviation about the mean is }6.2.
\displaystyle \\

\displaystyle \textbf{Question 56: }\text{Find mean deviation about the mean for the following data:}
\displaystyle \begin{array}{|c|c|c|c|c|c|c|} \hline x_i&2&5&6&8&10&12\\ \hline f_i&2&8&10&7&8&5 \\ \hline \end{array}
\displaystyle \text{Answer:}
\displaystyle \sum f_i=2+8+10+7+8+5=40.
\displaystyle \sum f_ix_i=4+40+60+56+80+60=300.
\displaystyle \bar{x}=\frac{\sum f_ix_i}{\sum f_i}=\frac{300}{40}=7.5.
\displaystyle \begin{array}{|c|c|c|c|} \hline x_i&f_i&|x_i-\bar{x}|&f_i|x_i-\bar{x}|\\ \hline 2&2&5.5&11\\5&8&2.5&20\\6&10&1.5&15\\8&7&0.5&3.5\\10&8&2.5&20\\12&5&4.5&22.5 \\ \hline \end{array}
\displaystyle \sum f_i|x_i-\bar{x}|=92.
\displaystyle \text{M.D. about mean}=\frac{\sum f_i|x_i-\bar{x}|}{\sum f_i}.
\displaystyle =\frac{92}{40}=2.3.
\displaystyle \therefore \text{The mean deviation about the mean is }2.3.
\displaystyle \\

\displaystyle \textbf{Question 57: }\text{Find the mean and variance of first 10 multiples of }3.
\displaystyle \text{Answer:}
\displaystyle \text{The first 10 multiples of }3\text{ are }3,6,9,\ldots,30.
\displaystyle \bar{x}=3\left(\frac{10+1}{2}\right)=\frac{33}{2}=16.5.
\displaystyle \text{Variance of first }n\text{ natural numbers}=\frac{n^2-1}{12}.
\displaystyle \text{Variance of first 10 natural numbers}=\frac{10^2-1}{12}=\frac{33}{4}.
\displaystyle \text{Since each observation is multiplied by }3,\text{ variance is multiplied by }3^2.
\displaystyle \therefore \text{Variance}=9\left(\frac{33}{4}\right)=\frac{297}{4}=74.25.
\displaystyle \therefore \text{Mean}=16.5\text{ and variance}=74.25.
\displaystyle \\

\displaystyle \textbf{Question 58: }\text{Find the mean and variance for the following frequency distribution:}
\displaystyle \begin{array}{|c|c|c|c|c|c|} \hline \text{Classes}&0-10&10-20&20-30&30-40&40-50\\ \hline \text{Frequencies}&5&8&15&16&6 \\ \hline \end{array}
\displaystyle \text{Answer:}
\displaystyle \begin{array}{|c|c|c|c|c|} \hline x_i&f_i&f_ix_i&x_i^2&f_ix_i^2\\ \hline 5&5&25&25&125\\15&8&120&225&1800\\25&15&375&625&9375\\35&16&560&1225&19600\\45&6&270&2025&12150 \\ \hline \end{array}
\displaystyle \sum f_i=50,\qquad\sum f_ix_i=1350,\qquad\sum f_ix_i^2=43050.
\displaystyle \bar{x}=\frac{\sum f_ix_i}{\sum f_i}=\frac{1350}{50}=27.
\displaystyle \sigma^2=\frac{\sum f_ix_i^2}{\sum f_i}-\bar{x}^{\,2}.
\displaystyle =\frac{43050}{50}-27^2=861-729=132.
\displaystyle \therefore \text{Mean}=27\text{ and variance}=132.
\displaystyle \\

\displaystyle \textbf{Question 59: }\text{Find the mean deviation about the median for the following data:}
\displaystyle \begin{array}{|c|c|c|c|c|c|c|c|c|} \hline x_i&3&6&9&12&13&15&21&22\\ \hline f_i&3&4&5&2&4&5&4&3 \\ \hline \end{array}
\displaystyle \text{Answer:}
\displaystyle \sum f_i=3+4+5+2+4+5+4+3=30.
\displaystyle \begin{array}{|c|c|c|} \hline x_i&f_i&\text{Cumulative frequency}\\ \hline 3&3&3\\6&4&7\\9&5&12\\12&2&14\\13&4&18\\15&5&23\\21&4&27\\22&3&30 \\ \hline \end{array}
\displaystyle \text{The }15\text{th and }16\text{th observations are both }13.
\displaystyle \therefore \text{Median}=13.
\displaystyle \begin{array}{|c|c|c|c|} \hline x_i&f_i&|x_i-13|&f_i|x_i-13|\\ \hline 3&3&10&30\\6&4&7&28\\9&5&4&20\\12&2&1&2\\13&4&0&0\\15&5&2&10\\21&4&8&32\\22&3&9&27 \\ \hline \end{array}
\displaystyle \sum f_i|x_i-13|=149.
\displaystyle \text{M.D. about median}=\frac{149}{30}\approx4.97.
\displaystyle \therefore \text{The mean deviation about the median is approximately }4.97.
\displaystyle \\

\displaystyle \textbf{Question 60: }\text{Find the mean deviation about the mean for the following data:}
\displaystyle \begin{array}{|c|c|c|c|c|c|c|c|} \hline \text{Marks obtained}&10-20&20-30&30-40&40-50&50-60&60-70&70-80\\ \hline \text{Number of students}&2&3&8&14&8&3&2 \\ \hline \end{array}
\displaystyle \text{Answer:}
\displaystyle \text{The class marks are }15,25,35,45,55,65,75.
\displaystyle \sum f_i=2+3+8+14+8+3+2=40.
\displaystyle \sum f_ix_i=30+75+280+630+440+195+150=1800.
\displaystyle \bar{x}=\frac{\sum f_ix_i}{\sum f_i}=\frac{1800}{40}=45.
\displaystyle \begin{array}{|c|c|c|c|} \hline x_i&f_i&|x_i-45|&f_i|x_i-45|\\ \hline 15&2&30&60\\25&3&20&60\\35&8&10&80\\45&14&0&0\\55&8&10&80\\65&3&20&60\\75&2&30&60 \\ \hline \end{array}
\displaystyle \sum f_i|x_i-\bar{x}|=400.
\displaystyle \text{M.D. about mean}=\frac{400}{40}=10.
\displaystyle \therefore \text{The mean deviation about the mean is }10.
\displaystyle \\

\displaystyle \textbf{Question 61: }\text{The A.M. and S.D. of 9 variates are }43\text{ and }5\text{ respectively. If a variate}
\displaystyle 63\text{ is added to the variates, find the mean and standard deviation of the 10 variates.}
\displaystyle \text{Answer:}
\displaystyle n=9,\quad \bar{x}=43,\quad \sigma=5.
\displaystyle \sum x=n\bar{x}=9(43)=387.
\displaystyle \sigma^2=\frac{\sum x^2}{n}-\bar{x}^{\,2}.
\displaystyle 25=\frac{\sum x^2}{9}-43^2.
\displaystyle \therefore \sum x^2=9(25+1849)=16866.
\displaystyle \text{After adding }63,
\displaystyle \sum x=387+63=450.
\displaystyle \therefore \bar{x}_{\text{new}}=\frac{450}{10}=45.
\displaystyle \sum x^2_{\text{new}}=16866+63^2=20835.
\displaystyle \sigma_{\text{new}}^2=\frac{20835}{10}-45^2.
\displaystyle =2083.5-2025=58.5.
\displaystyle \therefore \sigma_{\text{new}}=\sqrt{58.5}\approx7.65.
\displaystyle \therefore \text{New mean}=45\text{ and new S.D.}\approx7.65.
\displaystyle \\

\displaystyle \textbf{Question 62: }\text{Find the mean and variance of the frequency distribution given below:}
\displaystyle \begin{array}{|c|c|c|c|c|} \hline x&1\le x<3&3\le x<5&5\le x<7&7\le x<10\\ \hline f&6&4&5&1 \\ \hline \end{array}
\displaystyle \text{Answer:}
\displaystyle \text{The class marks are }2,\ 4,\ 6,\ 8.5.
\displaystyle \begin{array}{|c|c|c|c|} \hline x_i&f_i&f_ix_i&f_ix_i^2\\ \hline 2&6&12&24\\4&4&16&64\\6&5&30&180\\8.5&1&8.5&72.25 \\ \hline \end{array}
\displaystyle \sum f_i=16,\qquad\sum f_ix_i=66.5,\qquad\sum f_ix_i^2=340.25.
\displaystyle \bar{x}=\frac{66.5}{16}=4.15625.
\displaystyle \sigma^2=\frac{\sum f_ix_i^2}{\sum f_i}-\bar{x}^{\,2}.
\displaystyle =\frac{340.25}{16}-(4.15625)^2.
\displaystyle =21.265625-17.2744140625.
\displaystyle \approx3.99.
\displaystyle \therefore \text{Mean}\approx4.16\text{ and variance}\approx3.99.
\displaystyle \\

\displaystyle \textbf{Question 63: }\text{For the following frequency distribution, find the standard deviation:}
\displaystyle \begin{array}{|c|c|c|c|c|c|c|} \hline x_i&2&3&4&5&6&7\\ \hline f_i&4&9&16&14&11&6 \\ \hline \end{array}
\displaystyle \text{Answer:}
\displaystyle \begin{array}{|c|c|c|c|} \hline x_i&f_i&f_ix_i&f_ix_i^2\\ \hline 2&4&8&16\\3&9&27&81\\4&16&64&256\\5&14&70&350\\6&11&66&396\\7&6&42&294 \\ \hline \end{array}
\displaystyle \sum f_i=60,\qquad\sum f_ix_i=277,\qquad\sum f_ix_i^2=1393.
\displaystyle \bar{x}=\frac{\sum f_ix_i}{\sum f_i}=\frac{277}{60}.
\displaystyle \sigma^2=\frac{\sum f_ix_i^2}{\sum f_i}-\bar{x}^{\,2}.
\displaystyle =\frac{1393}{60}-\left(\frac{277}{60}\right)^2.
\displaystyle =\frac{83580-76729}{3600}=\frac{6851}{3600}.
\displaystyle \therefore \sigma=\sqrt{\frac{6851}{3600}}=\frac{\sqrt{6851}}{60}\approx1.38.
\displaystyle \therefore \text{The standard deviation is approximately }1.38.
\displaystyle \\

\displaystyle \textbf{Question 64: }\text{The following frequency distribution, where }A\text{ is a positive integer,}
\displaystyle \text{has a variance of }160.\text{ Determine the value of }A.
\displaystyle \begin{array}{|c|c|c|c|c|c|c|} \hline x_i&A&2A&3A&4A&5A&6A\\ \hline f_i&2&1&1&1&1&1 \\ \hline \end{array}
\displaystyle \text{Answer:}
\displaystyle \sum f_i=7.
\displaystyle \sum f_ix_i=2A+2A+3A+4A+5A+6A=22A.
\displaystyle \sum f_ix_i^2=2A^2+4A^2+9A^2+16A^2+25A^2+36A^2=92A^2.
\displaystyle \bar{x}=\frac{22A}{7}.
\displaystyle \sigma^2=\frac{92A^2}{7}-\left(\frac{22A}{7}\right)^2.
\displaystyle =\frac{644A^2-484A^2}{49}=\frac{160A^2}{49}.
\displaystyle \text{Given, }\sigma^2=160.
\displaystyle \therefore \frac{160A^2}{49}=160.
\displaystyle A^2=49.
\displaystyle \therefore A=7,\text{ since }A\text{ is positive.}
\displaystyle \therefore \text{The required value of }A\text{ is }7.
\displaystyle \\

\displaystyle \textbf{Question 65: }\text{Find the mean deviation about the mean of the distribution:}
\displaystyle \begin{array}{|c|c|c|c|c|c|} \hline \text{Size}&20&21&22&23&24\\ \hline \text{Frequency}&6&4&5&1&4 \\ \hline \end{array}
\displaystyle \text{Answer:}
\displaystyle \sum f_i=6+4+5+1+4=20.
\displaystyle \sum f_ix_i=120+84+110+23+96=433.
\displaystyle \bar{x}=\frac{\sum f_ix_i}{\sum f_i}=\frac{433}{20}=21.65.
\displaystyle \begin{array}{|c|c|c|c|} \hline x_i&f_i&|x_i-\bar{x}|&f_i|x_i-\bar{x}|\\ \hline 20&6&1.65&9.90\\21&4&0.65&2.60\\22&5&0.35&1.75\\23&1&1.35&1.35\\24&4&2.35&9.40 \\ \hline \end{array}
\displaystyle \sum f_i|x_i-\bar{x}|=25.
\displaystyle \text{M.D. about mean}=\frac{25}{20}=1.25.
\displaystyle \therefore \text{The mean deviation about the mean is }1.25.
\displaystyle \\

\displaystyle \textbf{Question 66: }\text{Find the mean deviation about the median of the following distribution:}
\displaystyle \begin{array}{|c|c|c|c|c|c|} \hline \text{Marks obtained}&10&11&12&14&15\\ \hline \text{No. of students}&2&3&8&3&4 \\ \hline \end{array}
\displaystyle \text{Answer:}
\displaystyle \sum f_i=2+3+8+3+4=20.
\displaystyle \begin{array}{|c|c|c|} \hline x_i&f_i&\text{Cumulative frequency}\\ \hline 10&2&2\\11&3&5\\12&8&13\\14&3&16\\15&4&20 \\ \hline \end{array}
\displaystyle \text{The }10\text{th and }11\text{th observations are both }12.
\displaystyle \therefore \text{Median}=12.
\displaystyle \begin{array}{|c|c|c|c|} \hline x_i&f_i&|x_i-12|&f_i|x_i-12|\\ \hline 10&2&2&4\\11&3&1&3\\12&8&0&0\\14&3&2&6\\15&4&3&12 \\ \hline \end{array}
\displaystyle \sum f_i|x_i-12|=25.
\displaystyle \text{M.D. about median}=\frac{25}{20}=1.25.
\displaystyle \therefore \text{The mean deviation about the median is }1.25.
\displaystyle \\

\displaystyle \textbf{Question 67: }\text{There are 60 students in a class. The following is the frequency}
\displaystyle \text{distribution of the marks obtained by the students in a test:}
\displaystyle \begin{array}{|c|c|c|c|c|c|c|} \hline \text{Marks}&0&1&2&3&4&5\\ \hline \text{Frequency}&x-2&x&x^2&(x+1)^2&2x&x+1 \\ \hline \end{array}
\displaystyle \text{where }x\text{ is a positive integer. Determine the mean and standard deviation of the marks.}
\displaystyle \text{Answer:}
\displaystyle (x-2)+x+x^2+(x+1)^2+2x+(x+1)=60.
\displaystyle 2x^2+7x=60.
\displaystyle 2x^2+7x-60=0.
\displaystyle (2x-8)(x+15/2)=0.
\displaystyle \therefore x=4,\text{ since }x\text{ is a positive integer.}
\displaystyle \begin{array}{|c|c|c|c|c|c|c|} \hline  x_i&0&1&2&3&4&5\\ \hline f_i&2&4&16&25&8&5 \\ \hline \end{array}
\displaystyle \sum f_i=60.
\displaystyle \sum f_ix_i=0+4+32+75+32+25=168.
\displaystyle \therefore \bar{x}=\frac{168}{60}=2.8.
\displaystyle \sum f_ix_i^2=0+4+64+225+128+125=546.
\displaystyle \sigma^2=\frac{\sum f_ix_i^2}{\sum f_i}-\bar{x}^{\,2}.
\displaystyle =\frac{546}{60}-(2.8)^2=9.1-7.84=1.26.
\displaystyle \therefore \sigma=\sqrt{1.26}\approx1.12.
\displaystyle \therefore \text{Mean}=2.8\text{ and standard deviation}\approx1.12.
\displaystyle \\

\displaystyle \textbf{Question 68: }\text{Calculate the mean deviation about the mean for the following}
\displaystyle \text{frequency distribution:}
\displaystyle \begin{array}{|c|c|c|c|c|c|} \hline \text{Class Interval}&0-4&4-8&8-12&12-16&16-20\\ \hline \text{Frequency}&4&6&8&5&2 \\ \hline  \end{array}
\displaystyle \text{Answer:}
\displaystyle \text{The class marks are }2,6,10,14,18.
\displaystyle \sum f_i=4+6+8+5+2=25.
\displaystyle \sum f_ix_i=8+36+80+70+36=230.
\displaystyle \bar{x}=\frac{230}{25}=9.2.
\displaystyle \begin{array}{|c|c|c|c|} \hline x_i&f_i&|x_i-\bar{x}|&f_i|x_i-\bar{x}|\\ \hline 2&4&7.2&28.8\\6&6&3.2&19.2\\10&8&0.8&6.4\\14&5&4.8&24\\18&2&8.8&17.6 \\ \hline \end{array}
\displaystyle \sum f_i|x_i-\bar{x}|=96.
\displaystyle \text{M.D. about mean}=\frac{96}{25}=3.84.
\displaystyle \therefore \text{The mean deviation about the mean is }3.84.
\displaystyle \\

\displaystyle \textbf{Question 69: }\text{The scores of a batsman in 10 innings are }48,80,58,44,52,65,73,
\displaystyle 56,64,54.\text{ Find the mean deviation from the median.}
\displaystyle \text{Answer:}
\displaystyle \text{Arranging the scores: }44,48,52,54,56,58,64,65,73,80.
\displaystyle \text{Median}=\frac{56+58}{2}=57.
\displaystyle \sum|x_i-57|=13+9+5+3+1+1+7+8+16+23=86.
\displaystyle \text{M.D. about median}=\frac{86}{10}=8.6.
\displaystyle \therefore \text{The mean deviation from the median is }8.6.
\displaystyle \\

\displaystyle \textbf{Question 70: }\text{The mean deviation of the data }3,10,10,4,7,10,5\text{ from the mean is }\underline{\hspace{1.5cm}}.
\displaystyle \text{Answer:}
\displaystyle \bar{x}=\frac{3+10+10+4+7+10+5}{7}=\frac{49}{7}=7.
\displaystyle \sum|x_i-\bar{x}|=4+3+3+3+0+3+2=18.
\displaystyle \text{M.D. about mean}=\frac{18}{7}\approx2.57.
\displaystyle \therefore \text{The mean deviation from the mean is }\frac{18}{7}\approx2.57.
\displaystyle \\

\displaystyle \text{LONG ANSWER QUESTIONS}

\displaystyle \textbf{Question 71: }\text{Calculate the mean deviation about median for the following data:}
\displaystyle \begin{array}{|c|c|c|c|c|c|c|}\hline \text{Class}&0-10&10-20&20-30&30-40&40-50&50-60\\ \hline \text{Frequency}&6&7&15&16&4&2\\ \hline \end{array}
\displaystyle \text{Answer:}
\displaystyle N=6+7+15+16+4+2=50.
\displaystyle \frac{N}{2}=25.
\displaystyle \begin{array}{|c|c|c|}\hline \text{Class}&f&\text{Cumulative frequency}\\ \hline 0-10&6&6\\ \hline 10-20&7&13\\ \hline 20-30&15&28\\ \hline 30-40&16&44\\ \hline 40-50&4&48\\ \hline 50-60&2&50\\ \hline \end{array}
\displaystyle \therefore \text{Median class}=20-30.
\displaystyle \text{Median}=l+\left(\frac{\frac{N}{2}-c}{f}\right)h.
\displaystyle =20+\left(\frac{25-13}{15}\right)10.
\displaystyle =20+8=28.
\displaystyle \text{The class marks are }5,15,25,35,45,55.
\displaystyle \begin{array}{|c|c|c|c|}\hline x_i&f_i&|x_i-28|&f_i|x_i-28|\\ \hline 5&6&23&138\\ \hline 15&7&13&91\\ \hline 25&15&3&45\\ \hline 35&16&7&112\\ \hline 45&4&17&68\\ \hline 55&2&27&54\\ \hline \end{array}
\displaystyle \sum f_i|x_i-28|=138+91+45+112+68+54=508.
\displaystyle \text{M.D. about median}=\frac{\sum f_i|x_i-M|}{\sum f_i}.
\displaystyle =\frac{508}{50}=10.16.
\displaystyle \therefore \text{The mean deviation about the median is }10.16.
\displaystyle \\

\displaystyle \textbf{Question 72: }\text{The mean of 5 observations is }4.4\text{ and their variance is }8.24.\text{ If three}
\displaystyle \text{of the observations are }1,2\text{ and }6,\text{ find the other two observations.}
\displaystyle \text{Answer:}
\displaystyle \text{Let the other two observations be }x\text{ and }y.
\displaystyle \bar{x}=4.4,\qquad n=5.
\displaystyle \therefore \sum x_i=5(4.4)=22.
\displaystyle 1+2+6+x+y=22.
\displaystyle \therefore x+y=13.\qquad\text{...(1)}
\displaystyle \sigma^2=\frac{\sum x_i^2}{n}-\bar{x}^{\,2}.
\displaystyle 8.24=\frac{\sum x_i^2}{5}-(4.4)^2.
\displaystyle 8.24=\frac{\sum x_i^2}{5}-19.36.
\displaystyle \therefore \frac{\sum x_i^2}{5}=27.6.
\displaystyle \therefore \sum x_i^2=138.
\displaystyle 1^2+2^2+6^2+x^2+y^2=138.
\displaystyle \therefore x^2+y^2=97.
\displaystyle (x+y)^2=x^2+y^2+2xy.
\displaystyle 13^2=97+2xy.
\displaystyle 169=97+2xy.
\displaystyle \therefore xy=36.
\displaystyle \therefore x\text{ and }y\text{ are the roots of }t^2-13t+36=0.
\displaystyle (t-4)(t-9)=0.
\displaystyle \therefore t=4\text{ or }9.
\displaystyle \therefore \text{The other two observations are }4\text{ and }9.
\displaystyle \\

\displaystyle \textbf{Question 73: }\text{The mean and standard deviation of 100 observations were calculated as }40
\displaystyle \text{and }5.1,\text{ respectively, by a student who took by mistake }50\text{ instead of }40\text{ for one}
\displaystyle \text{observation. What are the correct mean and standard deviation?}
\displaystyle \text{Answer:}
\displaystyle \text{Incorrect }\sum x=100(40)=4000.
\displaystyle \text{Correct }\sum x=4000-50+40=3990.
\displaystyle \therefore \text{Correct mean}=\frac{3990}{100}=39.9.
\displaystyle \text{Also, incorrect variance}=(5.1)^2=26.01.
\displaystyle 26.01=\frac{\sum x^2}{100}-40^2.
\displaystyle \therefore \sum x^2=100(26.01+1600)=162601.
\displaystyle \text{Correct }\sum x^2=162601-50^2+40^2.
\displaystyle =162601-2500+1600=161701.
\displaystyle \text{Correct variance}=\frac{161701}{100}-(39.9)^2.
\displaystyle =1617.01-1592.01=25.
\displaystyle \therefore \text{Correct S.D.}=\sqrt{25}=5.
\displaystyle \therefore \text{Correct mean}=39.9\text{ and correct standard deviation}=5.
\displaystyle \\

\displaystyle \textbf{Question 74: }\text{The mean and standard deviation of six observations are }8\text{ and }4,
\displaystyle \text{respectively. If each observation is multiplied by }3,\text{ find the new mean and standard}
\displaystyle \text{deviation of the resulting observations.}
\displaystyle \text{Answer:}
\displaystyle \text{If each observation is multiplied by }k,\text{ the mean and S.D. are multiplied by }|k|.
\displaystyle \therefore \text{New mean}=3(8)=24.
\displaystyle \text{New S.D.}=3(4)=12.
\displaystyle \therefore \text{The new mean is }24\text{ and the new standard deviation is }12.
\displaystyle \\

\displaystyle \textbf{Question 75: }\text{The mean and standard deviation of a group of 100 observations}
\displaystyle \text{were found to be }20\text{ and }3,\text{ respectively. Later on, it was found that three}
\displaystyle \text{observations were incorrect, which were recorded as }21,21\text{ and }18.\text{ Find the mean and}
\displaystyle \text{standard deviation if the incorrect observations are omitted.}
\displaystyle \text{Answer:}
\displaystyle n=100,\quad\bar{x}=20,\quad\sigma=3.
\displaystyle \sum x=n\bar{x}=100(20)=2000.
\displaystyle \sigma^2=\frac{\sum x^2}{n}-\bar{x}^{\,2}.
\displaystyle 9=\frac{\sum x^2}{100}-400.
\displaystyle \therefore \sum x^2=40900.
\displaystyle \text{After omitting }21,21\text{ and }18,
\displaystyle n'=100-3=97.
\displaystyle \sum x'=2000-(21+21+18)=1940.
\displaystyle \therefore \bar{x}'=\frac{1940}{97}=20.
\displaystyle \sum x'^2=40900-(21^2+21^2+18^2).
\displaystyle =40900-(441+441+324)=39694.
\displaystyle \sigma'^2=\frac{39694}{97}-20^2.
\displaystyle =\frac{39694-38800}{97}=\frac{894}{97}.
\displaystyle \therefore \sigma'=\sqrt{\frac{894}{97}}\approx3.04.
\displaystyle \therefore \text{The new mean is }20\text{ and the new standard deviation is approximately }3.04.
\displaystyle \\

\displaystyle \textbf{Question 76: }\text{Calculate mean, variance and standard deviation for the following distribution.}
\displaystyle \begin{array}{|c|c|c|c|c|c|c|c|}\hline \text{Class}&30-40&40-50&50-60&60-70&70-80&80-90&90-100\\ \hline \text{Frequency}&3&7&12&15&8&3&2\\ \hline \end{array}
\displaystyle \text{Answer:}
\displaystyle \text{The class marks are }35,45,55,65,75,85,95.
\displaystyle \begin{array}{|c|c|c|c|c|}\hline x_i&f_i&f_ix_i&x_i^2&f_ix_i^2\\ \hline 35&3&105&1225&3675\\ \hline 45&7&315&2025&14175\\ \hline 55&12&660&3025&36300\\ \hline 65&15&975&4225&63375\\ \hline 75&8&600&5625&45000\\ \hline 85&3&255&7225&21675\\ \hline 95&2&190&9025&18050\\ \hline \end{array}
\displaystyle \sum f_i=50,\qquad \sum f_ix_i=3100,\qquad \sum f_ix_i^2=202250.
\displaystyle \bar{x}=\frac{\sum f_ix_i}{\sum f_i}=\frac{3100}{50}=62.
\displaystyle \sigma^2=\frac{\sum f_ix_i^2}{\sum f_i}-\bar{x}^{\,2}.
\displaystyle =\frac{202250}{50}-62^2.
\displaystyle =4045-3844=201.
\displaystyle \therefore \sigma=\sqrt{201}\approx14.18.
\displaystyle \therefore \text{Mean}=62,\text{ variance}=201\text{ and standard deviation}\approx14.18.
\displaystyle \\

\displaystyle \textbf{Question 77: }\text{Calculate the mean deviation about median for the following data:}
\displaystyle \begin{array}{|c|c|c|c|c|c|c|c|c|}\hline \text{Age (in years)}&16-20&21-25&26-30&31-35&36-40&41-45&46-50&51-55\\ \hline \text{Number of persons}&5&6&12&14&26&12&16&9\\ \hline \end{array}
\displaystyle \text{Answer:}
\displaystyle N=5+6+12+14+26+12+16+9=100.
\displaystyle \frac{N}{2}=50.
\displaystyle \begin{array}{|c|c|c|}\hline \text{Class}&f&\text{Cumulative frequency}\\ \hline 16-20&5&5\\ \hline 21-25&6&11\\ \hline 26-30&12&23\\ \hline 31-35&14&37\\ \hline 36-40&26&63\\ \hline 41-45&12&75\\ \hline 46-50&16&91\\ \hline 51-55&9&100\\ \hline \end{array}
\displaystyle \therefore \text{Median class}=36-40.
\displaystyle \text{Using class boundaries, }l=35.5,\quad c=37,\quad f=26,\quad h=5.
\displaystyle \text{Median}=l+\left(\frac{\frac{N}{2}-c}{f}\right)h.
\displaystyle =35.5+\left(\frac{50-37}{26}\right)5.
\displaystyle =35.5+2.5=38.
\displaystyle \text{The class marks are }18,23,28,33,38,43,48,53.
\displaystyle \begin{array}{|c|c|c|c|}\hline x_i&f_i&|x_i-38|&f_i|x_i-38|\\ \hline 18&5&20&100\\ \hline 23&6&15&90\\ \hline 28&12&10&120\\ \hline 33&14&5&70\\ \hline 38&26&0&0\\ \hline 43&12&5&60\\ \hline 48&16&10&160\\ \hline 53&9&15&135\\ \hline \end{array}
\displaystyle \sum f_i|x_i-38|=735.
\displaystyle \text{M.D. about median}=\frac{\sum f_i|x_i-M|}{\sum f_i}.
\displaystyle =\frac{735}{100}=7.35.
\displaystyle \therefore \text{The mean deviation about the median is }7.35.
\displaystyle \\

\displaystyle \textbf{Question 78: }\text{The diameters of circles (in mm) drawn in a design are given below:}
\displaystyle \begin{array}{|c|c|c|c|c|c|}\hline \text{Diameters}&33-36&37-40&41-44&45-48&49-52\\ \hline \text{No. of circles}&15&17&21&22&25\\ \hline \end{array}
\displaystyle \text{Calculate the standard deviation and mean diameter of the circles.}
\displaystyle \text{Answer:}
\displaystyle \text{The class marks are }34.5,\ 38.5,\ 42.5,\ 46.5,\ 50.5.
\displaystyle \begin{array}{|c|c|c|c|c|}\hline x_i&f_i&f_ix_i&x_i^2&f_ix_i^2\\ \hline 34.5&15&517.5&1190.25&17853.75\\ \hline 38.5&17&654.5&1482.25&25198.25\\ \hline 42.5&21&892.5&1806.25&37931.25\\ \hline 46.5&22&1023&2162.25&47569.50\\ \hline 50.5&25&1262.5&2550.25&63756.25\\ \hline \end{array}
\displaystyle \sum f_i=100,\qquad\sum f_ix_i=4350,\qquad\sum f_ix_i^2=192309.
\displaystyle \bar{x}=\frac{\sum f_ix_i}{\sum f_i}=\frac{4350}{100}=43.5.
\displaystyle \sigma^2=\frac{\sum f_ix_i^2}{\sum f_i}-\bar{x}^{\,2}.
\displaystyle =\frac{192309}{100}-(43.5)^2.
\displaystyle =1923.09-1892.25=30.84.
\displaystyle \therefore \sigma=\sqrt{30.84}\approx5.55.
\displaystyle \therefore \text{Mean diameter}=43.5\text{ mm and standard deviation}\approx5.55\text{ mm.}
\displaystyle \\

\displaystyle \textbf{Question 79: }\text{The mean and standard deviation of 20 observations are found to be}
\displaystyle 10\text{ and }2,\text{ respectively. On rechecking, it was found that an observation }8\text{ was incorrect.}
\displaystyle \text{Calculate the correct mean and standard deviation if it is replaced by }12.
\displaystyle \text{Answer:}
\displaystyle n=20,\qquad \bar{x}=10,\qquad \sigma=2.
\displaystyle \sum x=n\bar{x}=20(10)=200.
\displaystyle \sigma^2=\frac{\sum x^2}{n}-\bar{x}^{\,2}.
\displaystyle 4=\frac{\sum x^2}{20}-100.
\displaystyle \therefore \sum x^2=20(104)=2080.
\displaystyle \text{Correct }\sum x=200-8+12=204.
\displaystyle \therefore \text{Correct mean}=\frac{204}{20}=10.2.
\displaystyle \text{Correct }\sum x^2=2080-8^2+12^2=2160.
\displaystyle \text{Correct variance}=\frac{2160}{20}-(10.2)^2.
\displaystyle =108-104.04=3.96.
\displaystyle \therefore \text{Correct S.D.}=\sqrt{3.96}\approx1.99.
\displaystyle \therefore \text{Correct mean}=10.2\text{ and correct standard deviation}\approx1.99.
\displaystyle \\

\displaystyle \textbf{Question 80: }\text{Determine mean and standard deviation of first }n\text{ terms of an A.P.}
\displaystyle \text{whose first term is }a\text{ and common difference is }d.
\displaystyle \text{Answer:}
\displaystyle \text{The terms are }a,\ a+d,\ a+2d,\ldots,a+(n-1)d.
\displaystyle \bar{x}=\frac{a+\{a+(n-1)d\}}{2}.
\displaystyle =a+\frac{(n-1)d}{2}.
\displaystyle \text{Since adding }a\text{ does not affect variance,}
\displaystyle \sigma^2=d^2\times\text{variance of }0,1,2,\ldots,n-1.
\displaystyle \text{Variance of }0,1,2,\ldots,n-1=\frac{n^2-1}{12}.
\displaystyle \therefore \sigma^2=\frac{d^2(n^2-1)}{12}.
\displaystyle \therefore \sigma=|d|\sqrt{\frac{n^2-1}{12}}.
\displaystyle \text{If }d>0,\text{ then }\sigma=d\sqrt{\frac{n^2-1}{12}}.
\displaystyle \therefore \text{Mean}=a+\frac{(n-1)d}{2},\quad \text{S.D.}=|d|\sqrt{\frac{n^2-1}{12}}.
\displaystyle \\

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