Class 11: Statistics – Exercise 32.6 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Question 1: Calculate the Mean and Standard Deviation for the following data:

$\displaystyle \begin{array} {|l | c| c| c| c | c |} \hline \text{Expenditure in Rs.: } & 0-10 & 10-20 & 20-30 & 30-40 & 40-50 \\ \hline \text{Frequency: } &14 &13 & 27& 21& 15 \\ \hline \end{array}$

Answer:

Calculation of Mean, Variance and Standard Deviation

$\displaystyle \begin{array} { | l | c| c| c| c | c | c |} \hline \text{Class} & \text{Frequency} & \text{Mid-values} & u_i = \frac{x_i - 25}{10} & f_iu_i & {u_i}^2 & f_i{u_i}^2 \\ \text{interval} & (f_i) & (x_i) & & & & \\ \hline 0-10 & 14 & 5 & -2 & -28 & 4 & 56 \\ \hline 10-20 & 13 & 15 & -2 & -13 & 1 & 13 \\ \hline 20-30 & 27 & 25 & 0 & 0 & 0 & 0 \\ \hline 30-40 & 21 & 35 &1 & 21 & 1 & 21 \\ \hline 40-50 & 15 & 45 & 2 & 30 & 4 & 60 \\ \hline & N = \sum f_i & & & \sum f_iu_i & & \sum f_i{u_i}^2 \\ & = 90 & & & = 10 & & = 150 \\ \hline \end{array}$

$\displaystyle \text{Here, } N = 90, \hspace{0.5cm} \sum f_iu_i = 10, \hspace{0.5cm} \sum f_i{u_i}^2= 150 \hspace{0.5cm} A = 25, \hspace{0.5cm} h = 10$

$\displaystyle \text{Mean } = \overline{X} = A + h \Bigg( \frac{1}{N} \sum f_i u_i \Bigg) = 25 + 10 \Big( \frac{10}{90} \Big) = 26.11$

$\displaystyle \text{and, } Var(X) = h^2 \Bigg[ \Bigg( \frac{1}{N} \sum f_i{u_i}^2 \Bigg) - \Bigg( \frac{1}{N} \sum f_i{u_i} \Bigg)^2 \Bigg] \\ \\ \\ { \hspace{2.5cm} = 100 \Bigg[ \Bigg( \frac{150}{90} \Bigg) - \Bigg( \frac{10}{90} \Bigg)^2 \Bigg] = 165.4 }$

$\displaystyle \therefore \text{ Standard Deviation } = \sqrt{ Var(X)} = \sqrt{165.4} = 12.86$

$\\$

Question 2: Calculate the Standard Deviation for the following data:

$\displaystyle \begin{array} {|l | c| c| c| c | c | c | c|} \hline \text{Class:} & 0-30 & 30-60 & 60-90 & 90-120 & 120-150 & 150-180 & 180-210 \\ \hline \text{Frequency:} & 9 & 17 & 43 & 82 & 81 & 44 & 24 \\ \hline \end{array}$

Answer:

Calculation of Mean, Variance and Standard Deviation

$\displaystyle \begin{array} { | l | c| c| c| c | c | c |} \hline \text{Class} & \text{Frequency} & \text{Mid-values} & u_i = \frac{x_i - 105}{30} & f_iu_i & {u_i}^2 & f_i{u_i}^2 \\ \text{interval} & (f_i) & (x_i) & & & & \\ \hline 0- 30 & 9 & 15 & -3 & -27 & 9 & 81 \\ \hline 30-60 & 17 & 45 & -2 & -34 & 4 & 68 \\ \hline 60-90 & 43 & 75 & -1 & -43 &1 & 43 \\ \hline 90-120 & 82 & 105 & 0 & 0 & 0 & 0 \\ \hline 120 - 150 & 81 & 135 & 1 & 1 &1 & 81 \\ \hline 150-180 & 44 & 165 & 2 & 4 &4 & 176 \\ \hline 180-210& 24 & 195 & 3 & 9 & 9 & 216 \\ \hline & N = \sum f_i & & & \sum f_iu_i & & \sum f_i{u_i}^2 \\ & = 300 & & & = 137 & & = 665 \\ \hline \end{array}$

$\displaystyle \text{Here, } N = 300, \hspace{0.5cm} \sum f_iu_i = 137, \hspace{0.5cm} \sum f_i{u_i}^2= 665 \hspace{0.5cm} A = 105, \hspace{0.5cm} h = 30$

$\displaystyle \text{Mean } = \overline{X} = A + h \Bigg( \frac{1}{N} \sum f_i u_i \Bigg) = 105 + 30 \Big( \frac{137}{300} \Big) = 118.7$

$\displaystyle \text{and, } Var(X) = h^2 \Bigg[ \Bigg( \frac{1}{N} \sum f_i{u_i}^2 \Bigg) - \Bigg( \frac{1}{N} \sum f_i{u_i} \Bigg)^2 \Bigg] \\ \\ \\ { \hspace{2.5cm} = 900 \Bigg[ \Bigg( \frac{665}{300} \Bigg) - \Bigg( \frac{137}{300} \Bigg)^2 \Bigg] = 1807.31 }$

$\displaystyle \therefore \text{ Standard Deviation } = \sqrt{ Var(X)} = \sqrt{1807.31} = 42.51$

$\\$

Question 3: Calculate the A.M and Standard Deviation for the following distribution:

$\displaystyle \begin{array} {|l | c| c| c| c | c | c | c | c | } \hline \text{Class:} & 0-10 & 10-20 & 20-30 & 30-40 & 40-50 & 50-60 & 60-70 & 70-80 \\ \hline \text{Frequency:} & 18 & 16 & 15 & 12 & 10 & 5 & 2 & 1 \\ \hline \end{array}$

Answer:

Calculation of Mean, Variance and Standard Deviation

$\displaystyle \begin{array} { | l | c| c| c| c | c | c |} \hline \text{Class} & \text{Frequency} & \text{Mid-values} & u_i = \frac{x_i - 35}{10} & f_iu_i & {u_i}^2 & f_i{u_i}^2 \\ \text{interval} & (f_i) & (x_i) & & & & \\ \hline 0-10 & 18 & 5 & -3 & -54 & 9 & 162 \\ \hline 10-20 & 16 & 15 & -2 & -32 & 4 & 64 \\ \hline 20-30 & 15 & 25 & -1 & -15 & 1& 15 \\ \hline 30-40 & 12 &35 &0 & 0 & 0 & 0 \\ \hline 40-50 & 10 & 45 & 1 & 10 & 1 & 10 \\ \hline 50-60 & 5 &55 & 2 &10 & 4 & 20 \\ \hline 60-70 & 2 & 65 & 3 & 6 & 9 & 18 \\ \hline 70-80 & 1 & 75 & 4 &4 & 16 & 16 \\ \hline & N = \sum f_i & & & \sum f_iu_i & & \sum f_i{u_i}^2 \\ & = 79 & & & = -71 & & = 305 \\ \hline \end{array}$

$\displaystyle \text{Here, } N = 79, \hspace{0.5cm} \sum f_iu_i = -71, \hspace{0.5cm} \sum f_i{u_i}^2= 305 \hspace{0.5cm} A = 35, \hspace{0.5cm} h = 10$

$\displaystyle \text{Mean } = \overline{X} = A + h \Bigg( \frac{1}{N} \sum f_i u_i \Bigg) = 35 + 10 \Big( \frac{-71}{79} \Big) = 26.01$

$\displaystyle \text{and, } Var(X) = h^2 \Bigg[ \Bigg( \frac{1}{N} \sum f_i{u_i}^2 \Bigg) - \Bigg( \frac{1}{N} \sum f_i{u_i} \Bigg)^2 \Bigg] \\ \\ \\ { \hspace{2.5cm} = 100 \Bigg[ \Bigg( \frac{305}{79} \Bigg) - \Bigg( \frac{-71}{79} \Bigg)^2 \Bigg] = 305.30 }$

$\displaystyle \therefore \text{ Standard Deviation } = \sqrt{ Var(X)} = \sqrt{305.30} = 17.47$

$\\$

Question 4: A student obtained mean and standard deviation of 100 observations as 40 and 5.1 respectively. It was later found that one observation was wrongly copied as 50, the correct figure being 40. Find the correct mean and standard deviation.

Answer:

$\displaystyle \text{Given: Mean } = \overline{X} =40, \hspace{1.0cm} n = 100, \hspace{1.0cm} \sigma = S.D. = 5.1$

$\displaystyle \text{Mean } \overline{X} = \frac{\sum x_i}{n}$

$\displaystyle \Rightarrow \frac{1}{100} \sum x_i = 40$

$\displaystyle \Rightarrow \sum x_i = 4000$

Variance $\displaystyle \sigma^2 = 26.01$

$\displaystyle \frac{1}{n} \sum {x_i}^2 - (\overline{X})^2 = 26.01$

$\displaystyle \frac{1}{100} \sum {x_i}^2 - 40^2 = 26.01$

$\displaystyle \frac{1}{100} \sum {x_i}^2 = 1626.01$

$\displaystyle \sum {x_i}^2 = 162601 \text{ ( This is incorrect because of incorrect values) }$

If 50 is replaced by 40

$\displaystyle \text{Corrected} \sum x_i = 4000-50+40 = 3990$

$\displaystyle \text{Corrected } \sum {x_i}^2 = 162601 - 50^2 + 40^2= 161701$

$\displaystyle \text{Therefore the correct mean } = \frac{\sum x_i}{n} = \frac{3990}{100} = 39.90$

We know,

$\displaystyle \text{Corrected Variance } = \frac{1}{n} \Big( \text{Corrected } \sum {x_i}^2 \Big) - ( \text{corrected mean })^2$

$\displaystyle = \frac{161701}{100} - \Big( \frac{3990}{100} \Big)^2 = 1617.01 - 1592.01 = 25$

$\displaystyle \text{Corrected } S.D. = \sqrt{\text{Variance}} = \sqrt{25} = 5$

$\\$

Question 5: Calculate the mean, median and standard deviation for the following distribution:

$\displaystyle \begin{array} {|l | c| c| c| c | c | c | c | c | } \hline \text{Class Interval:} & 31-35 & 36-40 & 41-45 & 46-50 & 51-55 & 56-60 & 61-65 & 66-70 \\ \hline \text{Frequency:} & 2 & 3 & 8 & 12 & 16 & 5 & 2 & 3 \\ \hline \end{array}$

Answer:

Calculation of Mean, Variance and Standard Deviation

$\displaystyle \begin{array} { | l | c| c| c| c | c | c |} \hline \text{Class} & \text{Frequency} & \text{Mid-values} & u_i = \frac{x_i - 53}{4} & f_iu_i & {u_i}^2 & f_i{u_i}^2 \\ \text{interval} & (f_i) & (x_i) & & & & \\ \hline 31-35 & 2 & 33 & -5 & 25 & -10 & 50 \\ \hline 36-40 & 3 & 38 & -3.75 & 14.06 & -11.25 & 42.18 \\ \hline 41-45 & 8 & 43 & -2.5 & 6.25 & -20 & 50 \\ \hline 46-50 & 12 & 48 & -1.25 & 1.56 & -15 & 18.72 \\ \hline 51-55 & 16 & 53 & 0 & 0 & 0 & 0 \\ \hline 56-60 & 5 & 58 & 1.25 & 1.56 & 6.25 & 7.8 \\ \hline 61-65 & 2 & 63 & 2.5 & 6.25 &5 & 12.5 \\ \hline 66-70 & 3 & 68 & 3.75 & 14.06 & 11.25 & 42.18 \\ \hline & N = \sum f_i & & & \sum f_iu_i & & \sum f_i{u_i}^2 \\ & = 51 & & & = -33.75 & & = 223.38 \\ \hline \end{array}$

$\displaystyle \text{Here, } N = 51, \hspace{0.5cm} \sum f_iu_i = -33.75, \hspace{0.5cm} \sum f_i{u_i}^2= 223.38 \hspace{0.5cm} A = 53, \hspace{0.5cm} h = 4$

$\displaystyle \text{Mean } = \overline{X} = A + h \Bigg( \frac{1}{N} \sum f_i u_i \Bigg) = 53 + 4 \Big( \frac{-33.75}{51} \Big) = 50.36$

$\displaystyle \text{and, } Var(X) = h^2 \Bigg[ \Bigg( \frac{1}{N} \sum f_i{u_i}^2 \Bigg) - \Bigg( \frac{1}{N} \sum f_i{u_i} \Bigg)^2 \Bigg] \\ \\ \\ { \hspace{2.5cm} = 16 \Bigg[ \Bigg( \frac{223.38}{51} \Bigg) - \Bigg( \frac{-33.75}{51} \Bigg)^2 \Bigg] = 63.07 }$

$\displaystyle \therefore \text{ Standard Deviation } = \sqrt{ Var(X)} = \sqrt{63.07} = 7.94$

$\displaystyle \text{Clearly, } N = 51 \Rightarrow \frac{N}{2} = \frac{51}{2} = 25.5$

$\displaystyle \text{The cumulative frequency just greater than } \frac{N}{2} = 25.5 \text{ is } 41$

$\displaystyle \text{The corresponding value of class } \text{ is } 51-55$

$\displaystyle \text{Therefore the median class } \text{ is } 51-55$

$\displaystyle \therefore l = 51, \hspace{0.5cm} f = 16, \hspace{0.5cm} F = 25, \hspace{0.5cm} N = 51, \hspace{0.5cm} h = 4$

$\displaystyle \therefore \text{ Median } = l + \frac{\Big( \frac{N}{2} - F \Big) }{f} \times h$

$\displaystyle = 51 + \frac{\Big( \frac{51}{2} - 25 \Big) }{16} \times 4$

$\displaystyle = 51 + \frac{\Big( 25.5 - 25 \Big) }{16} \times 4$

$\displaystyle = 51.125$

$\\$

Question 6: Calculate the mean and variance of frequency distribution given below:

$\displaystyle \begin{array} { | l | c| c| c| c | } \hline x_i: & 1 \leq x < 3 & 3 \leq x < 5 & 5 \leq x < 7 & 7 \leq x < 10 \\ \hline f_i: & 6 & 4 & 5 & 1 \\ \hline \end{array}$

Answer:

Calculation of Mean, Variance and Standard Deviation

$\displaystyle \begin{array} { | l | c| c| c| c | c | c |} \hline \text{Class} & \text{Frequency} & \text{Mid-values} & u_i = \frac{x_i - 6}{2} & f_iu_i & {u_i}^2 & f_i{u_i}^2 \\ \text{interval} & (f_i) & (x_i) & & & & \\ \hline 1-3 & 6 & 2 & -2 & -12 & 4 & 24 \\ \hline 3-5 & 4 & 4 & -1 & -4 & 1 & 4 \\ \hline 5-7 & 5 & 6& 0 & 0 & 0 & 0 \\ \hline 7-10 & 1 & 8.5 & 1.25 & 1.25 & 1.5625 &1.5625 \\ \hline & N = \sum f_i & & & \sum f_iu_i & & \sum f_i{u_i}^2 \\ & = 16 & & & = -14.75 & & = 29.5625 \\ \hline \end{array}$

$\displaystyle \text{Here, } N = 16, \hspace{0.5cm} \sum f_iu_i = -14.75, \hspace{0.5cm} \sum f_i{u_i}^2= 29.5625 \hspace{0.5cm} A = 6, \hspace{0.5cm} h = 2$

$\displaystyle \text{Mean } = \overline{X} = A + h \Bigg( \frac{1}{N} \sum f_i u_i \Bigg) = 6 + 2 \Big( \frac{-14.75}{16} \Big) = 4.15625$

$\displaystyle \text{and, } Var(X) = h^2 \Bigg[ \Bigg( \frac{1}{N} \sum f_i{u_i}^2 \Bigg) - \Bigg( \frac{1}{N} \sum f_i{u_i} \Bigg)^2 \Bigg] \\ \\ \\ { \hspace{2.5cm} = 4 \Bigg[ \Bigg( \frac{29.5625}{16} \Bigg) - \Bigg( \frac{-14.75}{16} \Bigg)^2 \Bigg] =3.9912 }$

$\displaystyle \therefore \text{ Standard Deviation } = \sqrt{ Var(X)} = \sqrt{3.9912} = 1.9978$

$\\$

Question 7: The weight of coffee in 70 jars is shown in the following table:

$\displaystyle \begin{array} { | l | c| c| c| c | c | c |} \hline \text{Weight in grams:} & 200-201 & 201-202 & 202-203 & 203-204 & 204-205 & 205-206 \\ \hline \text{Frequency:} & 13 & 27 & 18 & 10 & 1 & 1 \\ \hline \end{array}$

Determine the variance and standard deviation of the above distribution.

Answer:

Calculation of Mean, Variance and Standard Deviation

$\displaystyle \begin{array} { | l | c| c| c| c | c | c |} \hline \text{Class} & \text{Frequency} & \text{Mid-values} & u_i = \frac{x_i - 203.5 }{1} & f_iu_i & {u_i}^2 & f_i{u_i}^2 \\ \text{interval} & (f_i) & (x_i) & & & & \\ \hline 200-201 & 13 & 200.5 & -3 & -39 & 9 & 117 \\ \hline 201-202 &27 & 201.5 & -2 & -54 & 9 & 108 \\ \hline 202-203 &18 & 202.5 & -1 & -18 & 1 & 18 \\ \hline 203-204 &10 & 203.5 & 0 & 0 &0 & 0 \\ \hline 204-205 & 1 & 204.5 & 1 & 1 & 1 & 1 \\ \hline 205-206 & 1 & 205.5 & 2 & 2 & 4 & 4 \\ \hline & N = \sum f_i & & & \sum f_iu_i & & \sum f_i{u_i}^2 \\ & = 70 & & & = -108 & & = 248 \\ \hline \end{array}$

$\displaystyle \text{Here, } N = 70, \hspace{0.5cm} \sum f_iu_i = -108, \hspace{0.5cm} \sum f_i{u_i}^2= 248 \hspace{0.5cm} A = 203.5, \hspace{0.5cm} h = 1$

$\displaystyle \text{Mean } = \overline{X} = A + h \Bigg( \frac{1}{N} \sum f_i u_i \Bigg) = 203.5 + 1 \Big( \frac{-108}{70} \Big) =201.95$

$\displaystyle \text{and, } Var(X) = h^2 \Bigg[ \Bigg( \frac{1}{N} \sum f_i{u_i}^2 \Bigg) - \Bigg( \frac{1}{N} \sum f_i{u_i} \Bigg)^2 \Bigg] \\ \\ \\ { \hspace{2.5cm} = 1 \Bigg[ \Bigg( \frac{248}{70} \Bigg) - \Bigg( \frac{-108}{70} \Bigg)^2 \Bigg] =1.162 }$

$\displaystyle \therefore \text{ Standard Deviation } = \sqrt{ Var(X)} = \sqrt{1.162} = 1.078$

$\\$

Question 8: The mean and standard deviation of 100 observations were found to be 40 and 10 respectively. If at the time of calculations two observations were wrongly taken as 30 and 70 in place of 3 and 27 respectively, find the correct standard deviation.

Answer:

$\displaystyle \text{Given: Mean } = \overline{X} =40, \hspace{1.0cm} n = 100, \hspace{1.0cm} \sigma = S.D. = 10$

$\displaystyle \text{Mean } \overline{X} = \frac{\sum x_i}{n}$

$\displaystyle \Rightarrow \frac{1}{100} \sum x_i = 40$

$\displaystyle \Rightarrow \sum x_i = 4000$

Variance $\displaystyle \sigma^2 = 100$

$\displaystyle \frac{1}{n} \sum {x_i}^2 - (\overline{X})^2 = 100$

$\displaystyle \frac{1}{100} \sum {x_i}^2 - 40^2 = 100$

$\displaystyle \frac{1}{100} \sum {x_i}^2 = 1700$

$\displaystyle \sum {x_i}^2 = 170000 \text{ ( This is incorrect because of incorrect values) }$

If 30 and 70 are replaced by 3 and 27

$\displaystyle \text{Corrected} \sum x_i = 4000-30-70+3+27 = 3930$

$\displaystyle \text{Corrected } \sum {x_i}^2 = 17000 - 900- 4900 + 9 + 729= 164938$

$\displaystyle \text{Therefore the correct mean } = \frac{\sum x_i}{n} = \frac{3930}{100} = 39.30$

We know,

$\displaystyle \text{Corrected Variance } = \frac{1}{n} \Big( \text{Corrected } \sum {x_i}^2 \Big) - ( \text{corrected mean })^2$

$\displaystyle = \frac{164938}{100} - \Big( \frac{3930}{100} \Big)^2 = 1649.38 - 1544.49 = 104.89$

$\displaystyle \text{Corrected } S.D. = \sqrt{\text{Variance}} = \sqrt{104.89} = 10.24$

$\\$

Question 9: While calculating the mean and variance of 10 readings, a student wrongly used the reading of 52 for the correct reading 25. He obtained the mean and variance as 45 and 16 respectively. Find the correct mean and variance.

Answer:

$\displaystyle \text{Given: Mean } = \overline{X} =45, \hspace{1.0cm} n = 10, \hspace{1.0cm} \text{Variance} = 16$

$\displaystyle \text{Mean } \overline{X} = \frac{\sum x_i}{n}$

$\displaystyle \Rightarrow \frac{1}{10} \sum x_i = 45$

$\displaystyle \Rightarrow \sum x_i = 450$

Variance $\displaystyle = 16$

$\displaystyle \frac{1}{n} \sum {x_i}^2 - (\overline{X})^2 = 16$

$\displaystyle \frac{1}{10} \sum {x_i}^2 - 45^2 = 16$

$\displaystyle \frac{1}{10} \sum {x_i}^2 = 2041$

$\displaystyle \sum {x_i}^2 = 20410 \text{ ( This is incorrect because of incorrect values) }$

If 52 is replaced by 25

$\displaystyle \text{Corrected} \sum x_i = 450-52+25 = 423$

$\displaystyle \text{Corrected } \sum {x_i}^2 = 20410-2704+625= 18331$

$\displaystyle \text{Therefore the correct mean } = \frac{\sum x_i}{n} = \frac{423}{10} = 42.3$

We know,

$\displaystyle \text{Corrected Variance } = \frac{1}{n} \Big( \text{Corrected } \sum {x_i}^2 \Big) - ( \text{corrected mean })^2$

$\displaystyle = \frac{18331}{10} - \Big( \frac{423}{10} \Big)^2 = 1833.10 - 1789.29 = 43.81$

$\displaystyle \text{Corrected } S.D. = \sqrt{\text{Variance}} = \sqrt{43.81} = 6.6189 = 6.62$

$\\$

Question 10: Calculate the mean, variance and standard deviation of the following frequency distribution:

$\displaystyle \begin{array} { | l | c| c| c| c | c | c |} \hline \text{Class:} & 1-10 & 10-20 & 20-3- & 30-40 & 40-50 & 50-60 \\ \hline \text{Frequency:} & 11 & 29 & 18 & 4 & 5 & 3 \\ \hline \end{array}$

Answer:

Calculation of Mean, Variance and Standard Deviation

$\displaystyle \begin{array} { | l | c| c| c| c | c | c |} \hline \text{Class} & \text{Frequency} & \text{Mid-values} & u_i = \frac{x_i - 35}{10} & f_iu_i & {u_i}^2 & f_i{u_i}^2 \\ \text{interval} & (f_i) & (x_i) & & & & \\ \hline 0-10 & 11 & 5 & -3 & -33 & 9 & 99 \\ \hline 10-20 & 29 & 15 & -2 & -58 & 4 & 116 \\ \hline 20-30 & 18 & 25 & -1 & -18 & 1& 18 \\ \hline 30-40 & 4 &35 &0 & 0 & 0 & 0 \\ \hline 40-50 & 5 & 45 & 1 & 5 & 1 & 5 \\ \hline 50-60 & 3 & 55 & 2 & 6 & 4 & 12 \\ \hline & N = \sum f_i & & & \sum f_iu_i & & \sum f_i{u_i}^2 \\ & = 70 & & & = -98 & & = 250 \\ \hline \end{array}$

$\displaystyle \text{Here, } N = 70, \hspace{0.5cm} \sum f_iu_i = -98, \hspace{0.5cm} \sum f_i{u_i}^2= 250 \hspace{0.5cm} A = 35, \hspace{0.5cm} h = 10$

$\displaystyle \text{Mean } = \overline{X} = A + h \Bigg( \frac{1}{N} \sum f_i u_i \Bigg) = 35 + 10 \Big( \frac{-98}{70} \Big) = 21$

$\displaystyle \text{and, } Var(X) = h^2 \Bigg[ \Bigg( \frac{1}{N} \sum f_i{u_i}^2 \Bigg) - \Bigg( \frac{1}{N} \sum f_i{u_i} \Bigg)^2 \Bigg] \\ \\ \\ { \hspace{2.5cm} = 100 \Bigg[ \Bigg( \frac{250}{70} \Bigg) - \Bigg( \frac{-98}{70} \Bigg)^2 \Bigg] = 161.14 }$

$\displaystyle \therefore \text{ Standard Deviation } = \sqrt{ Var(X)} = \sqrt{161.14} = 12.694$

ICSE / ISC / CBSE Mathematics Portal for K12 Students

Mathematics Made Easy for School Students

Class 11: Statistics – Exercise 32.6

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