Class 11: Statistics – Exercise 32.4 – ICSE / ISC / CBSE Mathematics Portal for K12 Students

Question 1: Find the mean, variance and standard deviation for the following data:

(i) 2, 4, 5, 6, 8, 17 (ii) 6, 7, 10, 12, 13, 4, 8, 12

(iii) 227, 235, 255, 269, 292, 299, 312, 333, 348 (iv) 15, 22, 27, 11, 9, 21, 14, 9

Answer:

(i) Given data : 2, 4, 5, 6, 8, 17

Let $\overline{X}$ be the mean of the given set of observations. Then,

$\displaystyle \overline{X} = \frac{2+ 4+ 5+ 6+ 8+ 17}{6} = \frac{42}{6} = 7$

Computation of Variance

$\displaystyle \begin{array} {|c | c| c|} \hline x_i & x_i - \overline{X} = x_i - 7 & \Big( x_i - \overline{X} \Big)^2 \\ \hline 2 & -5 &25 \\ \hline 4 & -3 &9 \\ \hline 5 & -2 & 4 \\ \hline 6 & -1 &1 \\ \hline 8 & 1 &1 \\ \hline 17 & 10 & 100 \\ \hline & & \sum_{i=1}^{6} \Big( x_i - \overline{X} \Big)^2 = 140 \\ \hline \end{array}$

$\displaystyle \text{Clearly, } n = 6 \text{ and } \sum \Big( x_i - \overline{X} \Big)^2 = 140$

$\displaystyle \text{Therefore Variance } = \frac{1}{n} \sum \Big( x_i - \overline{X} \Big)^2 = \frac{140}{6} = 23.33$

$\displaystyle \text{Hence, Standard deviation } (\sigma) = \sqrt{\text{Variance}} = \sqrt{23.33} = 4.83$

$\\$

(ii) Given data : 6, 7, 10, 12, 13, 4, 8, 12

Let $\overline{X}$ be the mean of the given set of observations. Then,

$\displaystyle \overline{X} = \frac{6+ 7+ 10+ 12+ 13+ 4+ 8+ 12}{8} = \frac{72}{8} = 9$

Computation of Variance

$\displaystyle \begin{array} {|c | c| c|} \hline x_i & x_i - \overline{X} = x_i - 9 & \Big( x_i - \overline{X} \Big)^2 \\ \hline 6 & -3 & 9 \\ \hline 7 & -2 & 4 \\ \hline 10 & 1 & 1 \\ \hline 12 & 3 & 9 \\ \hline 13 & 4 &16 \\ \hline 4 & -5 & 25 \\ \hline 8 & -1 & 1 \\ \hline 12 & 3 & 9 \\ \hline & & \sum_{i=1}^{6} \Big( x_i - \overline{X} \Big)^2 = 74 \\ \hline \end{array}$

$\displaystyle \text{Clearly, } n = 8 \text{ and } \sum \Big( x_i - \overline{X} \Big)^2 = 74$

$\displaystyle \text{Therefore Variance } = \frac{1}{n} \sum \Big( x_i - \overline{X} \Big)^2 = \frac{74}{8} = 9.25$

$\displaystyle \text{Hence, Standard deviation } (\sigma) = \sqrt{\text{Variance}} = \sqrt{9.25} = 3.04$

$\\$

(iii) Given data : 227, 235, 255, 269, 292, 299, 312, 333, 348

Let $\overline{X}$ be the mean of the given set of observations. Then,

$\displaystyle \overline{X} = \frac{227+ 235+ 255+ 269+ 292+ 299+ 312+ 333+348}{10} = \frac{2891}{10} = 289.1$

Computation of Variance

$\displaystyle \begin{array} {|c | c| c|} \hline x_i & x_i - \overline{X} = x_i - 289.1 & \Big( x_i - \overline{X} \Big)^2 \\ \hline 227 & -62.1& 3856.41 \\ \hline 235 & -54.1 & 2926.81 \\ \hline 255 & -34.1 & 1162.81 \\ \hline 269 & -20.1 & 404.01 \\ \hline 292 & 2.9 & 8.41 \\ \hline 299 & 9.9 & 98.01 \\ \hline 312 & 22.9 & 524.41 \\ \hline 321 &31.9 & 1017.61 \\ \hline 333 & 43.9 & 1927.21 \\ \hline 348 & 58.9 & 3469.21 \\ \hline & & \sum_{i=1}^{10} \Big( x_i - \overline{X} \Big)^2 = 15394.9 \\ \hline \end{array}$

$\displaystyle \text{Clearly, } n = 10 \text{ and } \sum \Big( x_i - \overline{X} \Big)^2 = 15394.9$

$\displaystyle \text{Therefore Variance } = \frac{1}{n} \sum \Big( x_i - \overline{X} \Big)^2 = \frac{15394.9}{10} = 1539.49$

$\displaystyle \text{Hence, Standard deviation } (\sigma) = \sqrt{\text{Variance}} = \sqrt{1539.49} = 39.24$

$\\$

(iv) Given data : 15, 22, 27, 11, 9, 21, 14, 9

Let $\overline{X}$ be the mean of the given set of observations. Then,

$\displaystyle \overline{X} = \frac{15+ 22+ 27+ 11+ 9+ 21+ 14+ 9}{8} = \frac{128}{8} = 16$

Computation of Variance

$\displaystyle \begin{array} {|c | c| c|} \hline x_i & x_i - \overline{X} = x_i - 7 & \Big( x_i - \overline{X} \Big)^2 \\ \hline 15 & -1 & 1 \\ \hline 22 & 6 & 36 \\ \hline 27 & 11 & 121 \\ \hline 11 & 5 & 25 \\ \hline 9 & -7 & 49 \\ \hline 21 & 5 & 25 \\ \hline 14 & -2 & 4 \\ \hline 9 & -7 & 49 \\ \hline & & \sum_{i=1}^{8} \Big( x_i - \overline{X} \Big)^2 = 310 \\ \hline \end{array}$

$\displaystyle \text{Clearly, } n = 8 \text{ and } \sum \Big( x_i - \overline{X} \Big)^2 = 310$

$\displaystyle \text{Therefore Variance } = \frac{1}{n} \sum \Big( x_i - \overline{X} \Big)^2 = \frac{310}{8} = 38.75$

$\displaystyle \text{Hence, Standard deviation } (\sigma) = \sqrt{\text{Variance}} = \sqrt{38.75} = 6.22$

$\\$

Question 2: The variance of 20 observations is 5. If each observation is multiplied by 2, find the variance of the resulting observations.

Answer:

$\displaystyle \text{Let } x_1, x_2, x_3, \cdots , x_{20} \text{ be the } 20 \text{ given observations.}$

$\displaystyle \text{We know, Mean } \overline{X} = \frac{ \sum_{i=1}^{20} x_i}{20}$

$\displaystyle \text{Given Variance } ( X ) = 5$

$\displaystyle \text{We know, Variance } = \frac{1}{n} \sum \Big( x_i - \overline{X} \Big)^2$

$\displaystyle \text{Therefore } \frac{1}{20} \Big( x_i - \overline{X} \Big)^2= 5$

$\displaystyle \text{Let } u_1, u_2, u_3, \cdots , u_{20} \text{ be the } 20 \text{ new observations.}$

$\displaystyle u_i = 2 x_i$

$\displaystyle \text{We know, Mean } \overline{U} = \frac{ \sum_{i=1}^{20} u_i}{20} = \frac{ \sum_{i=1}^{20} 2x_i}{20} = 2 \frac{ \sum_{i=1}^{20} x_i}{20} = 2 \overline{X}$

$\displaystyle u_i - \overline{U} = 2x_i - 2 \overline{X}$

$\displaystyle \Rightarrow u_i - \overline{U} = 2(x_i - \overline{X})$

Squaring both sides

$\displaystyle \Rightarrow \Big( u_i - \overline{U} \Big)^2 = \Big( 2(x_i - \overline{X}) \Big)^2$

$\displaystyle \Rightarrow \Big( u_i - \overline{U} \Big)^2 = 4 \Big( x_i - \overline{X} \Big)^2$

$\displaystyle \Rightarrow \frac{1}{20} \sum_{i=1}^{20} \Big( u_i - \overline{U} \Big)^2 = 4 \times \frac{1}{20} \sum_{i=1}^{20} \Big( x_i - \overline{X} \Big)^2$

$\displaystyle \text{Variance } (U) = 4 \times \text{Variance } (X) = 4 \times 5 = 20$

Therefore the variance of the new observations is 20.

$\\$

Question 3: The variance of 15 observations is 4. If each observation is increased by 9, find the variance of the resulting observations.

Answer:

$\displaystyle \text{Let } x_1, x_2, x_3, \cdots , x_{15} \text{ be the } 15 \text{ given observations.}$

$\displaystyle \text{We know, Mean } \overline{X} = \frac{ \sum_{i=1}^{15} x_i}{15}$

$\displaystyle \text{Given Variance } ( X ) = 4$

$\displaystyle \text{We know, Variance } = \frac{1}{n} \sum \Big( x_i - \overline{X} \Big)^2$

$\displaystyle \text{Therefore } \frac{1}{15} \Big( x_i - \overline{X} \Big)^2= 4$

$\displaystyle \text{Let } u_1, u_2, u_3, \cdots , u_{15} \text{ be the } 15 \text{ new observations.}$

$\displaystyle u_i = x_i + 9$

$\displaystyle \text{We know, Mean } \overline{U} = \frac{ \sum_{i=1}^{15} u_i}{15} = \frac{ \sum_{i=1}^{15} (x_i+9) }{15} = \frac{ \sum_{i=1}^{15} x_i}{15} + \frac{9 \times 15}{15}= \overline{X} + 9$

$\displaystyle u_i - \overline{U} = (x_i+9) - ( \overline{X} + 9)$

$\displaystyle \Rightarrow u_i - \overline{U} = x_i - \overline{X}$

Squaring both sides

$\displaystyle \Rightarrow \Big( u_i - \overline{U} \Big)^2 = \Big( x_i - \overline{X} \Big)^2$

$\displaystyle \Rightarrow \frac{1}{15} \sum_{i=1}^{15} \Big( u_i - \overline{U} \Big)^2 = \frac{1}{15} \sum_{i=1}^{15}\Big( x_i - \overline{X} \Big)^2$

$\displaystyle \text{Variance } (U) = \text{Variance } (X) = 4$

Therefore the variance of the new observations is 4.

$\\$

Question 4: The mean of 5 observations is 4.4 and their variance is 8.24. If three of the observations are 1, 2 and 6, find the two other observations.

Answer:

Let $x$ and $y$ be the other two observations.

$\displaystyle \text{Given Mean }= 4.4$

$\displaystyle \Rightarrow \frac{1+2 + 6 + x + y}{5} = 4.4$

$\displaystyle \Rightarrow 9 + x + y = 22$

$\displaystyle \Rightarrow x + y = 13 \text{ ... ... ... ... ... i)}$

$\displaystyle \text{We know, Variance } = \frac{1}{n} \sum \Big( x_i - \overline{X} \Big)^2$

$\displaystyle \text{Therefore } \frac{1}{5} \Big( x_i - \overline{X} \Big)^2= 8.24$

$\displaystyle ( 1 - 4.4)^2 + ( 2- 4.4)^2 + ( 6-4.4)^2 + ( x - 4.4)^2 + ( y - 4.4)^2 = 41.2$

$\displaystyle \Rightarrow 3.4^2 + 2.4^2 + 1.6^2 + ( x - 4.4)^2 + ( y - 4.4)^2 = 41.2$

$\displaystyle \Rightarrow ( x - 4.4)^2 + ( y - 4.4)^2 = 21.32$

$\displaystyle \Rightarrow x^2 + 19.36- 8.8 x + y^2 + 19.36 - 8.8y = 21.32$

$\displaystyle \Rightarrow x^2 + y^2 - 8.8 ( x + y) = - 17.4$

$\displaystyle \Rightarrow x^2 + y^2 = 97 \text{ ... ... ... ... ... ii)}$

$\displaystyle \text{We know, } ( x+y)^2 + ( x-y)^2 =2 ( x^2 + y^2)$

$\displaystyle \Rightarrow 13^2 + (x-y)^2 = 2 \times 97$

$\displaystyle \Rightarrow (x-y)^2 = 194-169=25$

$\displaystyle \Rightarrow x-y = \pm 5 \text{ ... ... ... ... ... iii)}$

$\displaystyle \therefore x-y = 5 \text{ ... ... ... ... ... iv)}$

$\text{or } \displaystyle x-y = -5 \text{ ... ... ... ... ... iv)}$

Solving i) and iv) we get $x = 9$ and $y = 4$

Solving i) and v) we get $x = 4$ and $y = 9$

Hence the other two observations are 4 and 9.

$\\$

Question 5: The mean and standard deviation of 6 observations are 8 and 4 respectively. If each observation is multiplied by 3, find the new mean and new standard deviation of the resulting observations.

Answer:

$\displaystyle \text{Given: Mean } = \overline{X} = 8, \hspace{1.0cm} n = 6, \hspace{1.0cm} \sigma = S.D. = 4$

$\displaystyle \text{If } x_1, x_2, \cdots , x_6 \text{ are the given observations. Then, }$

$\displaystyle \text{Mean } \overline{X} = \frac{\sum_{i=1}^{6} x_i}{n}$

$\displaystyle \Rightarrow \frac{\sum_{i=1}^{6} x_i}{6} = 8$

$\displaystyle \text{Let } u_1, u_2, \cdots , u_6 \text{ are the given observations. }$

$\displaystyle \text{also } u_i = 3 x_i$

$\displaystyle \text{Mean of new observations } \overline{X} = \frac{\sum_{i=1}^{6} ux_i}{n}$

$\displaystyle \Rightarrow \frac{\sum_{i=1}^{6} u_i}{6} = \frac{\sum_{i=1}^{6} 3x_i}{6} = 3\frac{\sum_{i=1}^{6} x_i}{6} = 3 \times 8 = 24$

$\displaystyle \text{Given S.D. } = \sigma_x = 4$

$\displaystyle \sigma^2_x =\text{ Variance }(X)$

$\displaystyle \therefore \text{Variance } X = 16$

$\displaystyle \Rightarrow \frac{1}{6} \sum_{i=1}^{6} \Big( x_i - \overline{X} \Big)^2 = 16$

$\displaystyle \text{Variance } (U) = \sigma^2_u = \sum_{i=1}^{6} \Big( u_i - \overline{U} \Big)^2$

$\displaystyle = \sum_{i=1}^{6} \Big( 3x_i - 3\overline{X} \Big)^2$

$\displaystyle = 3^2 \sum_{i=1}^{6} \Big( x_i - \overline{X} \Big)^2$

$\displaystyle = 9 \times 16 = 144$

$\displaystyle \sigma_u = \sqrt{\text{Variance } (U)} = \sqrt{144} =12$

Therefore the new mean and new standard deviation of the resulting observations are 24 and 12respectively.

$\\$

Question 6: The mean and the variance of 8 observations are 9 and 9.25 respectively. If six of the observations are 6, 7, 10, 12, 12 and 13, find the remaining two observations.

Answer:

Let $x$ and $y$ be the other two observations.

$\displaystyle \text{Given Mean }= 9$

$\displaystyle \Rightarrow \frac{6+7+10+12+12+13 + x + y}{8} = 9$

$\displaystyle \Rightarrow 60 + x + y = 72$

$\displaystyle \Rightarrow x + y = 12 \text{ ... ... ... ... ... i)}$

$\displaystyle \text{We know, Variance } = \frac{1}{n} \sum \Big( x_i - \overline{X} \Big)^2$

$\displaystyle \text{Therefore } \frac{1}{8} \Big( x_i - \overline{X} \Big)^2 = 9.25$

$\displaystyle ( 6 - 9)^2 + ( 7- 9)^2 + ( 10-9)^2 + ( 12-9)^2+ ( 12-9)^2+ ( 13-9)^2 + ( x - 9)^2 + ( y - 9)^2 = 74$

$\displaystyle \Rightarrow 3^2 + 2^2 + 1^2 + 3^2 + 3^2 + 4^2 + ( x - 9)^2 + ( y - 9)^2 = 74$

$\displaystyle \Rightarrow ( x - 9)^2 + ( y - 9)^2 = 26$

$\displaystyle \Rightarrow x^2 + 81- 18 x + y^2 + 81 - 18y = 26$

$\displaystyle \Rightarrow x^2 + y^2 - 18 ( x + y) = - 136$

$\displaystyle \Rightarrow x^2 + y^2 = 80 \text{ ... ... ... ... ... ii)}$

$\displaystyle \text{We know, } ( x+y)^2 + ( x-y)^2 =2 ( x^2 + y^2)$

$\displaystyle \Rightarrow 12^2 + (x-y)^2 = 2 \times 80$

$\displaystyle \Rightarrow (x-y)^2 = 160-144 =16$

$\displaystyle \Rightarrow x-y = \pm 4 \text{ ... ... ... ... ... iii)}$

$\displaystyle \therefore x-y = 4 \text{ ... ... ... ... ... iv)}$

$\text{or } \displaystyle x-y = -4 \text{ ... ... ... ... ... iv)}$

Solving i) and iv) we get $x = 8$ and $y = 4$

Solving i) and v) we get $x = 4$ and $y = 8$

Hence the other two observations are 4 and 9.

$\\$

Question 7: For a group of 200 candidates, the mean and standard deviations of scores were found to be 40 and 15 respectively. Later on it was discovered that the scores of 43 and 35 were misread as 34 and 53 respectively. Find the correct mean and standard deviation.

Answer:

$\displaystyle \text{Given: Mean } = \overline{X} = 40, \hspace{1.0cm} n = 200, \hspace{1.0cm} \sigma = S.D. = 15$

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

$\displaystyle \Rightarrow \frac{1}{200} \sum x_i = 40$

$\displaystyle \Rightarrow \sum x_i = 8000$

Since the scores were misread, this sum is incorrect. The correct sum would be

$\displaystyle \sum x_i = 8000-34-53+43+35 = 8000-9 = 7991$

$\displaystyle \text{Therefore the correct mean } = \frac{\sum x_i}{200} = \frac{7991}{200} = 39.955$

$\displaystyle \sigma = 15 \Rightarrow \sigma^2 = 225$

We know,

$\displaystyle \frac{1}{n} \sum \Big( x_i - \overline{X} \Big)^2 = 225$

$\displaystyle \Rightarrow \frac{1}{200} \sum \Big( x_i - 40 \Big)^2 = 225$

$\displaystyle \Rightarrow \sum \Big( x_i - 40 \Big)^2 = 45000$

$\displaystyle \Rightarrow \sum \Big( {x_i}^2 + 1600 - 80 x_i \Big) = 45000$

$\displaystyle \Rightarrow \Bigg[ \sum \Big( {x_i}^2 - 80 x_i \Big) \Bigg] + 320000= 45000$

$\displaystyle \Rightarrow \sum \Big( {x_i}^2 - 80 x_i \Big) = - 275000$

$\displaystyle \Rightarrow \sum {x_i}^2 - 80 \sum x_i = - 275000$

$\displaystyle \Rightarrow \sum {x_i}^2 - 80 (8000) = - 275000$

$\displaystyle \Rightarrow \sum {x_i}^2 = 365000$

$\displaystyle \text{Correct } \sum {x_i}^2 = \text{ Incorrect } \sum {x_i}^2 - \text{ ( sum of squares of incorrect values ) } \\ { \hspace{3.0cm}+ \text{ ( sum of squares correct values) } }$

$\displaystyle = 36500 - 34^2 - 54^2 + 43^2+ 35^2 = 364109$

$\displaystyle S.D. = \sqrt{ \frac{1}{n} \sum_{i=1}^{200} {x_i}^2 - \Bigg( \frac{1}{200} \sum_{i=1}^{200} x_i \Bigg)^2 } \\ \\ \\ = \sqrt{\frac{364109}{200} - \Bigg( \frac{7991}{200}\Bigg)^2 } = \sqrt{ 1820.545-1596.402 } = 14.97$

$\\$

Question 8: The mean and standard deviation of 100 observations were calculated as 40 and 5.1 respectively by a student who took by mistake 50 instead of 40 for one observation. What are the correct mean and standard deviation?

Answer:

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

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

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

$\displaystyle \Rightarrow \sum x_i = 4000$

Since the score was misread, this sum is incorrect. The correct sum would be

$\displaystyle \sum x_i = 48000-50+40 = 4000-910 = 3990$

$\displaystyle \text{Therefore the correct mean } = \frac{\sum x_i}{100} = \frac{3990}{100} = 39.9$

$\displaystyle \sigma = 5.1 \Rightarrow \sigma^2 = 26.01$

We know,

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

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

$\displaystyle \Rightarrow \sum \Big( x_i - 40 \Big)^2 = 2601$

$\displaystyle \Rightarrow \sum \Big( {x_i}^2 + 1600 - 80 x_i \Big) = 2601$

$\displaystyle \Rightarrow \Bigg[ \sum \Big( {x_i}^2 - 80 x_i \Big) \Bigg] + 160000= 2601$

$\displaystyle \Rightarrow \sum \Big( {x_i}^2 - 80 x_i \Big) = - 157399$

$\displaystyle \Rightarrow \sum {x_i}^2 - 80 \sum x_i = - 157399$

$\displaystyle \Rightarrow \sum {x_i}^2 - 80 (4000) = - 157399$

$\displaystyle \Rightarrow \sum {x_i}^2 = 162601$

$\displaystyle \text{Correct } \sum {x_i}^2 = \text{ Incorrect } \sum {x_i}^2 - \text{ ( sum of squares of incorrect values ) } \\ { \hspace{3.0cm}+ \text{ ( sum of squares correct values) } }$

$\displaystyle = 162601 -50^2 + 40^2 = 161701$

$\displaystyle S.D. = \sqrt{ \frac{1}{n} \sum_{i=1}^{100} {x_i}^2 - \Bigg( \frac{1}{100} \sum_{i=1}^{100} x_i \Bigg)^2 } \\ \\ \\ = \sqrt{\frac{161701}{100} - \Bigg( \frac{3990}{100} \Bigg)^2 } = \sqrt{ 1617.01-1592.01 } =5$

$\\$

Question 9: The mean and standard deviation of 20 observations are found to be 10 and 2 respectively. On rechecking it was found that an observation 8 was incorrect. Calculate the correct mean and standard deviation in each of the following cases.

(i) If wrong item is omitted (ii) if it is replaced by 12

Answer:

$\displaystyle \text{Given: Mean } = \overline{X} =10, \hspace{1.0cm} n = 20, \hspace{1.0cm} \sigma = S.D. = 2$

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

$\displaystyle \Rightarrow \frac{1}{20} \sum x_i = 10$

$\displaystyle \Rightarrow \sum x_i = 200$

Variance $\displaystyle \sigma^2 = 4$

$\displaystyle \frac{1}{n} \sum {x_i}^2 - (\overline{X})^2 = 4$

$\displaystyle \frac{1}{20} \sum {x_i}^2 - 10^2 = 4$

$\displaystyle \frac{1}{20} \sum {x_i}^2 = 104$

$\displaystyle \sum {x_i}^2 = 2080 \text{ ( This is incorrect because of incorrect values) }$

(i) If wrong item is omitted

$\displaystyle \text{Corrected} \sum x_i = 200-8 = 192$

$\displaystyle \text{Corrected } \sum {x_i}^2 = 2080 - 8^2 = 2080 - 64 = 2016$

$\displaystyle \text{Therefore the correct mean } = \frac{\sum x_i}{19} = \frac{192}{19} = 10.10$

We know,

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

$\displaystyle = \frac{2016}{19} - \Big( \frac{192}{19} \Big)^2 = \frac{1440}{361}$

$\displaystyle \text{Corrected } S.D. = \sqrt{\text{Variance}} = \sqrt{\frac{1440}{361}} = 1.997$

(ii) if it is replaced by 12

$\displaystyle \text{Corrected} \sum x_i = 200-8+ 12 = 204$

$\displaystyle \text{Corrected } \sum {x_i}^2 = 2080 - 8^2 + 12^2= 2160$

$\displaystyle \text{Therefore the correct mean } = \frac{\sum x_i}{20} = \frac{204}{20} = 10.2$

We know,

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

$\displaystyle = \frac{2160}{20} - \Big( \frac{204}{20} \Big)^2 = \frac{1584}{400}$

$\displaystyle \text{Corrected } S.D. = \sqrt{\text{Variance}} = \sqrt{\frac{1584}{400}} = 1.9899$

$\\$

Question 10: The mean and standard deviation of 20 observations are found to be 20 and 3 respectively. Later on, it was found that three observations were incorrect, which were recorded as 21, 21, and 18. Find the mean and standard deviation if the incorrect observations were omitted.

Answer:

$\displaystyle \text{Given: Mean } = \overline{X} =20, \hspace{1.0cm} n = 100, \hspace{1.0cm} \sigma = S.D. = 3$

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

$\displaystyle \Rightarrow \frac{1}{100} \sum x_i = 20$

$\displaystyle \text{Incorrect} \Rightarrow \sum x_i = 2000$

Variance $\displaystyle \sigma^2 = 9$

$\displaystyle \frac{1}{n} \sum {x_i}^2 - (\overline{X})^2 = 9$

$\displaystyle \frac{1}{100} \sum {x_i}^2 - 20^2 = 9$

$\displaystyle \frac{1}{100} \sum {x_i}^2 = 409$

$\displaystyle \text{Incorrect} \sum {x_i}^2 = 40900$

When the incorrect observations 21, 12, 18 are removed, n = 97

$\displaystyle \text{Corrected} \sum x_i = 2000 - 21-21-18 = 1940$

$\displaystyle \text{Therefore the correct mean } = \frac{\sum x_i}{97} = \frac{1940}{20} = 20$

$\displaystyle \text{Corrected } \sum {x_i}^2 = 40900 - 21^2 - 21^2 - 18^2= 39694$

We know,

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

$\displaystyle = \frac{39694}{97} - \Big( \frac{1940}{97} \Big)^2 = 409.216- 400 = 9.216$

$\displaystyle \text{Corrected } S.D. = \sqrt{\text{Variance}} = \sqrt{9.216} = 3.0357$

$\\$

Question 11: Show that the two formulae for the standard deviation of ungrouped date

$\displaystyle \sigma = \sqrt{ \frac{1}{n} \sum ( x_i - \overline{X})^2 } \text{ and } \sigma' = \sqrt{ \frac{1}{n} \sum {x_i}^2 - (\overline{X})^2 } \text{ are equivalent, where } \\ \\ \overline{X} = \frac{1}{n} \sum x_i$