Question1: Find the mean and variance for each of the data.

$6,7,10,12,13,4,8,12$

Answer:

Mean ( $\bar{x}$ ) of the given data:

$\bar{x}=\frac{1}{8}\sum _{i=1}^8{x}_i=\frac{6+7+10+12+13+4+8+12}{8}=\frac{72}{8}=9$

The respective values of the deviations from mean, $(x_i-\bar{x})$ are

-3, -2, 1 3 4 -5 -1 3

$\therefore$ $\sum _{i=1}^8(x_i-10{)}^2=74$

$\therefore$ ${\sigma }^2=\frac{1}{n}\sum _{i=1}^n(x_i-\bar{x}{)}^2$

$\frac{1}{8}\sum _{i=1}^8(x_i-\bar{x}{)}^2=\frac{74}{8}=9.25$

Hence, Mean = 9 and Variance = 9.25

Question 2: Find the mean and variance for each of the data.

First n natural numbers.

Answer:

Mean ( $\bar{x}$ ) of first n natural numbers:

$\bar{x}=\frac{1}{n}\sum _{i=1}^n{x}_i=\frac{\frac{n(n+1)}{2}}{n}=\frac{n+1}{2}$

We know, Variance ${\sigma }^2=\frac{1}{n}\sum _{i=1}^n(x_i-\bar{x}{)}^2$

${\sigma }^2=\frac{1}{n}\sum _{i=1}^n{(x_i-\frac{n+1}{2})}^2$

We know that $(a-b{)}^2=a^2-2ab+b^2$

$$\begin{gathered} \therefore n{\sigma }^2=\sum _{i=1}^n{x}_i^2+\sum _{i=1}^n(\frac{n+1}{2}{)}^2-2\sum _{i=1}^n{x}_i\frac{n+1}{2} \\ =\frac{n(n+1)(2n+1)}{6}+\frac{(n+1{)}^2}{4}\times n-2.\frac{(n+1)}{2}.\frac{n(n+1)}{2} \end{gathered}$$

$⟹{\sigma }^2=\frac{(n+1)(2n+1)}{6}+\frac{(n+1{)}^2}{4}-\frac{(n+1{)}^2}{2}$

$$\begin{gathered} =\frac{(n+1)(2n+1)}{6}-\frac{(n+1{)}^2}{4} \\ =(n+1)\left[\frac{4n+2-3n-3}{12}\right] \\ =(n+1).\frac{(n-1)}{12} \\ =\frac{n^2-1}{12} \end{gathered}$$

Hence, Mean = $\frac{n+1}{2}$ and Variance = $\frac{n^2-1}{12}$

Question 3: Find the mean and variance for each of the data

First 10 multiples of 3

Answer:

First 10 multiples of 3 are:

3, 6, 9, 12, 15, 18, 21, 24, 27, 30

Mean ( $\bar{x}$ ) of the above values:

$$\begin{gathered} \bar{x}=\frac{1}{10}\sum _{i=1}^{10}{x}_i=\frac{3+6+9+12+15+18+21+24+27+30}{10} \\ =3.\frac{\frac{10(10+1)}{2}}{10}=16.5 \end{gathered}$$

The respective values of the deviations from mean, $(x_i-\bar{x})$ are

-13.5, -10.5, -7.5, -4.5, -1.5, 1.5, 4.5, 7.5, 10.5, 13.5

$\therefore$ $\sum _{i=1}^{10}(x_i-16.5{)}^2=742.5$

$\therefore$ ${\sigma }^2=\frac{1}{n}\sum _{i=1}^n(x_i-\bar{x}{)}^2$

$\frac{1}{10}\sum _{i=1}^{10}(x_i-\bar{x}{)}^2=\frac{742.5}{10}=74.25$

Hence, Mean = 16.5 and Variance = 74.25

Question 4: Find the mean and variance for each of the data.

Answer:

$N=\sum _{i=1}^7{f}_i=40;\sum _{i=1}^7{f}_i{x}_i=760$

$\bar{x}=\frac{1}{N}\sum _{i=1}^n{f}_i{x}_i=\frac{760}{40}=19$

We know, Variance, ${\sigma }^2=\frac{1}{N}\sum _{i=1}^n(x_i-\bar{x}{)}^2$

$⟹{\sigma }^2=\frac{1736}{40}=43.4$

Hence, Mean = 19 and Variance = 43.4

Question 5: Find the mean and variance for each of the data.

Answer:

$N=\sum _{i=1}^7{f}_i=22;\sum _{i=1}^7{f}_i{x}_i=2200$

$\bar{x}=\frac{1}{N}\sum _{i=1}^n{f}_i{x}_i=\frac{2200}{22}=100$

We know, Variance, ${\sigma }^2=\frac{1}{N}\sum _{i=1}^n(x_i-\bar{x}{)}^2$

$⟹{\sigma }^2=\frac{640}{22}=29.09$

Hence, Mean = 100 and Variance = 29.09

Question 6: Find the mean and standard deviation using short-cut method.

Answer:

Let the assumed mean, A = 64 and h = 1

$N=\sum _{i=1}^9{f}_i=100;\sum _{i=1}^9{f}_i{y}_i=0$

Mean,

$\bar{x}=A+\frac{1}{N}\sum _{i=1}^n{f}_i{y}_i\times h=64+\frac{0}{100}=64$

We know, Variance, ${\sigma }^2=\frac{1}{N^2}[N\sum f_i{y}_i^2-(\sum f_i{y}_i{)}^2]\times h^2$

$$\begin{gathered} ⟹{\sigma }^2=\frac{1}{(100{)}^2}[100(286)-(0{)}^2] \\ =\frac{28600}{10000}=2.86 \end{gathered}$$

We know, Standard Deviation = $\sigma =\sqrt{Variance}$

$\therefore \sigma =\sqrt{2.86}=1.691$

Hence, Mean = 64 and Standard Deviation = 1.691

Question 7: Find the mean and variance for the following frequency distributions.

Answer:

$\bar{x}=\frac{1}{N}\sum _{i=1}^n{f}_i{x}_i=\frac{3210}{30}=107$

We know, Variance, ${\sigma }^2=\frac{1}{N}\sum _{i=1}^n(x_i-\bar{x}{)}^2$

$⟹{\sigma }^2=\frac{68280}{30}=2276$

Hence, Mean = 107 and Variance = 2276

Question 8: Find the mean and variance for the following frequency distributions.

Answer:

$\bar{x}=\frac{1}{N}\sum _{i=1}^n{f}_i{x}_i=\frac{1350}{50}=27$

We know, Variance, ${\sigma }^2=\frac{1}{N}\sum _{i=1}^n(x_i-\bar{x}{)}^2$

$⟹{\sigma }^2=\frac{6600}{50}=132$

Hence, Mean = 27 and Variance = 132

Question 9: Find the mean, variance and standard deviation using short-cut method.

Answer:

Let the assumed mean, A = 92.5 and h = 5

Mean,

$\bar{y}=A+\frac{1}{N}\sum _{i=1}^n{f}_i{y}_i\times h=92.5+\frac{6}{60}\times 5=93$

We know, Variance, ${\sigma }^2=\frac{1}{N^2}[N\sum f_i{y}_i^2-(\sum f_i{y}_i{)}^2]\times h^2$

$$\begin{gathered} ⟹{\sigma }^2=\frac{1}{(60{)}^2}[60(254)-(6{)}^2] \\ =\frac{1}{(60{)}^2}[15240-36] \\ =\frac{15204}{144}=105.583 \end{gathered}$$

We know, Standard Deviation = $\sigma =\sqrt{Variance}$

$\therefore \sigma =\sqrt{105.583}=10.275$

Hence, Mean = 93, Variance = 105.583 and Standard Deviation = 10.275

Question 10: The diameters of circles (in mm) drawn in a design are given below:

Calculate the standard deviation and mean diameter of the circles.
[ Hint First make the data continuous by making the classes as $32.5-36.5,36.5-40.4,40.5-44.5,44.5-48.5,48.5-52.5$ and then proceed.]

Answer:

Let the assumed mean, A = 92.5 and h = 5

Mean,

$\bar{x}=A+\frac{1}{N}\sum _{i=1}^n{f}_i{y}_i\times h=42.5+\frac{25}{100}\times 4=43.5$

We know, Variance, ${\sigma }^2=\frac{1}{N^2}[N\sum f_i{y}_i^2-(\sum f_i{y}_i{)}^2]\times h^2$

$$\begin{gathered} ⟹{\sigma }^2=\frac{1}{(100{)}^2}[100(199)-(25{)}^2]\times 4^2 \\ =\frac{1}{625}[19900-625] \\ =\frac{19275}{625}=30.84 \end{gathered}$$

We know, Standard Deviation = $\sigma =\sqrt{Variance}$

$\therefore \sigma =\sqrt{30.84}=5.553$

Hence, Mean = 43.5, Variance = 30.84 and Standard Deviation = 5.553