Answer :

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

Mean, $\overline{x}=\dfrac{\sum _{i=1}^{8} x_i}{n}=\dfrac{6+7+10+12+13+4+8+12}{8}=\dfrac{72}{8}=9$

The following table is obtained.

Variance $(\sigma^{2})=\dfrac{1}{n} \sum _{i=1}^{8}(x_i-\bar{{}x})^{2}=\dfrac{1}{8} \times 74=9.25$

Answer :

The mean of first $n$ natural numbers is calculated as follows.

Mean $=\dfrac{\text{ Sum of all observations }}{\text{ Number of observations }}$ $\therefore$ Mean $=\dfrac{\dfrac{n(n+1)}{2}}{n}=\dfrac{n+1}{2}$

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

$=\dfrac{1}{n} \sum _{i=1}^{n}[x_i-(\dfrac{n+1}{2})]^{2}$

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

$=\dfrac{1}{n} \dfrac{n(n+1)(2 n+1)}{6}-(\dfrac{n+1}{n})[\dfrac{n(n+1)}{2}]+\dfrac{(n+1)^{2}}{4 n} \times n$

$=\dfrac{(n+1)(2 n+1)}{6}-\dfrac{(n+1)^{2}}{2}+\dfrac{(n+1)^{2}}{4}$

$=\dfrac{(n+1)(2 n+1)}{6}-\dfrac{(n+1)^{2}}{4}$

$=(n+1)[\dfrac{4 n+2-3 n-3}{12}]$

$=\dfrac{(n+1)(n-1)}{12}$

$=\dfrac{n^{2}-1}{12}$

Answer :

The first 10 multiples of 3 are

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

Here, number of observations, $n=10$

Mean, $\bar{{}x}=\dfrac{\sum _{i=1}^{10} x_i}{10}=\dfrac{165}{10}=16.5$

The following table is obtained.

Variance $(\sigma^{2})=\dfrac{1}{n} \sum _{i=1}^{10}(x_i-\overline{x})^{2}=\dfrac{1}{10} \times 742.5=74.25$

Answer :

The data is obtained in tabular form as follows.

Here, $N=40, \quad \sum _{i=1}^{7} f_i x_i=760$

$\therefore \overline{x}=\dfrac{\sum _{i=1}^{7} f_i x_i}{N}=\dfrac{760}{40}=19$

Variance $=(\sigma^{2})=\dfrac{1}{N} \sum _{i=1}^{7} f_i(x_i-\bar{{}x})^{2}=\dfrac{1}{40} \times 1736=43.4$

Answer :

The data is obtained in tabular form as follows.

Here, $N=22$

$ \sum _{i=1}^{7} f_i x_i=2200 $

$\therefore \overline{x}=\dfrac{1}{N} \sum _{i=1}^{7} f_i x_i=\dfrac{1}{22} \times 2200=100$

$Variance(\sigma^{2})=\dfrac{1}{N} \sum _{i=1}^{7} f_i(x_i-\bar{{}x})^{2}=\dfrac{1}{22} \times 640=29.09$

Answer :

The data is obtained in tabular form as follows.

$\overline{x}=A \dfrac{\sum _{i=1}^{9} f_i y_i}{N} \times h=64+\dfrac{0}{100} \times 1=64+0=64$

Variance, $\sigma^{2}=\dfrac{h^{2}}{N^{2}}[N \sum _{i=1}^{9} f_i y_i{ }^{2}-(\sum _{i=1}^{9} f_i y_i)^{2}]$

$ \begin{aligned} & =\dfrac{1}{100^{2}}[100 \times 286-0] \\ & =2.86 \end{aligned} $

$\therefore S \tan$ dard deviation $(\sigma)=\sqrt{2.86}=1.69$

Find the mean and variance for the following frequency distributions in Exercises 7 and 8.

Answer :

Mean,

$ \begin{aligned} Variance(\sigma^{2}) & =\dfrac{h^{2}}{N^{2}}[N \sum _{i=1}^{7} f_i y_i^{2}-(\sum _{i=1}^{7} f_i y_i)^{2}] \\ & =\dfrac{(30)^{2}}{(30)^{2}}[30 \times 76-(2)^{2}] \\ & =2280-4 \\ & =2276 \end{aligned} $

Answer :

$\overline{x}=A+\dfrac{\sum _{i=1}^{5} f_i y_i}{N} \times h=25+\dfrac{10}{50} \times 10=25+2=27$ $Variance(\sigma^{2})=\dfrac{h^{2}}{N^{2}}[N \sum _{i=1}^{5} f_i y_i{ }^{2}-(\sum _{i=1}^{5} f_i y_i)^{2}]$

$ \begin{aligned} & =\dfrac{(10)^{2}}{(50)^{2}}[50 \times 68-(10)^{2}] \\ & =\dfrac{1}{25}[3400-100]=\dfrac{3300}{25} \\ & =132 \end{aligned} $

Answer :

$\begin{aligned} & \therefore \text { Mean }=a+\left(\frac{\sum f_i . d_i}{N}\right) \times h \\ & =92 \cdot 5+\left(\frac{6}{60}\right) \times 5 \\ & =92 \cdot 5+0 \cdot 5=93 \\ & \text { Variance }=h^2\left[\frac{1}{N} \sum f_i . d_i^2-\left(\frac{\sum f_i . d_i}{N}\right)^2\right] \\ & =(5)^2\left[\frac{254}{60}-\left(\frac{6}{60}\right)^2\right] \\ & =25\left[\frac{127}{30}-\frac{1}{100}\right]=105-58 \\ & \text { S.D. }=\sqrt{\text { Variance }} \\ & =\sqrt{105-58}=10.27\end{aligned}$

Answer :

Here, $N=100, h=4$

Let the assumed mean, $A$, be 42.5 .

$ \bar{{}x}=A+\dfrac{\sum _{i=1}^{5} f_i y_i}{N} \times h=42.5+\dfrac{25}{100} \times 4=43.5 $

$Variance(\sigma^{2})=\dfrac{h^{2}}{N^{2}}[N \sum _{i=1}^{5} f_i y_i{ }^{2}-(\sum _{i=1}^{5} f_i y_i)^{2}]$

$ \begin{aligned} & =\dfrac{16}{10000}[100 \times 199-(25)^{2}] \\ & =\dfrac{16}{10000}[19900-625] \\ & =\dfrac{16}{10000} \times 19275 \\ & =30.84 \end{aligned} $

$\therefore S \tan$ dard deviation $(\sigma)=5.55$