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

$\hspace{1.9cm}=\dfrac{1}{n} \sum _ {i=1}^{n}\left[x_i-(\dfrac{n+1}{2})\right]^{2}$

$\hspace{1.9cm}=\dfrac{1}{n} \sum _ {i=1}^{n} x_i{ }^{2}-\dfrac{1}{n} \sum _ {i=1}^{n} 2(\dfrac{n+1}{2}) x_i+\dfrac{1}{n} \sum _ {i=1}^{n}(\dfrac{n+1}{2})^{2}$

$\hspace{1.9cm}=\dfrac{1}{n} \dfrac{n(n+1)(2 n+1)}{6}-(\dfrac{n+1}{n})\left[\dfrac{n(n+1)}{2}\right]+\dfrac{(n+1)^{2}}{4 n} \times n$

$\hspace{1.9cm}=\dfrac{(n+1)(2 n+1)}{6}-\dfrac{(n+1)^{2}}{2}+\dfrac{(n+1)^{2}}{4}$

$\hspace{1.9cm}=\dfrac{(n+1)(2 n+1)}{6}-\dfrac{(n+1)^{2}}{4}$

$\hspace{1.9cm}=(n+1)\left[\dfrac{4 n+2-3 n-3}{12}\right]$

$\hspace{1.9cm}=\dfrac{(n+1)(n-1)}{12}$

$\hspace{1.9cm}=\dfrac{n^{2}-1}{12}$

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, $

$ \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 $

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

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

Answer :

The data is obtained in tabular form as follows.

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

$ \begin{aligned} \text{Variance,} \ \sigma^{2} & =\dfrac{h^{2}}{N^{2}}\left[N \sum _ {i=1}^{9} f_i y_i{ }^{2}-(\sum _ {i=1}^{9} f_i y_i)^{2}\right] \\ \\ & =\dfrac{1}{100^{2}}\left[100 \times 286-0\right] \\ \\ & =2.86 \end{aligned} $

$\therefore \ \ $ Standard deviation $(\sigma)=\sqrt{2.86}=1.69$

Answer :

$\begin{aligned} \text { Here, } N & =30, h=30 \\ \\ \text { Mean, } \bar{x} & =A+\frac{\sum _ {i=1}^7 f_i y_i}{N} \times h \\ \\ & =105+\frac{2}{30} \times 30 \\ \\ & =105+2=107 \end{aligned}$

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

Answer :

$\overline{x}=A+\dfrac{\sum _ {i=1}^{5} f_i y_i}{N} \times h=25+\dfrac{10}{50} \times 10=25+2=27$

$ \begin{aligned} \text{Variance} \ (\sigma^{2}) & =\dfrac{h^{2}}{N^{2}}\left[N \sum _ {i=1}^{5} f_i y_i{ }^{2}-(\sum _ {i=1}^{5} f_i y_i)^{2}\right] \\ \\ & =\dfrac{(10)^{2}}{(50)^{2}}\left[50 \times 68-(10)^{2}\right] \\ \\ & =\dfrac{1}{25}\left[3400-100\right]=\dfrac{3300}{25} \\ \\ & =132 \end{aligned} $

Answer :

We have $N=\sum f_i=60$

Width of class $\mathrm{h}=5$

Assumed mean $\mathrm{A}=92.5$ (say)

$ \begin{aligned} & \text { Mean, } \overline{\mathrm{x}}=\mathrm{A}+\frac{\sum _ {\mathrm{i}=1}^9 \mathrm{f} _ {\mathrm{i}} \mathrm{u} _ {\mathrm{i}}}{\mathrm{ ~ N}} \times \mathrm{h} \\ \\ & =92.5+\frac{6}{60} \times 5=92.5+0.5=93 \\ \\ & \Rightarrow \overline{\mathrm{x}}=93 \end{aligned} $

$ \begin{aligned} \text { Variance } \sigma^2 & =h^2\left[\frac{\sum _ {i=1}^9 f_i u_i^2}{N}-\left(\frac{\sum _ {i=1}^9 f_i u_i}{N}\right)^2\right] \\ \\ \sigma^2 & =5^2\left[\frac{254}{60}-\left(\frac{6}{60}\right)^2\right] \\ \\ & =5^2 \frac{\left[60 \times 254-6^2\right]}{(60)^2} \\ \\ & =\frac{25}{3600}[15240-36] \\ \\ & =\frac{25}{3600} \times 15204 \\ \\ \sigma^2 & =105.58 \end{aligned} $

Standard Deviation $=\sqrt{\text { Variance }}$

$ \begin{aligned} \hspace{2.1cm} \sigma & =\sqrt{105 \cdot 5^8} \\ \\ \hspace{2.1cm} \sigma & =10.27 \end{aligned} $

Answer :

Since, the given frequency distribution is not continuous, so to make it continuous, we will subtract 0.5 from lower limit and add 0.5 to the upper limit of each class.

Here, $\mathrm{N}=100, \mathrm{ ~ h}=4$

Let the assumed mean $\mathrm{A}=42.5$

$ \begin{aligned} \text { Mean } \bar{x} & =A+\frac{\sum _ {i=1}^5 f_ i u_ i}{N} \times h \\ \\ & =42.5+\frac{25}{100} \times 4=43.5 \end{aligned} $

$ \begin{aligned} \text { Variance }\left(\sigma^2\right) & =\frac{\mathrm{h}^2}{\mathrm{ ~ }^2}\left[\mathrm{ ~ N} \sum _ {\mathrm{i}=1}^5 \mathrm{f} _ {\mathrm{i}} \mathrm{u} _ {\mathrm{i}}^2-\left(\sum _ {\mathrm{i}=1}^5 \mathrm{f} _ {\mathrm{i}} \mathrm{u} _ {\mathrm{i}}\right)^2\right] \\ \\ & =\frac{16}{10000}\left[100 \times 199-(25)^2\right] \\ \\ & =\frac{16}{10000}[19900-625] \\ \\ & =\frac{16}{10000} \times 19275 \\ \\ & =30.84 \end{aligned} $

$ \therefore \text { Standard deviation }(\sigma) =5.55$