$1$ . Calculate the first four moments about of the following distribution about the mean and hence find $\beta_{1}$ and $\beta_{2}$ $\newline$ \begin{tabular}{|c|c|c|c|c|c|c|c|c|c|} $\newline$ \hline $x$ & $0$ & $1$ & $2$ & $3$ & $4$ & $5$ & $6$ & $7$ & $8$ \\ $\newline$ \hline $f$ & $1$ & $8$ & $28$ & $56$ & $70$ & $56$ & $28$ & $8$ & $1$ \\ $\newline$ \hline $\newline$ \end{tabular} $\newline$ Also find the Karl Pearson's coefficient of skewness and Bowley's coefficient of skewness.

Full solution

1

. Calculate the first four moments about of the following distribution about the mean and hence find

\beta_{1}

and

\beta_{2}

\newline

\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|}

\newline

\hline

x

0

1

2

3

4

5

6

7

8

\newline

\hline

f

1

8

28

56

70

56

28

8

1

\newline

\hline

\newline

\end{tabular}

\newline

Also find the Karl Pearson's coefficient of skewness and Bowley's coefficient of skewness.

Calculate Mean: First, let's calculate the mean ( $\mu$ ). To do this, we multiply each value of $x$ by its frequency $f$ and sum them up, then divide by the total frequency. $\newline$ $\mu = \frac{(0\times1 + 1\times8 + 2\times28 + 3\times56 + 4\times70 + 5\times56 + 6\times28 + 7\times8 + 8\times1)}{(1+8+28+56+70+56+28+8+1)}$ $\newline$ $\mu = \frac{(0 + 8 + 56 + 168 + 280 + 280 + 168 + 56 + 8)}{256}$ $\newline$ $\mu = \frac{1024}{256}$ $\newline$ $\mu = 4$
Calculate First Moment: Now, we calculate the first moment about the mean $\mu_1$ , which is always $0$ because it's the mean itself. $\newline$ $\mu_1 = 0$
Calculate Second Moment: Next, we calculate the second moment about the mean $\mu_2$ . This is the variance, which is the sum of each $(x - \mu)^2 \times f$ , divided by the total frequency. $\mu_2 = \frac{[(0-4)^2\times1 + (1-4)^2\times8 + (2-4)^2\times28 + (3-4)^2\times56 + (4-4)^2\times70 + (5-4)^2\times56 + (6-4)^2\times28 + (7-4)^2\times8 + (8-4)^2\times1]}{256}$ $\mu_2 = \frac{[16\times1 + 9\times8 + 4\times28 + 1\times56 + 0\times70 + 1\times56 + 4\times28 + 9\times8 + 16\times1]}{256}$ $\mu_2 = \frac{[16 + 72 + 112 + 56 + 0 + 56 + 112 + 72 + 16]}{256}$ $\mu_2 = \frac{512}{256}$ $\mu_2 = 2$