Question

Calculate karl Pearson's co-efficient of skewness from the following data.
X
1
2
3
4
5
6
7
Frequency
10
18
30
25
12
3
2

Answer 1

$\large\underline{\underline\color{red}{⧪{\pmb{\sf{ \: Y◍ur \: Answer :-}}}}}$

Here, n = 100

Mean -:

$\huge{\underline{\boxed{\sf{\red{A \: + \frac{∑fd}{n} }}}}}$

Here,

A= 4

∑fd=1706

n = 100

Hence,

$4 + \frac{1706}{100} \\ \\ 4 + 1.706 \\ \\ 5.706$

☞Mean=5.706

Mode:-

{Observation With Highest Frequency}

So, here

Highest frequency is 30

So it's observation is 3

Hence,

$\huge\mathcal\colorbox{navy}{{\color{cyan}{(☞Mode=3࿐)}}}$

Standard Deviation:-

$\sqrt{\frac{ {∑fd}^{2} }{n} - ({ \frac{∑fd}{n} )}^{2} }$

Here,

∑fd² = 35972

∑fd=1706

n =100

$\sqrt{ \frac{35972}{100} - ({ \frac{1706}{100} })^{2} } \\ \\ \sqrt{359.72 - \frac{291.0436}{10000} } \\ \\ \sqrt{359.72 - 291.0436} \\ \\ \sqrt{68.6764} \\ \\ 8.29$

Skewness:-

Mean- Mode

5.706 - 3

Skewness:- 2.706

Coefficient of Skewness,j:-

$\frac{Mean- Mode}{s}$

$\frac{5.706 - 3}{8.29} \\ \\ \frac{2.706}{8.29} \\ \\ 0.32641737 \\ \\ 0.33$

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