Question

Calculate pearson's coefficient of skewness

Answer 1

Step 1: Subtract the median from the mean: 70.5 – 80 = -9.5. Step 2: Divide by the standard deviation: -28.5 / 19.33 = -1.47. Caution: Pearson's first coefficient of skewness uses the mode. Therefore, if the mode is made up of too few pieces of data it won't be a stable measure of central tendency.

Answer 2

Answer:

A) The mean is $41.7$

B) The mode is $42.5$

C) The standard deviation is $15.43$

The Pearson's coefficient of skewness is $-0.0518$

Step-by-step explanation:

$\begin{array}{lccccc}\text { Class interval } & \mathbf{f} & \mathbf{x} & \mathbf{d}=\frac{x-A}{10} & \boldsymbol{\Sigma} f d & f d^{2} \\0-10 & 6 & 5 & -3 & -18 & 54 \\10-20 & 12 & 15 & -2 & -24 & 48 \\20-30 & 22 & 25 & -1 & -22 & 22 \\30-40 & 48 & (\mathbf{A}) 35 & 0 & 0 & 0 \\40-50 & 56 & 45 & 1 & 56 & 56 \\50-60 & 32 & 55 & 2 & 64 & 128 \\60-70 & 18 & 65 & 3 & 54 & 162 \\70-80 & 6 & 75 & 4 & \underline{24} & \underline{96}\end{array}$

$\begin{aligned}\sigma &=\sqrt{\frac{\sum f d^{2}}{\sum f}-\left(\frac{\sum f d}{\sum f}\right)^{2}} \times C \\&=\sqrt{\frac{566}{200}-\left(\frac{134}{200}\right)^{2}} \times 10 \\&=\sqrt{\frac{566}{200}-\frac{17956}{40000}} \times 10\end{aligned}$

$\begin{aligned}&=\sqrt{2.83-0.4489} \times 10 \\&=\sqrt{2.3811} \times 10 \\\sigma &=15.43 \\S_{k p} &=\frac{\text { Mean }-\text { Mode or } \frac{\bar{x}-M_{0}}{\sigma}}{\sigma} \\&=\frac{41.7-42.5}{15.43} \\&=\frac{-0.8}{15.43} \\S_{k p} &=-0.0518\end{aligned}$

$\begin{aligned}\bar{x} &=A+\frac{\sum f_{d}}{\sum f_{34}} \times 10 \\&=35+\frac{1}{200} \times h \\\bar{x} &=41.7 \\M_{0} &=l_{1}+\left[\frac{f_{1}-f_{0}}{2 f_{1}-f_{0}-f_{2}}\right] \times i \\&=40+\left[\frac{56-48}{2(56)-48-32}\right] \times 10 \\M_{0} &=42.5 \\\end{aligned}$