Answer to Question #122728 in Statistics and Probability for David

Question

The data below represent statistics and probability end of semester examination marks of 30
students randomly selected from the population of students who registered for probability and
statistics during 2018/2019 academic year.
16 51 36 45 23 48
37 19 28 28 25 36
76 22 27 18 28 42
38 47 44 29 37 42
27 33 35 46 28 27
Using sturge’s approximation rule with all your answers in one decimal place,
construct a frequency distribution table as follows
Class
Boundary
Tally Frequency Class
midpoint
CF FX FX^2
Calculate the coefficient of skewness for the data set and interpret our result.

Answer 1

$16, 18, 19,22, 23, 25,$

$27,27,27,28,28,28,$

$28,29,33,35,36,36,$

$37,37,38,42,42,44,$

$45,46,47,48,51,76$

Sturge’s Rule: $k=1+3.322(\log n), k$ is the number of classes, $n$ is the size of the data. 

Given $n=30$

$$k=1+3.322(\log 30)\approx6$$

1. $6$ classes, approximately.

2. Find the class width

$${Range\over number\ of\ classes}={76-16\over 6}=10$$

$$\def\arraystretch{1.5} \begin{array}{c:c} Class & \Large Class \atop Boundaries & \Large Class \atop Midpoints & Tally & Frequency\\ \hline 13-23 & 12.5-23.5 & 18 & |\ |\ |\ |\ | & 5 \\ \hdashline 24-34 & 23.5-34.5 & 29 & |\ |\ | \ |\ |\ |\ |\ |\ |\ |& 10 \\ 35-45 & 34.5-45.5 & 40 & |\ |\ | \ |\ |\ |\ | \ | \ | \ |& 10\\ 46-56 & 45.5-56.5 & 51 & |\ |\ | \ |& 4\\ 57-67 & 56.5-67.5 & 62 & & 0 \\ 68-78 & 67.5-78.5 & 73 & | & 1 \end{array}$$

$$\def\arraystretch{1.5} \begin{array}{c:c} Class & \Large Class \atop Midpoints & Frequency & \Large Relative \atop Frequency & \Large Cumulative \atop Rel.Frequency \\ \hline 13-23 & 18 & 5 & 1/6& 1/6 \\ \hdashline 24-34 & 29 & 10 & 1/3& 1/2 \\ 35-45 & 40 & 10 & 1/3 & 5/6\\ 46-56 & 51 & 4 & 2/15 & 29/30\\ 57-67 & 62 & 0 & 0 & 29/30 \\ 68-78 & 73 & 1 & 1/30 & 1 \end{array}$$

$$Mean=34.6$$

$$Median=34$$

$$\sigma_2^2={1\over n}\displaystyle\sum_{i=1}^n(\bar{x}-x_i)^2=152.110345$$

$$s^2={1\over n-1}\displaystyle\sum_{i=1}^n(\bar{x}-x_i)^2=152.110345$$

$$standard\ deviation=s=\sqrt{s^2}=12.3333$$

$$Skewness={\displaystyle\sum_{i=1}^n(\bar{x}-x_i)^3\over (n-1)\sigma^2}$$

$$={62103.36\over (30-1)(\dfrac{4411.2}{30})}=14.564$$

$$Median\ skewness={3(Mean-Median)\over s}=$$

$$={3(34.6-34)\over 12.3333}\approx0.146$$