Chapter 3, Problem 4DA

Randomly select 10 values from the number of suspensions in the local school districts in southwestern Pennsylvania in Data Set V in Appendix B. Find the mean, median, mode, range, variance, and standard deviation of the number of suspensions by using the Pearson coefficient of skewness.

To determine

The mean, median, mode, range, variance and standard deviation of suspension.

Answer to Problem 4DA

The mean, median, mode, range, variance and the standard deviation are 41.6, 46.5, no mode, 55, 17.65, and −0.83, respectively.

Explanation of Solution

Given info:

The randomly selected 10 number of suspensions of the enrolled in 41 local school districts in southwestern Pennsylvania is as follows:

37

29

47

51

46

65

10

60

15

56

Calculation:

Answers will vary. One of the possible answers is given below:

Mean:

Formula to calculate the sample mean is,

$\text{Mean}=\frac{{\displaystyle \sum X}}{n}$

Where,

• $n$ is the total number.
• $X_1,X_2\cdots X_{10}$ are values.

Substitute the value of $X_1,X_2\cdots X_{10}$ and 10 for $n$ in the above formula,

$\begin{array}{c}\text{Mean}=\frac{{\displaystyle \sum X}}{n}\\ =\frac{\text{37}+\text{29}+\text{47}+\text{51}+\text{46}+\text{65}+\text{10}+\text{60}+\text{15}+\text{56}}{10}\\ =\frac{416}{10}\\ =41.6\end{array}$

Thus, the mean of the data is 41.6.

Median:

Arrange the data in ascending order shown below,

10

15

29

37

46

47

51

56

60

65

Total number of the data is 10 which is even so the average of two middle value will be the median of the data.

The middle two data values are $46$ and $47$ so the median is,

$\begin{array}{c}\text{Median}=\frac{46+47}{2}\\ =\frac{93}{2}\\ =46.5\end{array}$

Thus, the median of the data is 46.5.

Mode:

Here, no value occurs more than once. Hence, the data has no mode.

Range:

Formula to calculate the range of the data is,

$\text{R}=\text{Highest}\text{value}-\text{Lowest}\text{value}$

Substitute 65 for the highest value and 10 for lowest value in the above formula to get range,

$\begin{array}{c}\text{R}=\text{Highest}\text{value}-\text{Lowest}\text{value}\\ \text{=65}-10\\ \text{=55}\end{array}$

Thus, the range of the data is 55.

Variance:

Formula to calculate the variance of the data is,

${\sigma }^2=\frac{{\displaystyle \sum {(X-\mu )}^2}}{N}$

Where,

• X is a individual value.
• $\mu$ is mean of the given data.
• $\text{N}$ is the total number of entries in given table.

Determine the mean for the data,

$\begin{array}{c}\text{Mean}=\frac{{\displaystyle \sum X}}{n}\\ =\frac{\text{37}+\text{29}+\text{47}+\text{51}+\text{46}+\text{65}+\text{10}+\text{60}+\text{15}+\text{56}}{10}\\ =\frac{416}{10}\\ =41.6\end{array}$

Determine the deviation for each data value and square each of the deviation,

X	Mean	Deviation for each data value $(X-\mu )$	Square of each of the deviation ${\left(X-\mu \right)}^2$
10	$41.6$	$10-41.6=-31.6$	998.56
15	$41.6$	$15-41.6=-26.6$	707.56
29	$41.6$	$29-41.6=-12.6$	158.76
37	$41.6$	$37-41.6=-4.6$	21.16
46	$41.6$	$46-41.6=4.4$	19.36
47	$41.6$	$47-41.6=5.4$	29.16
51	$41.6$	$51-41.6=9.4$	88.36
56	$41.6$	$56-41.6=14.4$	207.36
60	$41.6$	$60-41.6=18.4$	338.56
65	$41.6$	$65-41.6=23.4$	547.56

Determine the sum of the squares,

$\begin{array}{c}{{\displaystyle \sum \left(X-\mu \right)}}^2=\left(\begin{array}{c}\text{998}\text{.56 }+\text{707}\text{.56}+\text{158}\text{.76}+\text{21}\text{.16}+\text{19}\text{.36}+\\ \text{29}\text{.16}+\text{88}\text{.36}+\text{207}\text{.36}+\text{338}\text{.56}+\text{547}\text{.56}\end{array}\right)\\ =3,116.4\end{array}$

Substitute 3116.4 for ${\left(X-\mu \right)}^2$ and 10 for $\text{N}$ in the above formula to calculate the variance,

$\begin{array}{c}{\sigma }^2=\frac{{\displaystyle \sum {(X-\mu )}^2}}{N}\\ =\frac{3,116.4}{10}\\ =311.64\end{array}$

Thus the variance of the given data is 311.64.

Formula to calculate the sample standard deviation is,

$\sigma =\sqrt{\frac{{\displaystyle \sum {(X-\mu )}^2}}{N}}$

Where,

• X is a individual value.
• $\mu$ is mean of the given data.
• $\text{N}$ is the total number of entries in given table.

Determine the standard deviation by taking square root of the variance,

$\begin{array}{c}\sigma =\sqrt{\frac{{\displaystyle \sum {(X-\mu )}^2}}{N}}\\ =\sqrt{311.64}\\ =17.65\end{array}$

Thus, the standard deviation of the data is 17.65.

Pearson coefficient of skewness:

The Pearson coefficient of skewness is a measure which can be used to determine skewness of distribution.

The formula for Pearson coefficient of skewness is,

$\text{PC}=\frac{\text{3}\left(\mu -\text{MD}\right)}{\text{σ}}$

Where,

• PC is Pearson coefficient of skewness
• µ is mean.
• MD is median.
• σ is standard deviation.

Substitute value of µ as $41.6$ MD as $46.5$, and σ as 17.65 in the above equation,

$\begin{array}{c}\text{PC}=\frac{\text{3}\left(41.6-46.5\right)}{\text{17}\text{.65}}\\ =\frac{\text{3}\left(-4.9\right)}{\text{17}\text{.65}}\\ =\frac{-14.7}{\text{17}\text{.65}}\\ =-0.83\end{array}$

Thus, the Pearson coefficient is −0.83.

The distribution is negatively skewed, as the Pearson coefficient of skewness is negative.