Chapter 3, Problem 1PS

The following set of data is from sample of $n=5:$

a. Compute the mean, median, and mode.

b. Compute the range, variance, standard deviation and coefficient of variation.

c. Compute the Z scores. Are there any outliers?

d. Describe the shape of the data set.

To determine

Find the mean, median and mode of the data.

Answer to Problem 1PS

The mean is 6, median is 7, and there is no mode.

Explanation of Solution

A sample of data of size $n=5$ is given.

The formula to calculate sample mean is,

$\overline{X}=\frac{{\displaystyle \sum _{i=1}^n{X}_i}}{n}$

Here n is the sample size, $X_i$ is the ith value of the variable X , and $\overline{X}$ is the sample mean.

Substitute the given values of the variable X in the formula. So, the mean is calculated as,

$\begin{array}{c}\overline{X}=\frac{7+4+9+8+2}{5}\\ =\frac{30}{5}\\ =6\end{array}$

Thus, the mean of the sample is 6.

The median is the $\frac{n+1}{2}$ ranked value. So, the median is calculated as,

$\begin{array}{c}\text{Median}=\frac{5+1}{2}\text{ ranked value}\\ \text{=}\frac{6}{2}\text{ ranked value}\\ =3^{rd}\text{ ranked value}\end{array}$

This means that by Rule 1, the median is the value associated with third-ranked value. So, arrange the values from the smallest to the largest and give ranks. So, the given data is ranked as,

$\begin{array}{cccccc}\text{Ranked values}& 2& 4& 7& 8& 9\\ \text{Rank}& 1& 2& 3& 4& 5\end{array}$

So, the median is 7.

The mode is the value which appears most frequently. In the given data set, none of the data are repeated. So, there is no mode for the data.

Therefore, the mean is 6, median is 7, and there is no mode for the data.

To determine

Find the range, variance, standard deviation, coefficient of variation.

Answer to Problem 1PS

The range is $7$ , variance is $8.5$ , standard deviation is $2.92$ , and coefficient of variation is $48.67\%$ .

Explanation of Solution

The formula for range is defined as,

$\text{Range}=X\text{largest}-X\text{smallest}$

In the given data, the largest value is 9, and the smallest value is 2. So, the range is calculated as,

$\begin{array}{c}\text{Range}=9-2\\ =7\end{array}$

Thus, the range is 7.

The formula for sample variance is defined as,

$S^2=\frac{{\displaystyle \sum _{i=1}^n{\left(X_i-\overline{X}\right)}^2}}{n-1}$

The sample mean of the data is obtained as 6 in part (a). So, $\overline{X}=6$ and $n=5$ . Use this to calculate the variance as shown below.

$\begin{array}{c}{S}^2=\frac{{\left(7-6\right)}^2+{\left(4-6\right)}^2+{\left(9-6\right)}^2+{\left(8-6\right)}^2+{\left(2-6\right)}^2}{5-1}\\ =\frac{34}{4}\\ =8.5\end{array}$

Thus, the variance is $8.5$ .

The standard deviation is the square root of the variance. So, the sample standard deviation is calculated as,

$\begin{array}{c}S=\sqrt{8.5}\\ \approx 2.92\end{array}$

Thus, the standard deviation is $2.92$ .

The formula for coefficient of variation is defined as,

$CV=\left(\frac{S}{\overline{X}}\right)100\%$

Substitute the obtained values of $S=2.92,\overline{X}=6$ in the formula to calculate the coefficient of variation. So,

$\begin{array}{c}CV=\left(\frac{2.92}{6}\right)100\%\\ =48.67\%\end{array}$

Thus, the coefficient of variation is $48.67\%$ .

Therefore, the range is $7$ , variance is $8.5$ , standard deviation is $2.92$ , and coefficient of variation is $48.67\%$ .

To determine

Find the Z scores and determine if there are any outliers.

Answer to Problem 1PS

The Z scores are,

$\begin{array}{cc}X& Z\text{ score}\\ 7& 0.34\\ 4& -0.68\\ 9& 1.03\\ 8& 0.68\\ 2& -1.37\end{array}$

There are no outliers in the data set.

Explanation of Solution

The formula for Z score is defined as,

$Z=\frac{X-\overline{X}}{S}$

Use this formula and the obtained values $S=2.92,\overline{X}=6$ to calculate the Z scores for all the values of the data set. So, the Z scores are calculated as,

$\begin{array}{cc}X& Z=\frac{X-6}{2.92}\\ 7& 0.34\\ 4& -0.68\\ 9& 1.03\\ 8& 0.68\\ 2& -1.37\end{array}$

The largest Z score is $1.08$ which corresponds to the value 9. The smallest Z score is $-1.37$ which corresponds to the value 2. The Z scores show that none of the Z scores is less than $-3.0$ or greater than $+3.0$ and, hence, there are no potential outliers.

To determine

Explain the shape of the data set.

Answer to Problem 1PS

The distribution of data is negative or left-skewed.

Explanation of Solution

The mean is obtained as $6$ , and the median is obtained as $7$ in part (a). The distribution of the data is said to be left skewed if mean is less than median.

Here the mean value (6) is less than the median value (7). So, the data distribution is negative or left-skewed distribution.