91影视

Arranging the data in ascending order,

(0, 2, 4, 5, 7, 8, 9, 10, 10, 11, 11, 12, 13, 14, 15, 15, 16, 16, 17, 17, 18, 18,19, 19, 20, 21, 22, 22, 24, 24, 26, 28, 29, 31, 34, 38, 47, 49, 50, 64)

$\begin{array}{rcl}\mathrm{Q}1& =& \frac{\mathrm{N}+1}{4}\\ & =& \frac{40+1}{4}\\ & =& \frac{41}{4}\\ & =& 10.25\mathrm{th}\end{array}$

Term =11

$\begin{array}{rcl}\mathrm{Q}2& =& \frac{\mathrm{N}+1}{2}\\ & =& \frac{40+1}{2}\\ & =& \frac{41}{2}\\ & =& 20.{25}^{\mathrm{th}}\end{array}$

Term = 17.5

$\begin{array}{rcl}\mathrm{Q}3& =& \frac{3\left(\mathrm{N}+1\right)}{4}\\ & =& \frac{3\left(40+1\right)}{4}\\ & =& \frac{123}{4}\\ & =& 30.{75}^{\mathrm{th}}\\ \mathrm{Term}& =& 25.5\end{array}$

$\begin{array}{rcl}IQR& =& Q_U-Q_L\\ & =& 25.5-11\\ & =& 14.5\end{array}$

$\begin{array}{rcl}\mathrm{Lower}\mathrm{Inner}\mathrm{Fence}& =& {\mathrm{Q}}_{\mathrm{L}}-1.5\left(\mathrm{IQR}\right)\\ & =& 11-1.5\left(14.5\right)\\ & =& 11-21.75\\ & =& -10.75\end{array}$

$\begin{array}{rcl}\mathrm{Upper}\mathrm{Inner}\mathrm{Fence}& =& {\mathrm{Q}}_{\mathrm{U}}+1.5\left(\mathrm{IQR}\right)\\ & =& 25.5+1.5\left(14.5\right)\\ & =& 25.5+21.75\\ & =& 47.25\end{array}$

The graph is given below:

The frequency distribution is skewed to the rightbecause the top whisker is long. The median downtime taken is 17.5 months.

The variation is low in the data set because 75% of data is very close to each other only 25% is widely spread. We can notice this from the boxplot as the 3^rd quartile is very close to 0 while the data is spread to 64.

Customers with unusually high downtime areCustomer Number 264 鈥� 64 hours, Customer Number 268 鈥� 49 hours, and Customer Number 269 鈥� 50 hours.

$\begin{array}{l}\text{z-scorefor268=}\frac{\text{Value}鈭�\text{Mean}}{\text{StandardDeviation}}\\ \text{=}\frac{49鈭�20.375}{13.8891}\\ =\frac{28.625}{13.8891}\\ =2.06\end{array}$

The z-score value,is 2.06<3, therefore; Customer 268 is not an outlier.

$\begin{array}{c}\text{z-scorefor269=}\frac{\text{Value}鈭�\text{Mean}}{\text{StandardDeviation}}\\ =\frac{50鈭�20.375}{13.8891}\\ =\frac{29.625}{13.8891}\\ =2.13\end{array}$

The z-score value for Customer 269 is 2.13. Therefore, it is

not an outlier.

$\begin{array}{c}\text{z-scorefor264=}\frac{\text{Value}鈭�\text{Mean}}{\text{StandardDeviation}}\\ =\frac{64鈭�20.375}{13.8891}\\ =\frac{43.625}{13.8891}\\ =3.14\end{array}$

As the z-score value for Customer 264 is greater than 3,therefore,264 is an outlier.