Network server downtime.A manufacturer of network computer server systems is interested in improving its customer support services. As a first step, its marketing department has been charged with the responsibility of summarizing the extent of customer problems in terms of system downtime. The 40 most recent customers were surveyed to determine the amount of downtime (in hours) they had experienced during the previous month. These data are listed in the table.

Customer Number	Downtime	Customer Number	Downtime
230	12	250	4
231	16	251	10
232	5	252	15
233	16	253	7
234	21	254	20
235	29	255	9
236	38	256	22
237	14	257	18
238	47	258	28
239	0	259	19
240	24	260	34
241	15	261	26
242	13	262	17
243	8	263	11
244	2	264	64
245	11	265	19
246	22	266	18
247	17	267	24
248	31	268	49
249	10	269	50

a.Construct a box plot for these data. Use the information reflected in the box plot to describe the frequency distribution of the data set. Your description should address central tendency, variation, and skewness.

b.Use your box plot to determine which customers are having unusually lengthy downtimes.

c.Find and interpret the z-scores associated with the customers you identified in part b.

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 – 64 hours, Customer Number 268 – 49 hours, and Customer Number 269 – 50 hours.

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

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

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

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

not an outlier.

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

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