Minitab was used to construct the five number | Complete Solution

DIRECTIONS: Show your work or explain how you arrived at your conclusion whenever appropriate.

    Minitab was used to construct the five number summary and boxplot below for a sample of checkout times (measured in seconds) taken at Corner Convenience Store last month. Use the Minitab output provided to answer the questions that follow on the page below. (32 points)

Data Display

Time (in seconds)

10aa10aa15aa18aa22aa25aa28aa29aa29aa30aa30aa30aa30

30aa31aa32aa32aa32aa35aa35aa35aa35aa35aa44aa45aa45

47aa48aa49aa51aa55aa55aa58aa59aa60aa60aa63aa63aa63

65aa66aa68aa69aa75aa75aa78aa79aa81aa82aa83aa83aa83

88aa89aa89aa90aa90aa90aa90aa90aa91aa92aa93aa94aa98

100aa115aa120aa120aa122aa123aa125aa145aa150aa175

Five Number Summary: Time (in seconds)

Total

Variableaaaaaaaa Count aMinimum aQ1aaaaMedian aaQ3aaaMaximum

Time (in seconds)a 75aaa10.00aa a35.00aaa63.00aaa90.00aa175.00

Boxplot of Time (in seconds)

1.(Cont’d.)

a. True or False? The value of P20 could be 31 seconds. __T  (2 points)

b. True or False? The value of P75 could be 99 seconds. _____F_  (2 points)

c. True or False? The data is somewhat skewed right. ____T__  (3 points)

d. True or False? The data set has an outlier. _____T_  (3 points)

e. What check out time corresponds to the point labeled C?  63 (3 points)

f. The median is equal to the ______50th___ percentile.  (2 points)

g. Compute the Interquartile Range (IQR).  = [90-10]  (3 points)

h. Give the values of the fences.  (6 points)

The values for Q1 – 1.5×IQR and Q3 + 1.5×IQR are the "fences"

Lower Fence = ____10 -1.5*80 = -110__

Upper Fence = _10+1.5*

i. Find the value of P88 and interpret its practical meaning. (5 points)

.9

85% of all customers took less than 93.9 seconds to check out.

j. Approximate the percentile rank for a customer who takes 59 seconds to check out. (3 points)

There are 25 % percentiles

2.  A random sample of 50 employees at a company was obtained. The one-way distance from home to work was recorded for each employee in the sample. Suppose the mean of the sample was 15.2 miles and the standard deviation was 4.1 miles. (26 points)

a.  Compute the standard score (z - score) for a distance of 27 miles. Round your answer to the nearest hundredth and interpret the meaning of your answer as it pertains to this problem. (6 points)

z score = [27-15.2]/4.1 .878

b.  What is the one-way distance from home to work for an employee in the sample who had a standard score (z - score) of -1.50? (5 points)

x= [-1.5*4.1] +15.2 .05

c.   At least what percent of the distances in this sample should we expect to find within 2.5 standard deviations of the mean? Give your answer to the nearest percent. (5points)

When z= 2.5 then P(z< -25) = 0.003467

so answer is 2* 0.006934 = 7%

d.  At most what percent of the distances in this sample should we expect to find less than 2.9 miles and more than 27.5 miles? Give your answer to the nearest percent. (5points)

P(2.9<x<27.5) =.9973 = 100%

e.  At least what percent of the distances in this sample should we expect to find between 9.9 and 20.5 miles? Give your answer to the nearest percent. (5points)

P(9.9<x<20.5) .%

3.  In a random sample of 750 toner cartridges, the mean number of pages a toner cartridge can print is 4302 and the standard deviation is 340 pages. Assume the distribution of data is normally distributed. (22points)

a.  Between what two values can you expect to find (the middle) 99.7% of the sample data? (5points)

b.  What percent of the sample data should you expect to be more than 1 standard deviation away from the mean? (5points)

c.   How many of the toner cartridges in the sample should you expect to print between 3962 pages and 4642 pages? (6points)

d.        The company that makes the toner cartridges guarantees to replace any cartridge that prints fewer than 3622 pages. Approximately how many of the cartridges in the sample would you expect to be replaced under the guarantee policy?