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Chapter 2: Q110E (page 125)

Rankings of research universities.Refer to the College Choice 2015 Rankings of National Research Universities, Exercise 2.43 (p. 95). Recall that data on academic reputation score, financial aid awarded, and net cost to attend for the top 50 research universities are saved in the TOPUNIV file. The 50 academic reputation scores are listed in the accompanying table.

99 92 94 95 97 91 91 92 92 89 84 85 100 87 83 83 89 79 94 79 79 87 76 67 76 76 76 70 74 64 74 69 66 72 65 76 64 65 61 69 62 69 52 64 64 47 60 57 63 62

a.Find the median, lower quartile, and upper quartile for the data.

b.Find IQR for the data.

c.Graph the data with a box plot.

d.Do you detect any outliers? Suspect outliers?

Short Answer

Expert verified

(a) Median = 76, lower quartile = 64.75, upper quartile = 89.25

(b) IQR = 24.5

(c)

(d) No

Step by step solution

01

Finding the median, lower quartile, and upper quartile

Arranging the data in ascending order,

(47,52,57,60,61,62,62,63,64,64,64,64,65,65,66,67,69,69,69,70,72,74,74,76,76,76,76,76,79,79,79,83,83,84,85,87,87,89,89,91,91,92,92,92,94,94,95,97,99,100)Q1=N+14=50+14=514=12.75thterm=64.75Q2=N+12=50+12=512=25.5thterm=76Q3=3N+14=350+14=1534=38.25thterm=89.25

Therefore, Median = 76, lower quartile = 64.75, upper quartile = 89.25

02

Computing the IQR

IQR = QU 鈥� QL

= 89.25 鈥� 64.75

= 24.5

03

Constructing the box plot

Lower Inner Fence = QL - 1.5(IQR)

= 64.75 鈥� 1.5(24.5)

= 64.75 鈥� 36.75

= 28

Upper Inner Fence = QU + 1.5(IQR)

= 89.25 + 1.5(24.5)

= 89.25 + 24.5

= 113.75

04

Detecting outliers and Suspect outliers

No, I do not detect any outliers or suspect outliers.

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Most popular questions from this chapter

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

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230

12

250

4

231

16

251

10

232

5

252

15

233

16

253

7

234

21

254

20

235

29

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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

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262

17

243

8

263

11

244

2

264

64

245

11

265

19

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22

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18

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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.

Discuss the conditions under which the median is preferred to the mean as a measure of central tendency.

Monitoring weights of flour bags.When it is working properly, a machine that fills 25-pound bags of flour dispenses an average of 25 pounds per fill; the standard deviation of the amount of fill is .1 pound. To monitor the performance of the machine, an inspector weighs the contents of a bag coming off the machine鈥檚 conveyor belt every half hour during the day. If the contents of two consecutive bags fall more than 2 standard deviations from the mean (using the mean and standard deviation given above), the filling process is said to be out of control, and the machine is shut down briefly for adjustments. The data

given in the following table are the weights measured by the inspector yesterday. Assume the machine is never shut down for more than 15 minutes at a time. At what times yesterday was the process shut down for adjustment? Justify your answer.

Time

Weight (pounds)

8:00 a.m

25.10

8:30

25.15

9:00

24.81

9:30

24.75

10:00

25.00

10:30

25.05

11:00

25.23

11:30

25.25

12:00

25.01

12:30 p.m

25.06

1:00

24.95

1:30

24.80

2:00

24.95

2:30

25.21

3:00

24.90

3:30

24.71

4:00

25.31

4:30

25.15

5:00

25.20

Calculate the mean and median of the following grade point averages:

3.2, 2.5, 2.1, 3.7, 2.8, 2.0

Cable TV subscriptions and 鈥渃ord cutters.鈥� Has the increasing popularity of smartphones and video streaming over the Internet affected cable and satellite TV subscriptions? This was one of the questions of interest in a recent Pew Research Center survey (December 2015). Telephone (both landline and cell phone) interviews were conducted on a representative sample of 2,001 adults living in the United States. For this sample, 1,521 adults reported that they currently receive cable or satellite TV service at home, 180 revealed that they have never subscribed to cable/satellite TV service at home, and the remainder (300 adults) admitted that they are 鈥渃ord cutters,鈥� i.e., they canceled the cable/satellite TV service. The results are summarized in the Minitab pie chart shown.

a. According to the pie chart, what proportion of the adults in the sample currently have a cable/satellite TV subscription at home? Verify the accuracy of this proportion using the survey results.

b. Now consider only the 1,821 adults in the sample that have at one time or another subscribed to cable/satellite TV service. Create a graph that compares the proportions of adults who currently subscribe to cable/satellite TV service with the proportion who are 鈥渃ord cutters.鈥�
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