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Old Exam Questions: Problem Solutions for Chapter 2 in MSC 287 by Dr. Stafford, Exams of Business Statistics

Solutions to old examination problems from chapter 2 of msc 287, taught by dr. Stafford. The problems revolve around concepts such as cumulative proportions, relative frequencies, pie charts, stem-and-leaf diagrams, ogives, and frequency distributions. Students can use this document to review and understand the concepts covered in the class.

Typology: Exams

Pre 2010

Uploaded on 07/23/2009

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Module 02.9 Old Exam Questions, Chapter 2 Dr. Stafford MSC 287 Revised: 08/11/04 Page 3 of 2 pages
KEY: OLD EXAMINATION QUESTIONS
Dr. Stafford, MSC 287, Spring 2001, Unit 02
Problem 1
Part I . N is given as 1000. Cumulative proportion is the same as cumulative relative frequency, so transfer this
column to CRF column. Since CRFi = CRFi-1 + ri, ri = CRFi - CRFi-1 with r1 = CRF1 . Further, fi = ri(1000) and cfi =
crfi(1000).
Number
of Shares
Cumulative
Proportion of
Shareholders Frequency Cumulative
Frequency Relative
Frequency
Cumulative
Relative
Frequency
1 - 500 0.30 300 300 .30 0.30
501 - 1000 0.65 350 650 .35 0.65
1001 - 1500 0.78 130 780 .13 0.78
1501 - 2000 0.88 100 880 .10 0.88
2001 - 2500 0.96 80 960 .08 0.96
2501 - 3000 1.00 40 1000 .04 1.00
Totals 1000 1.00
Part II
350 1 From table, 650 own 1000 or less shares, so 1000 - 650 = 350 own more than 1000 shares.
350 2 From table above, frequency of class “501-1000" = 350 shares.
0.88 3 crf(1501-2000) = 0.88.
0.10 4 r(1501-2000) = 0.10.
Problem 2
O 1 Pie chart is a pie chart, not a bar chart.
T 2 Stem-and-leaf includes data values while histogram merges them into frequency distributions.
O 3 Just the opposite – stem-and-leaf shows rank ordering.
T 4 This we emphasized thrice in class.
T 5 Just use the formulas for r, crf, f, and cf.
T 6 The tables are hard to interpret, but the diagrams show a picture of what the data is doing.
O 7 Ogive is used for cumulative distributions: cf, crf, c%f.
T 8 For k classes, sum of frequencies does equal n.
O 9 As we learned in class, there are three ways to compute crf.
T 10 Need number of classes, class widths, and the limits.
Problem 3
Class fiABC
1 D 40 N S
2 E J 0.32 T
3 F K P 0.68
4 G L 0.12 V
5 H M Q W
Total 250 R X
pf2

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Module 02.9 Old Exam Questions, Chapter 2 Dr. Stafford MSC 287 Revised: 08/11/04 Page 3 of 2 pages

KEY: OLD EXAMINATION QUESTIONS

Dr. Stafford, MSC 287, Spring 2001, Unit 02

Problem 1

Part I. N is given as 1000. Cumulative proportion is the same as cumulative relative frequency, so transfer this

column to CRF column. Since CRFi = CRFi-1 + r i , ri = CRFi - CRF i-1 with r 1 = CRF 1. Further, fi = r i (1000) and cfi =

crf i (1000).

Number of Shares

Cumulative Proportion of Shareholders Frequency Cumulative Frequency Relative Frequency

Cumulative Relative Frequency 1 - 500 0.30 300 300 .30 0. 501 - 1000 0.65 350 650 .35 0. 1001 - 1500 0.78 130 780 .13 0. 1501 - 2000 0.88 100 880 .10 0. 2001 - 2500 0.96 80 960 .08 0. 2501 - 3000 1.00 40 1000 .04 1. Totals 1000 1.

Part II

350 1 From table, 650 own 1000 or less shares, so 1000 - 650 = 350 own more than 1000 shares.

350 2 From table above, frequency of class “501-1000" = 350 shares.

0.88 3 crf(1501-2000) = 0..

0.10 4 r(1501-2000) = 0..

Problem 2

O 1 Pie chart is a pie chart, not a bar chart.

T 2 Stem-and-leaf includes data values while histogram merges them into frequency distributions.

O 3 Just the opposite – stem-and-leaf shows rank ordering.

T 4 This we emphasized thrice in class.

T 5 Just use the formulas for r, crf, f, and cf.

T 6 The tables are hard to interpret, but the diagrams show a picture of what the data is doing.

O 7 Ogive is used for cumulative distributions: cf, crf, c%f.

T 8 For k classes, sum of frequencies does equal n.

O 9 As we learned in class, there are three ways to compute crf.

T 10 Need number of classes, class widths, and the limits.

Problem 3

Class f (^) i A B C 1 D 40 N S 2 E J 0.32 T 3 F K P 0. 4 G L 0.12 V 5 H M Q W Total 250 R X

Module 02.9 Old Exam Questions, Chapter 2 Dr. Stafford MSC 287 Revised: 08/11/04 Page 4 of 2 pages

Problem 3 {continued}

Class fi cfi A r (^) i B crf (^) i C Answer Letter Answer Letter 1 40 D 40 0.16 N 0.16 S r (^) i 31 B 0.20 36 P 2 80 E 120 J 0.32 0.48 T 80 32 E 1.00 37 R 3 50 F 170 K 0.20 P 0.68 50 33 F 0.16 38 S 4 30 G 200 L 0.12 .80 V 170 34 K 0.48 39 T 5 50 H 250 M 0.20 Q 1.00 W 250 35 M 1.00 40 W Total 250 1.00 R NONE X

1. Since col. 2 is freq dist., and col. 3 has whole numbers, it must be the cum freq dist.

2. Since the entry to row 2, col. 4 is > row 4, col. 4, col. 4 must be the rel freq dist.

3. This leaves col. 5 as the cum rel freq. Dist. No % dists are given.

4. D = 40 because cf 1 = f 1. Then N = S = 40/250 = 0..

5. E = (.32)(250) = 80. J = E+40 = 120. T = .16+.32 = .48. T+P = .68, or P = .20.

6. Then F=P(250) = (.2)(250) = 50. K=J+F = 120+50 = 170. G = (.12)(250) = 30.

7. L=G+K = 170+30 = 200. V=.68+.12 = .80. H=250-L = 50.

8. M = L+H = 250. Q = 50/250 = 0.20. And W = .8 + .2 = 1..

Problem 4

Class f M cf r crf Mf

0 up to 10 5 5 5 .0625 .0625 25

10 up to 20 15 15 20 .1875 .2500 225

20 up to 30 20 25 40 .2500 .5000 500

30 up to 40 30 35 70 .3750 .8750 1050

40 up to 50 10 45 80 .1250 1.0000 450

60 1 How many observations have a value of 20 or greater?

.3750 2 What proportion of the observations have a value of 30 up to 40?

.5000 3 What proportion of the observations have a value less than 30?

20 4 How many of the observations have a value less than 20?

80 5 What is the sample size for this problem?

35 6 How many of the observations have a value of 10 up to 30? 40 - 5

.7500 7 What proportion of the observations have a value of 20 or more? 1.00 -.

28.125 8 What is the value of the grouped mean for this problem? (Mf)/n = 2250/80 = 28.

10 9 What is the value of the class width for this problem?

O 10 (T/O) We do not need the class midpoint values to compute the answers to some of the above

nine questions for this problem. Need it for the grouped mean.