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Chapter 2: Descriptive Statistics

CHAPTER 2: Descriptive Statistics

2.3

[LO 1]
28 2007 #1 28 71,273.93 58,069,987.70 7,620.37 59490 87970 28480

Distribution is skewed right.

Descriptive statistics
count mean sample variance sample standard deviation minimum maximum range

Stem and Leaf plot for stem unit = leaf unit = Frequency 2 9 13 4 28

#1 10000 1000 Stem 5 6 7 8 Leaf 99 123446677 0000112444447 1377

Distribution is more normally shaped in 2007.
2.5 [LO 2] a. We have 2 6  64 and 2 7  128. Because 26 < n = 100 and 27 > n = 100, we use K = 7 classes.

Class length =

largest measurement  smallest measurement 11.6  0.4   1.6 K 7

Class .4 – 1.9 2.0 – 3.5 3.6 – 5.1 5.2 – 6.7

Frequency 8 13 27 24

Relative Frequency .08 .13 .27 .24

Boundaries .35, 1.95 1.95, 3.55 3.55, 5.15 5.15, 6.75

Midpoint 1.15 2.75 4.35 5.95

Student Solutions Manual Business Statistics in Practice, Second Canadian Edition © 2011 McGraw-Hill Ryerson Limited. All rights reserved. 1

Chapter 2: Descriptive Statistics

6.8 – 8.3 8.4 – 9.9 10.0 – 11.5 11.6 – 13.1*

14 10 3 1

.14 .10 .03 .01

6.75, 8.35 8.35, 9.95 9.95, 11.5 11.55, 13.15

7.55 9.15 10.75 12.35

* Since the largest measurement is not within the seventh class, we add an eighth class. b. 2.7

The population of all possible customer waiting times during peak business hours is somewhat positively skewed with a tail to the right.

[LO 1, 4]
Roger Maris 8 4 3 6 3 8 6 3 9 0 0 1 1 2 2 3 3 4 4 5 5 6 Babe Ruth

1

2 5 4 5 1 6 4 9 0

1 6 4

6

7

9

The 61 home runs hit by Maris would be considered an outlier, although an exceptional individual achievement.

2.13[LO 5] a b N 10 7 MEAN 20 503 MEDIAN 20 501 MODE 20 501

2.15[LO 5] a. b.

Yes, because x  6 .
5 .2  5 .3  5.25 2 The mean is slightly larger than the median. The stem-and-leaf display is somewhat skewed right. median 

2.17

[LO 3, 5] a.

mean = 272.333; median = 68 (Canada’s value)
Student Solutions Manual Business Statistics in Practice, Second Canadian Edition © 2011 McGraw-Hill Ryerson Limited. All rights reserved. 2

Chapter 2: Descriptive Statistics b. c.

USA numbers skew the distribution Histogram is probably best plot for the data: Air Fatalities

800 600 400 200 0 Air Canada USA Mexico Air Fatalities

2.19[LO 3, 5] a. b. c. 2.21

mean = 2201.0000; median = 1,478 (Mexico’s value) mean>median (again because of USA values) histogram probably best for plotting.

[LO 3, 5] a. b. c.

here the trick is that Mexico does not have available data, so mean = 467.5 = median only two values so mean is the median histogram probably best for plotting.

2.29 [LO 6]

Descriptive statistics
count mean sample variance sample standard deviation minimum maximum range #1 10 20,167.50 705,139,612.50 26,554.47 625 79000 78375

Student Solutions Manual Business Statistics in Practice, Second Canadian Edition © 2011 McGraw-Hill Ryerson Limited. All rights reserved. 3

Chapter 2: Descriptive Statistics

2.31[LO 5, 6] a.

Also

 x  157  132  109  145  125  139  109,925  x   (157  132  109  145  125  139)  (807)  651,249 2 i 2 2 2 2 2 2

157  132  109  145  125  139  134.5 6 (157 – 134.5) 2  (132 – 134.5) 2  (139 – 134.5) 2 s2  = 276.7 6 –1 s  276.7  16.63 x 2 2 2

i

s2 

1  n –1 

x

2 i



1 n

 x    6 1 1 109,925 – 1 (651,249)  276.7    –  6   2 i

b.

[ x  s]  [134.5  16.63]  [117.87,151.13] [ x  2 s ]  [134.5  2(16.63)]  [101.24,167.76] [ x  3s ]  [134.5  3(16.63)]  [84.61,184.39] Yes, because $190 is not within the 99.73% interval. z 157 = 157  134.5 = 1.353 16.63

c. d.

z 132 = –0.150 z 109 = –1.533 z 145 = 0.631 z 125 = –0.571 z 139 = 0.271 2.33 [LO 2, 6] a. b.

It is somewhat reasonable. [ x  s]  [5.46  2.475]  [2.985,7.935] [ x  2 s]  [5.46  2(2.475)]  [.51,10.41] [ x  3s ]  [5.46  3( 2.475)]  [–1.965,12.885] Yes,...
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