Solutions Manual for Statistical Methods 4th Edition by Donna L. Mohr William J. Wilson Rudolf J. Freund

,CHAPTER 1


EXERCISE 1 a) Mean = 17.00, SD = 5.53, median = 17.00,
Range = 22,
IQR = 20 – 13 = 7.
b) See boxplot.




c) A typical value for X is 17 which is both the mean and the
median. About half the values are between 13 and 20. The boxplot
shows that the upper tail is slightly longer, but difference is small
and mean and median are equal. This distribution is roughly
symmetric.




Chapter 1 – Page 1

,EXERCISE 2

Variable Mean Median Variance Standard Shape
Deviation

WATER 7.125 1.500 452.864 21.28 extremely positively
skewed

VEG 1.120 0.00 4.327 2.080 extremely positively
skewed

FOWL 75.635 11.5 42197.33 205.420 extremely positively
skewed



WATER
Stem Leaf # Boxplot

14 9 1 *
13
12
11
10
9
8
7
6
5
4
3 13 2 *
2
1 05667 5 0
0 00000000000001111111111112222222222345556779 44 +--+--+
+ + + + + + + +
Multiply Stem.Leaf by 10**+1




VEG
Stem Leaf # Boxplot
9 0 1 *
8
8 0 1 *
7
7 0 1 *
6
6
5
5 2 1 0
4
4 0 1 |
3 |
3 0 1 |
2 8 1 |
2 00002 5 |
1 58 2 +-----+
1 0002 4 | + |
0 58 2 | |
0 00000000000000000000000000000002 32 *-----*
+ + + + + +



Chapter 1 – Page 2

, FOWL

Stem Leaf # Boxplot
14 1 1 *
13
12
11
10
9
8
7
6
5
4
36 1 *
2124 3 0
1227888 6 0
000000000000000000000001111112222223335678 41 +--+--+
+ + + + + + + +-
Multiply Stem.Leaf by 10**+2




b) Frequency distribution for FOWL

Fowl Frequency Midpoint approximate mean = 5500/52
0 ≤ x < 100 41 50.0 = 105.8
100 ≤ x < 200 6 150.0
200 ≤ x < 300 3 250.0 approximate variance = 50746
300 ≤ x < 400 1 350.0
1400 ≤ x < 1500 1 1450.0



c) Scatterplots


fowl
1500

1400

1300

1200

1100

1000

900

800

700

600

500

400

300

200

100

0
0 1 2 3 4 5 6 7 8 9

veg




Aside from the single point with extraordinary number of waterfowl, there is little relationship.




Chapter 1 – Page 3