Homework Assignment 2 Solutions
2.76
a.
Mean
±
St. Dev is 7
±
1.7, or 5.3 to 8.7.
b.
Mean
±
2 St. Dev is 7
±
(2)(1.7), or 3.6 to 10.4.
c.
Mean
±
3 St. Dev is 7
±
(3)(1.7), or 1.9 to 12.1.
2.77
Figure for Exercise 2.77
2.78
a.
z
=
(200

170)/20 = 1.5.
b.
z
=
(140

170)/20 =

1.5.
c.
z
=
(170

170)/20 = 0.
d.
z
=
(230

170)/20 = 3.
2.79
a.
Mean =20; Standard deviation = 1.581.
b.
Mean =20; Standard deviation = 0.
c.
Mean = 20; Standard deviation = 33.09.
2.80
a.
98–41 = 57.
b.
Standard deviation
≈
Range/6
= 57/6 = 9.5.
2.81
The interval is

6.1 to 22.7 hours.
The interval includes negative values, which are impossible
times. Thus, the interval based an assumption of a bellshaped curve would not reflect reality.
2.91
Outliers affect the standard deviation.
This happens because the calculation uses the deviation
from the mean for every value. An outlier has a large deviation from the mean, so it inflates the
standard deviation.
Extreme values generally do not affect the quartiles, and consequently they
generally don't affect the interquartile range. Remember that a quartile is determined by counting
through the ordered data to a particular location, so the exact size of the largest or smallest
observations doesn't matter.
2.96
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
This is the end of the preview.
Sign up
to
access the rest of the document.
 Spring '08
 BARROSO,JOAOR
 Statistics, Normal Distribution, Standard Deviation, 0.3%, 75 mph, 22.7 hours

Click to edit the document details