STATISTICS 101 L - Homework 3
Due Friday, February 6, 2004
• Homework is due by 4:00 PM on the due date at 327 Snedecor. You can always hand in your homework
at the end of lecture on Friday.
• You may talk with others about the homework problems but please write your solutions up independently.
• Please answer homework questions in complete sentences. Make sure to
assignment together.
staple the pages of your
• Normally you will have an opportunity to get help on homework during lab.
Reading:
Jan. 26
Feb. 2
- Jan. 30
- Feb. 4
Section 1.3
Sections 2.1, 2.2
Assignment:
1. Read pages 63-84 of the text and do problems 1.82, 1.84, 1.86, 1.88, 1.90, 1.92, and 1.100.
2. Data was collected on the body temperature (o F ) for a sample of 65 males. The data are summarized
below in a frequency table.
Body
96.0
96.5
97.0
97.5
98.0
98.5
99.0
99.5
100.0
100.5
Temperature
≤ T emp <
≤ T emp <
≤ T emp <
≤ T emp <
≤ T emp <
≤ T emp <
≤ T emp <
≤ T emp <
≤ T emp <
≤ T emp <
(o F )
96.5
97.0
97.5
98.0
98.5
99.0
99.5
100.0
100.5
101.0
Count
0
1
5
10
17
16
10
4
1
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(a) Construct a histogram for body temperatures for the sample of 65 males. Describe the shape of
the histogram. Where is the approximate center of the histogram?
(b) What is the relative frequency for:
i. a temperature less than 98.0?
ii. a temperature greater than or equal to 99.5?
iii. a temperature greater than or equal to 97.0 but less than 100.0?
(c) It seems reasonable to model the distribution of body temperature for the male population as a
normal distribution. Assuming this normal (bell) shape with a population mean, µ = 98.5 and a
population standard deviation, σ = 0.75, what is the probability for:
i. a temperature less than 98.0?
ii. a temperature greater than or equal to 99.5?
iii. a temperature greater than or equal to 97.0 but less than 100.0?
(d) How close are the normal probabilities computed in (c) to the relative frequencies in (b)? What
does this say about the appropriateness of the normal model for body temperatures?
(e) From the frequency table can you determine the relative frequency for a male body temperature
greater than or equal to 98.6?
(f) Using the normal model, compute the probability that a male body temperature is greater than
or equal to 98.6.
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(g) Using the normal model, what is the temperature such that only 1% of males would have temperatures greater than or equal to this value?
3. Read pages 102-131 of the text and do problem 2.26.
4. An educational foundation would like to give scholarships to high school seniors who will be successful
in college. The foundation wishes to see if there is a relationship between the score on a verbal
aptitude test and the grade point average (GPA) in college. If there is, the foundation could use the
verbal aptitude test as a predictor of success in college and thus help them decide who should get the
scholarships. The verbal aptitude test is on a scale of 200 to 800 and GPA is on a scale from 0 to 4.
Below is a plot GPA versus the verbal aptitude test score for 50 students randomly selected from all
students at a large public university.
(a) From the plot, what is the lowest GPA? What verbal aptitude score is associated with the lowest
GPA?
(b) From the plot, what is the highest GPA? What verbal aptitude score is associated with the highest
GPA?
(c) Describe the general pattern of the relationship between verbal aptitude score and GPA.
(d) The value of the correlation coefficient for these 50 pairs of verbal aptitude score and GPA is
0.516. However, there appears to be an unusual pair or outlier. What are the verbal aptitude
score and GPA for that apparent outlier? If this apparent outlier were removed, would the
correlation coefficient calculated using the remaining 49 students be smaller than, about the same
as or larger than the 0.516? Explain briefly.
(e) Using the following summary information (verbal aptitude test score is X and GPA is Y) for the
49 students excluding the outlier, calculate the value of the correlation coefficient, r. Does this
agree with your assessment in (d)?
n = 49
X
(X − X̄)(Y − Ȳ ) = 1083.0
X
X
Y = 135.3
X = 29, 674
X
X
(Y − Ȳ )2 = 16.71
(X − X̄)2 = 246, 074.88
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