SOLUTIONS TO CHAPTER 6 HOMEWORK
Exercise 2. IQ is normally distributed with a mean of 100 and a standard deviation
of 15. Suppose one individual is randomly chosen. Let X = IQ of an individual.
a. X ∼ N (100, 15)
b. Find the probability that the person has an IQ greater than 120. Include
a sketch of the graph and write a probability statement.
solution: The graph is the usual bell curve, centered at 100, with a region
past 120 shaded in.
TI Calculator Method: We can calculate the area of the shaded region
via
normalcdf(120, 10000, 100, 15) = 0.091
Table Method: Find the z-score: z = (120 − 100)/15 = 1.333 and use a
table to find that P (0 < z < 1.33) = 0.408; from which we get
P (X > 120) = P (Z > 1.33) = 0.5 − 0.408 = 0.092
The slight difference in values should be due to the rounding of 1.333 to
1.33 for the table.
c. Mensa is an organization whose members have the top 2% of all IQs. Find
the minimum IQ needed to qualify for the Mensa organization. Sketch the
graph and write the probability statement.
solution: We want to find the value X = xM for which P (X > xM ) =
0.02. Two methods below both show that xM = 131, i.e., that an IQ of 131
is needed to join Mensa.
TI Calculator Method: invNorm(0.98, 100, 15) = 130.8
Table Method: We find that the z-score of z = 2.06 corresponds to the
probability statement P (0 < Z < 2.06) = 0.4803, so we see that less than
2% of IQ’s exceed 100 + 2.06(15) = 130.9.
d. The middle 50% of IQs fall between what two values? Sketch the graph
and write the probability statement.
solution: We are finding Q1 and Q3 , i.e., the box ends on a boxplot.
Either of the methods below shows that the middle 50% of IQ’s lie in the
interval [90,110].
TI Calculator Method: Q1 = invNorm(0.25, 100, 15) = 89.9
Table Method: We find that the z-score of z = 0.68 corresponds to
the probability statement P (0 < Z < 0.68) = 0.2517, so we see that a
little more than 25% of IQ’s are within 0.68 standard deviations above the
mean IQ. Thus Q3 is about 0.68(15) = 10.2 deviations from the mean. By
symmetry Q1 deviates by the same amount.
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Exercise 10. Suppose that Ricardo and Anita attend different colleges. Ricardos
GPA is the same as the average GPA at his school. Anitas GPA is 0.70 standard
deviations above her school average. In complete sentences, explain why each of
the following statements may be false.
a. Ricardos actual GPA is lower than Anitas actual GPA.
solution: The GPA at Ricardo’s school could be considerably higher than
the GPA at Anita’s school.
b. Ricardo is not passing since his z-score is zero.
solution: This almost certainly not true. A z-score of zero corresponds
with the mean and median of a normal distribution. Thus, if Ricardo is
failing, then so are 50% of the students at his school.
c. Anita is in the 70th percentile of students at her college.
solution: For a normal distribution, being 0.70 standard deviations above
the mean is a z-score of 0.70 and from our table we see that this is actually
the 76th percentile because 0.5 + .258 = .758.
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