Homework 1
2.1 The random variable X has a normal distribution with mean 60 and standard
deviation 12.
(a) The probability is 0.10 that this random variable takes a value less than what
number? ( 5 points)
(b) Find the number K such that the probability is 0.90 that the random variable takes
a value between 60-K and 60+K. ( 5 points)
(c) The random variable Y is defined by Y=200-2X. Find the mean and standard
deviation of Y. ( 10 points)
------ Answer to 2.1---------(a)
[From Section 2.2, subsection The Normal Distribution]
α = P(Z < −zα)
α = P((X−µ)/σ< − z α )
α = P(X < µ − σz α )
0.10 = P ( X < 60 − 12 × 1.282)
0.10 = P ( X < 44.616)
Answer to (a) is 44.616.
b)
[From Section 2.2, subsection The Normal Distribution]
1 − α = P( − z α / 2 < Z < + z α / 2 )
0.90 = 1 − 0.1 = P(−z 0 . 0 5 < Z < +z 0 . 0 5 )
0.90 = P(- 1.645 <
X − 60
< +1.645)
12
0.90 = P(60− 1.645×12 < X < 60+1.645×12)
0.90 = P(60−19.74 < X < 60+19.74)
Answer to (b) K = 19.74
(c)
[From Section 2.2, subsection Expected Values, Mean, and Variance]
Since Y = 200 − 2X, then µ Y = 200 − 2 µ X = 200 − 2 × 60 = 80, and σ Y
= 2 σ X = 2 × 12 = 24.
1
2.2 A particular brand of tie has a lifetime whose distribution is normal with mean
40,000 miles and standard deviation 10,000 miles.
(a) The probability is 0.95 that one of these tires has a lifetime of more than how
many miles? ( 10 points)
(b) Four of these tires are chosen at random. The probability is 0.95 that their average
lifetime is more than how many miles? ( 10 points)
------ Answer to 2.2---------(a)
[From Section 2.2, subsection The Normal Distribution]
α = P(Z > zα)
0.05= P(Z> z α = 1 . 6 4 5 )
and so
0.05= P(Z< − 1 . 6 4 5 )
0.95 = P (( X − µ ) / σ > −1.645)
0.95 = P ( X > 40, 000 − 1.645 ×10, 000)
Answer to (a) is 23,550.
(b)
[From Section 2.3, Sampling and Sampling Distributions]
α = P(Z > zα)
0.05= P(Z > z α = 1 . 6 4 5 )
and so
0.95= P(Z > − 1 . 6 4 5 )
0.95 = P(( X − µ ) / σ x > −1.645)
where σ x = σ / n = 10, 000 / 4 = 5, 000
0.95 = P( X > 40, 000 − 1.645 × 5, 000)
Answer to (b) is 31,775.
2
2.3 A mutual fund has a very large portfolio of stocks. A random sample of nine of
these showed the following percentage returns over a year. [See text for data].
Assume that the population distribution is normal. The annual percentage returns
of nine stocks from a portfolio: 8.2 15.7 3.1 4.7 12.8 9.6 18.3 11.4
10.2
(a) Find the sample mean. ( 5 points)
(b) Find the sample variance. ( 5 points)
(c) Find the sample standard deviation. ( 5 points)
(d) Find a 99% confidence interval for the population mean. ( 10 points)
(e) Find a 95% confidence interval for the population variance. ( 10 points)
(f) Find a 90% confidence interval for the population standard deviation. ( 10 points)
------ Answer to 2.3----------
[Parts (d), (e), and (f) are from Section 2.4, subsection Confidence Intervals]
3
2.4 Carefully explain why a 95% confidence interval for a population parameter will be
wider than a 90% confidence interval for that parameter based on the same information.
(15 points)
------ Answer to 2.4---------[From Section 2.4, subsection Confidence Intervals]
Quoting the text — "Notice that the 95% confidence interval is wider than the 90%
confidence interval. This result is quite general. Based on the same information, the
greater the probability content the wider will be the confidence interval for any population
parameter. This is to be expected; the surer we want to be that a computed interval will
contain the parameter, the wider the interval that will be required."
4