CH. 4 Class examples
I. Consider the discrete probability distribution when answering the following question.
X

1

2

3

6

8

P (x)

.10

?

.30

.20

.3


Calculate the mean and variance for this distribution.
II. Current estimates suggest that only 90% of the homebased computers have access to online services. Suppose 20 people with homebased PC were randomly and independently sampled.

Find the probability that at least 15 but at most 19 currently have access to online services.

Find the probability that fewer than quarter of those sampled currently have no access to online services.

Find the probability that at least 18 currently have access to online services.

According to a recent study, 1 in every 20 women has been a victim of domestic abuse at some point in their lives. Suppose we have randomly and independently sampled 20 women and asked whether they have been a victim of domestic abuse at some point in their lives.

Find the probability that at most 2 of the sampled women have been the victim of domestic abuse at some point in their lives.

Find the probability that more than 17 of the sampled women have not been the victim of domestic abuse at some point in their lives.

How many of the 20 women do we expect have been the victim of domestic abuse?
IV. Current estimates suggest that only 45% of entering college students will graduate within 4 years. Suppose 25 freshmen were randomly and independently sampled. Find the probability that fewer than three of those sampled will graduate within 4 years.
A) 0.000065 B) 0.000071 C) 0.000072 D) 0.000144
V. We believe that 77% of the population of all Business Statistics I students consider statistics an
exciting subject. Suppose we randomly and independently selected 32 students from the
population. If the true percentage is really 77%, find the probability of observing 31 or more of the
students who consider statistics to be an exciting subject in our sample of 32.
A) 0.002229 B) 0.002462 C) 0.000233 D) 0.997538
VI. Ann bakes six pies a day that cost $2 each to produce. On 21% of the days she sells only
two pies. On 19% of the days, she sells 4 pies, and on the remaining 60% of the days, she sells all six pies. If Ann sells her pies for $6 each, what is her expected profit for a day’s worth of pies?
X P(X)
3 0.21
5 0.19
6 0.60
E(x) = 3*.21+ 5*.19 + 6*.60 = 5.18
Profit = $5.18*6 – 6*2 = $31.08  12 = $19.08
VII. A local bakery has determined a probability distribution for the number of cheesecakes that they
sell in a given day. The distribution is as follows:
Number sold in a day 0 5 10 15 20
Prob. (Number sold) 0.2 0.22 0.13 0.08 0.37
Find the number of cheesecakes that this local bakery expects to sell in a day.
A) 10 B) 11 C) 11.2 D) 12
VIII. The chance that a Montesuma rose will blossom during the first year after planting is 80%. Let say, you bought 20 bushes of Montesuma rose and planted them in your garden. What is the probability that less than 75% of your roses will blossom the first year?
Answer: 0.196
IX. According to an article in the February 2008 issue of Reader’s Digest, Americans face a 1 in 20 chance of acquiring an infection while hospitalized. If the records of 15 randomly selected hospitalized patients are examined, find the probability that:

at least two develop an infection?
b. none develop an infection
1. The online access computer service industry is growing at an extraordinary rate. Current estimates suggest that only 45% of the homebased computers have access to online services. This number is expected to grow quickly over the next five years. Suppose 15 people with homebased computers were randomly and independently sampled. Find the probability that more than two of those sampled currently have access to online services.
n = 15

[the total number of people with computers]



p = .45

[the probability that they have access]

q = .55

[the probability that they do not have access]
Formula: P(x > 2) = 1  [P(x = 0) + P(x = 1) + P(x = 2)]

P(x = 0) =

15!
0!(15)!

(.45^{0})(.55^{15}) = 0.00013

P(x = 1) =

15!
1!(151)!

(.45^{1})(.55^{151}) = 0.00156

P(x = 2) =

15!
2!(152)!

(.45^{2})(.55^{152}) = 0.00896

Calculations: P(x > 2) = 1  0.016065 = 0.989349
2. According to a recent study, 1 in every 4 women has been a victim of domestic abuse at some point in their lives. Suppose we have randomly and independently sampled twenty women and asked whether they have been a victim of domestic abuse at some point in their lives. Find the probability that at least 2 of the sampled women have been the victim of domestic abuse at some point in their lives.
n = 20

[the total number of women]

k = 2

[the number of women you are looking for]

p = .25

[the probability that they are victims]

q = .75

[the probability that they are not victims]
P(x ≥ 2) = 1 – P(x < 2) = 1 – [P(x = 0) + P(x = 1)] = 0.976

P_{(1)} =

20!
1!(201)!

(.25^{1})(.75^{201}) =


P_{(0)} =

20!
1!(20)!

(.25^{0})(.75^{20}) =

3. The online access computer service industry is growing at an extraordinary rate. Current estimates suggest that only 45% of the homebased computers have access to online services. This number is expected to grow quickly over the next five years. Suppose 15 people with homebased computers were randomly and independently sampled. Find the probability that fewer than two of those sampled currently have access to online services.
n = 15

[the total number of people with computers]



p = .45

[the probability that they have access]

q = .55

[the probability that they do not have access]
P(x < 2) = [P(x = 0) + P(x = 1)] = 0.00169

P_{(1)} =

15!
1!(151)!

(.45^{1})(.55^{151}) = 0.00156

P_{(0)} =

15!
0!(15)!

(.45^{0})(.55^{15}) = 0.00013

Share with your friends: 