Question 1

Historical data show that the average number of patient arrivals at the intensive care unit of General Hospital is 3 patients every 2 hours. Assume that the patient arrivals are distributed according to a Poisson distribution. Determine the probability of at least 4 but no more than 8 patients arriving in a three-hour period.

Accepted Answer

P(4 ≤ X ≤ 8) = P(X = 8) + P(X = 7) + P(X = 6) + P(X = 5) + P(X = 4)
= (.0463) + (.0824) + (.1281) + (.1708) + (.1898) = .6174

Question 2

A multiple-choice test has 30 questions and each one has five possible answers, of which only one is correct. If all answers were guesses, find the probability of getting exactly four correct answers.

Accepted Answer

Probability of getting question correct = .2. Four successes in 30 trials. Treat as a binomial. P(X) = n!/[x!(n − x)!] × p^x(1 − p)ⁿ⁻^x, for x = 4, when p = .20 and n = 30.

Question 3

The probability that a given computer chip will fail is .02. Find the probability that of 5 delivered chips, exactly 2 chips will fail.

Accepted Answer

P(X) = n!/[x!(n − x)!] × p^x(1 − p)^n−x, for x = 2, n = 5, p = .02

Question 4

An appliance manufacturer gives a warranty and 95 percent of its appliances do not require repair before the warranty expires. An organization buys 10 of these appliances. Calculate an interval that contains 95.44 percent of all the appliances that will not require repair.

Accepted Answer

μ = np = 10(.95) = 9.5
σ === .6892
95.44% is a 2-standard-deviation interval: [μ ± 2σ] = [9.5 ± 2(.6892)]
= [8.12, 10.88]

Question 5

Of all individual tax returns, 37 percent include errors made by the taxpayer. If IRS examiners are assigned randomly selected returns in batches of 12, find the mean and standard deviation for the number of erroneous returns per batch.

Accepted Answer

Mean = np = 12(.37) = 4.44
Standard Deviation === 1.67

Question 6

A study conducted by a local university found that 25 percent of college freshmen support increased spending on environmental issues. If 6 college freshmen are randomly selected, find the probability that fewer than 4 support increased spending on environmental issues.

Accepted Answer

Σ P(X) = n!/[x!(n − x)!] × p^x(1 − p)^n−x, for x = 0, 1, 2, 3 when p = .25 and n = 6.

Question 7

A study conducted by a local university found that 25 percent of college freshmen support increased spending on environmental issues. If 6 college freshmen are randomly selected, find the probability that exactly 3 support increased spending on environmental issues.

Accepted Answer

P(X) = n!/[x!(n − x)!] × p^x(1 − p)^n−x, for x = 3 when p = .25 and n = 6.

Question 8

The J.O. Supplies Company buys calculators from a non-U.S. supplier. The probability of a defective calculator is 10 percent. If 10 calculators are selected at random, what is the probability that 3 or more of the calculators will be defective?

Accepted Answer

P(X ≥ 3) = 1 − P(X ≤ 2) = 1 − .9298 = .0702

Question 9

In the most recent election, 19 percent of all eligible college students voted. If a random sample of 20 students were surveyed, find the probability that exactly half voted in the election.

Accepted Answer

P(X) = n!/[x!(n − x)!] × p^x(1 − p)^n−x, for x = 10, when p = .19 and n = 20.

Question 10

The manager of the local grocery store has determined that, on average, 4 customers use the service desk every half-hour. Assume that the number of customers using the service desk has a Poisson distribution. What is the probability that during a randomly selected half-hour period, exactly 2 customers use the service desk?

Accepted Answer

P(2) = e⁻⁴(4)²/2! (e = 2.71828)

Question 11

The manager of the local grocery store has determined that, on average, 4 customers use the service desk every half-hour. Assume that the number of customers using the service desk has a Poisson distribution. What is the probability that during a randomly selected half-hour period, no more than 2 customers use the service desk?

Accepted Answer

P(x < 3) = Σ e^−μ(μ)^x/x!, for x = 0, 1, 2 (e = 2.71828)

Question 12

In a study conducted for the state Department of Education, 30 percent of the teachers who left teaching did so because they were laid off. Assume that we randomly select 10 teachers who have recently left their profession. Find the probability that exactly 4 of them were laid off.

Accepted Answer

P(X) = n!/[x!(n − x)!] × p^x(1 − p)ⁿ⁻^x, for x = 4, when p = .3 and n = 10

Question 13

Historical data show that the average number of patient arrivals at the intensive care unit of General Hospital is 3 patients every two hours. Assume that the patient arrivals are distributed according to a Poisson distribution. Determine the probability of 6 patients arriving in a five-hour period.

Accepted Answer

P(X = 6) == .1359

Question 14

The J.O. Supplies Company buys calculators from a non-U.S. supplier. The probability of a defective calculator is 10 percent. If 100 calculators are selected at random, what is the standard deviation of the number of defective calculators?

Accepted Answer

== 3

Question 15

The J.O. Supplies Company buys calculators from a non-U.S. supplier. The probability of a defective calculator is 10 percent. If 100 calculators are selected at random, what is the expected number of defectives?

Accepted Answer

E[X] = μ_x = (.10)(100) = 10

Question 16

A manufacturer tested a sample of semiconductor chips and found that 35 were defective and 190 were good. If additional tests are to be conducted with random samples of 160 semiconductor chips, find the mean for the number of defects in these groups of 160 (rounded to the nearest whole number).

Accepted Answer

Sample size = 35 + 190 = 225
p = 35/225 = .156
Mean of 160 groups = 160(.156) = 24.96, rounded to 25

Question 17

A study conducted by a local university found that 25 percent of college freshmen support increased spending on environmental issues. If 6 college freshmen are randomly selected, find the probability that only 1 supports increased spending on environmental issues.

Accepted Answer

P(X) = n!/[x!(n − x)!] × p^x(1 − p)^n−x, for x = 1 when p = .25 and n = 6.

Question 18

In the most recent election, 19 percent of all eligible college students voted. If a random sample of 20 students were surveyed, find the probability that none of the students voted.

Accepted Answer

P(X) = n!/[x!(n − x)!] × p^x(1 − p)^n−x, for x = 0 when p = .19 and n = 20.

Question 19

The J.O. Supplies Company buys calculators from a non-U.S. supplier. The probability of a defective calculator is 10 percent. If 3 calculators are selected at random, what is the probability that one of the calculators will be defective?

Accepted Answer

= .243

Question 20

According to a survey of adults, 64 percent have money in a bank savings account. If we were to survey 50 randomly selected adults, find the mean number of adults who would have bank savings accounts.

Accepted Answer

Mean = np = 50(.64) = 32

Quiz 6: Discrete Random Variables