Question 1

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

The answer of A study conducted by a local university...

Question 2

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

The probability can be calculated using the binomial probability formula: P(X=k) = (n choose k) * (p^k) * ((1-p)^(n-k)), where n=10, k=4, and p=0.3. This results in approximately 0.2001.

Question 3

The J.O.Supplies Company buys calculators from a non-US 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

The probability of getting 3 or more defective calculators is 1 - the probability of getting 0, 1, or 2 defective calculators. The probability of getting 0 defective calculators is (0.9)^10 = 0.3487. The probability of getting 1 defective calculator is 10 * (0.1) * (0.9)^9 = 0.2439. The probability of getting 2 defective calculators is 45 * (0.1)^2 * (0.9)^8 = 0.1350. Therefore, the probability of getting 3 or more defective calculators is 1 - 0.3487 - 0.2439 - 0.1350 = 0.2724.

Question 4

The J.O.Supplies Company buys calculators from a non-US 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

The expected number of defectives can be found by multiplying the total number of calculators (100) by the probability of a defective calculator (10%). 
Expected number of defectives = 100 x 0.10 = 10. Therefore, the best choice is C.

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

We know that the probability of errors on an individual tax return is 0.37. Let X be the number of erroneous returns in a batch of 12 returns. Then X follows a binomial distribution with n=12 and p=0.37.

The mean of a binomial distribution is μ = np = 12 x 0.37 = 4.44

The standard deviation of a binomial distribution is σ = sqrt(np(1-p)) = sqrt(12 x 0.37 x (1-0.37)) = 1.67

Therefore, the best choice is B with μ = 4.44 and σ = 1.67.

Question 6

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 7

The J.O.Supplies Company buys calculators from a non-US 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

We can solve this problem by finding the probability of selecting one defective calculator and two good calculators, and then multiplying by the number of ways to arrange these selections. 
The probability of selecting one defective calculator and two good calculators is: 
(0.1)(0.9)(0.9) + (0.9)(0.1)(0.9) + (0.9)(0.9)(0.1) = 0.243 
There are three ways to arrange these selections (defective calculator in position 1, 2, or 3), so we multiply by 3: 
0.243 * 3 = 0.729 
Therefore, the probability that one of the calculators will be defective is 0.729 or approximately 0.73. None of the other answer choices match this value, so the best choice is C.

Question 8

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

This is a binomial distribution problem, where the number of trials (selecting 6 freshmen) is known, the probability of success (supporting increased spending on environmental issues) is given, and we need to find the probability of a specific number of successes (exactly 3). The formula for the probability of exactly x successes in n trials is:

P(x) = (nCx) * p^x * q^(n-x)

where nCx is the number of ways to choose x items out of n, p is the probability of success, q is the probability of failure (1-p), and ^(n-x) means to raise q to the power of (n-x).

Using this formula with n=6, p=0.25, and x=3:

P(3) = (6C3) * (0.25)^3 * (0.75)^3
= (20) * 0.015625 * 0.421875
= 0.1318359375

Rounding to four decimal places, we get 0.1318, which corresponds to choice B.

Question 9

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

The problem follows a binomial distribution with n = 6 trials and probability of success p = 0.25. We need to find the probability that fewer than 4 support increased spending, which means calculating P(X < 4) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3). Using the binomial probability formula, this sum equals approximately 0.9624.

Question 10

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 11

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

The answer of A multiple-choice test has 30 questions and...

Quiz 5: Discrete Random Variables