Question

A survey conducted for the Northwestern National Life Insurance Company revealed that 70% of American workers say job stress caused frequent health problems. 33 % said they expected to burn out in the job in the near future. Thirty-four percent said they thought seriously about quitting their job last year because of work-place stress. Fifty-three percent said they were required to work more than 40 hours a week very often or somewhat often.
a. Suppose a random sample of 15 American workers is selected. What is the expected number of these sampled workers who say they will burn out in the near future?

b. What is the probability that none of the workers say they will burn out in the near future?

c. What is the probability that at least two of the workers say they will burn out in the near future?

Answer 1

Answer:

a) The expected number is 5.

b) 0.25% probability that none of the workers say they will burn out in the near future

c) 97.93% probability that at least two of the workers say they will burn out in the near future

Step-by-step explanation:

For each person, there are only two possible outcomes. Either they said they will burnout in the near future, or they did not. The probability of a person saying that they will burnout in the near future is independent of any other person. So the binomial probability distribution is used to solve this question.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

33 % said they expected to burn out in the job in the near future.

This means that $p = 0.33$

15 workers

This means that $n = 15$

a. Suppose a random sample of 15 American workers is selected. What is the expected number of these sampled workers who say they will burn out in the near future?

The expected number of the binomial distribution is given by the following formula:

$E(X) = np$

So

$E(X) = 15*0.33 = 5$

The expected number is 5.

b. What is the probability that none of the workers say they will burn out in the near future?

This is P(X = 0).

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{15,0}.(0.33)^{0}.(0.67)^{15} = 0.0025$

0.25% probability that none of the workers say they will burn out in the near future

c. What is the probability that at least two of the workers say they will burn out in the near future?

Either less than two say this, or at least two do. The sum of the probabilities of these events is decimal 1. So

$P(X < 2) + P(X \geq 2) = 1$

We want $P(X \geq 2)$. So

$P(X \geq 2) = 1 - P(X < 2)$

In which

$P(X < 2) = P(X = 0) + P(X = 1)$

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{15,0}.(0.33)^{0}.(0.67)^{15} = 0.0025$

$P(X = 1) = C_{15,1}.(0.33)^{1}.(0.67)^{14} = 0.0182$

$P(X < 2) = P(X = 0) + P(X = 1) = 0.0025 + 0.0182 = 0.0207$

$P(X \geq 2) = 1 - P(X < 2) = 1 - 0.0207 = 0.9793$

97.93% probability that at least two of the workers say they will burn out in the near future