Question

A survey conducted for the Northwestern National Life Insurance Company revealed that 70% of American workers say job stress caused frequent health problems. One in three 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 10 American workers is selected.What is the probability that more than seven of them say job stress caused frequent health problems? What is the expected number of workers who say job stress caused frequent health problems? b. 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? What is the probability that none of the workers say they will burn out in the near future? c. Suppose a sample of seven workers is selected randomly. What is the probability that all seven say they are asked very often or somewhat often to work more than 40 hours a week? If this outcome actually happened, what might you conclude?

Answer 1

Probability that american workers say job stress caused frequent health problems = 0.70

Probability that the expected to burn out in the job in the near future = 0.33

Probability for quitting their job = 0.34

Probability that the required to work more than 40 hours a week very often or somewhat often = 0.53

a)

To determine the probability that more than seven of them say job stress caused frequent health problems & the expected number of workers who say job stress caused frequent health problems

Given,

n = 10

p = 70%

p = 0.70

Let us consider,

Using binomial distribution,

P(X) = nCx p^x ( 1 - p)^n-x

i.e., X ~ bin (10,0.70)

P(X > 7) = 0.3830

Now expected number of workers = np

= 10*0.70

= 7

E(X) = 7

b)

To determine the expected number & required probability

Given,

n = 15

p = 1 out of 3 = 0.33

Expected number E(X) = np

substitute given values

= 15*0.33

= 4.95

E(X) = 5

X bin(15,0.33)

Using binomial distribution,

P(X) = nCx p^x ( 1 - p)^n-x

Probability that none of the workers say they will burn out in the near future = P(X = 0)

P(X = 0) = 15C0 * 0.33^0 * (1-0.33)^15

= 1*1*0.67^15

= 0.0025

Required probability = 0.0025

c)

To determine the the probability that all seven say they are asked very often or somewhat often to work more than 40 hours a week

Given,

Sample n = 7

p = 0.53

Using binomial distribution,

P(X) = nCx p^x ( 1 - p)^n-x

X ~ bin(7,0.53)

P(X = 7) = 0.0117

Required probability = 0.0117