Suppose a survey reveals that 69% of Canadian workers say job stress causes frequent health problems. One in four said they expected to burn out on the job in the near future. Thirty-two percent said they thought seriously about quitting their job last year because of workplace stress. Forty-nine 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 Canadian 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 b. Suppose a random sample of 15 Canadian 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 used very often or somewhat often to work more than 40 hours a week? check_circle Expert Answer thumb_up thumb_down Step 1 Binomial distribution: The binomial distribution gives the probability of number of successes out of n trials in a series of Bernoulli trials. The probability mass function (pmf) of a binomial random variable X is given as: P(X=x)=nCx×px×qn-x Step 2 a. It is given that 69% of Canadian workers say job stress causes frequent health problems. A random sample of 10 Canadian workers is selected. Thus, n=10, p=0.69, q=0.31 Let the random variable X denote the number of Canadian workers who say job stress causes frequent health problems. X~ Binomial (10,0.69). The probability that more than 7 of them say job stress causes frequent health problems is calculated as follows: P(X>7)=P(X=8)+P(X=9)+P(X=10) =10C8×0.698×0.312+10C9×0.699×0.311+10C10×0.6910×0.310 =0.2222+0.1099+0.0245 =0.3566. The probability that more than 7 of them say job stress causes frequent health problems is 0.3566. The expected value is given and calculated as follows: µ=np = 10(0.69) = 6.9. Thus, the expected value is 6.9. Step 3 b. It is given that one in four(1/4=0.25) said they expected to burn out on the job in the near future. A random sample of 15 Canadian workers is selected. Thus, n=15, p=0.25, q=0.75. Let the random variable X denote the number of Canadian workers who said they expected to burn out on the job in the near future. X~ Binomial (15,0.25). The expected value is given and calculated as follows: µ=np = 15(0.25) = 3.75. Thus, the expected value is 3.75. The probability that none of them say they burn out on the job in the near future is calculated as follows: P(X=0)=15C0×0.250×0.7515 =0.0134 The probability that none of them say they burn out on the job in the near future is 0.0134. Step 4 c. It is given that 49% said they were required to work more than 40 hours a week very often or somewhat often. A random sample of 7 Canadian workers is selected. Thus, n=7, p=0.49, q=0.51. Let the random variable X denote the number of Canadian workers who said they were required to work more than 40 hours a week very often or somewhat often. X~ Binomial (7,0.49). The probability that all seven say they are used very often or somewhat often to work more than 40 hours a week is calculated as follows: P(X=7) =7C7×0.497×0.510 =0.0068 The probability that all seven say they are used very often or somewhat often to work more than 40 hours a week is 0.0068. Are you able to tell me how to do this is exel?

Suppose a survey reveals that 69% of Canadian workers say job stress causes frequent health problems. One in four said they expected to burn out on the job in the near future. Thirty-two percent said they thought seriously about quitting their job last year because of workplace stress. Forty-nine 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 Canadian 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 b. Suppose a random sample of 15 Canadian 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 used very often or somewhat often to work more than 40 hours a week? check_circle Expert Answer thumb_up thumb_down Step 1 Binomial distribution: The binomial distribution gives the probability of number of successes out of n trials in a series of Bernoulli trials. The probability mass function (pmf) of a binomial random variable X is given as: P(X=x)=nCx×px×qn-x Step 2 a. It is given that 69% of Canadian workers say job stress causes frequent health problems. A random sample of 10 Canadian workers is selected. Thus, n=10, p=0.69, q=0.31 Let the random variable X denote the number of Canadian workers who say job stress causes frequent health problems. X~ Binomial (10,0.69). The probability that more than 7 of them say job stress causes frequent health problems is calculated as follows: P(X>7)=P(X=8)+P(X=9)+P(X=10) =10C8×0.698×0.312+10C9×0.699×0.311+10C10×0.6910×0.310 =0.2222+0.1099+0.0245 =0.3566. The probability that more than 7 of them say job stress causes frequent health problems is 0.3566. The expected value is given and calculated as follows: µ=np = 10(0.69) = 6.9. Thus, the expected value is 6.9. Step 3 b. It is given that one in four(1/4=0.25) said they expected to burn out on the job in the near future. A random sample of 15 Canadian workers is selected. Thus, n=15, p=0.25, q=0.75. Let the random variable X denote the number of Canadian workers who said they expected to burn out on the job in the near future. X~ Binomial (15,0.25). The expected value is given and calculated as follows: µ=np = 15(0.25) = 3.75. Thus, the expected value is 3.75. The probability that none of them say they burn out on the job in the near future is calculated as follows: P(X=0)=15C0×0.250×0.7515 =0.0134 The probability that none of them say they burn out on the job in the near future is 0.0134. Step 4 c. It is given that 49% said they were required to work more than 40 hours a week very often or somewhat often. A random sample of 7 Canadian workers is selected. Thus, n=7, p=0.49, q=0.51. Let the random variable X denote the number of Canadian workers who said they were required to work more than 40 hours a week very often or somewhat often. X~ Binomial (7,0.49). The probability that all seven say they are used very often or somewhat often to work more than 40 hours a week is calculated as follows: P(X=7) =7C7×0.497×0.510 =0.0068 The probability that all seven say they are used very often or somewhat often to work more than 40 hours a week is 0.0068. Are you able to tell me how to do this is exel?

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Expert Answer

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Step 1

Binomial distribution:

The binomial distribution gives the probability of number of successes out of n trials in a series of Bernoulli trials.

The probability mass function (pmf) of a binomial random variable X is given as:

P(X=x)=nCx×p^x×q^n-x

Step 2

It is given that 69% of Canadian workers say job stress causes frequent health problems.

A random sample of 10 Canadian workers is selected.

Thus, n=10, p=0.69, q=0.31

Let the random variable X denote the number of Canadian workers who say job stress causes frequent health problems.

X~ Binomial (10,0.69).

The probability that more than 7 of them say job stress causes frequent health problems is calculated as follows:

P(X>7)=P(X=8)+P(X=9)+P(X=10)

=10C8×0.69⁸×0.31²+10C9×0.69⁹×0.31¹+10C10×0.69¹⁰×0.31⁰

=0.2222+0.1099+0.0245

=0.3566.

The probability that more than 7 of them say job stress causes frequent health problems is 0.3566.

The expected value is given and calculated as follows:

µ=np = 10(0.69) = 6.9.

Thus, the expected value is 6.9.

Step 3

It is given that one in four(1/4=0.25) said they expected to burn out on the job in the near future.

A random sample of 15 Canadian workers is selected.

Thus, n=15, p=0.25, q=0.75.

Let the random variable X denote the number of Canadian workers who said they expected to burn out on the job in the near future.

X~ Binomial (15,0.25).

The expected value is given and calculated as follows:

µ=np = 15(0.25) = 3.75.

Thus, the expected value is 3.75.

The probability that none of them say they burn out on the job in the near future is calculated as follows:

P(X=0)=15C0×0.25⁰×0.75¹⁵

=0.0134

The probability that none of them say they burn out on the job in the near future is 0.0134.

Step 4

It is given that 49% said they were required to work more than 40 hours a week very often or somewhat often.

A random sample of 7 Canadian workers is selected.

Thus, n=7, p=0.49, q=0.51.

Let the random variable X denote the number of Canadian workers who said they were required to work more than 40 hours a week very often or somewhat often.

X~ Binomial (7,0.49).

The probability that all seven say they are used very often or somewhat often to work more than 40 hours a week is calculated as follows:

P(X=7) =7C7×0.49⁷×0.51⁰

=0.0068

The probability that all seven say they are used very often or somewhat often to work more than 40 hours a week is 0.0068.

Are you able to tell me how to do this is exel?

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