Question

According to a survey, 1.22% of factory employees in a country suffered from job related stress.

Supposed 100 employees are selected at random. We are interested to find the number of employees amongst these 100 that suffers from job-related stress.

b.(i) State the assumptions needed so that the situation can be modelled as a binomial experiment.

(ii) Hence, find the probability that at most 2 out of these 100 employees suffers from job-related stress.

(iii) Calculate the mean and standard deviation for the number of employees amongst these 100 that suffers from job-related stress.   

Suppose we are interested to use Poisson approximation for the problem in part (b).

c.(i) State the assumptions needed so that binomial distribution can be approximated by the Poisson distribution.

(ii) Assume that number of employees that suffers from job-related stress has a Poisson distribution with a mean which is equal to the mean calculated in part (b)(iii). Hence, find the probability that at most 2 of these employees suffers from job-related stress.

(iii) Compare your answer in part (c)(ii) to your answer in part (b)(ii). Is the Poisson distribution a good approximation in this case? Explain. (Note that an error of 0.005 or less would indicate a good approximation.)

(d) Suppose we are interested to use normal approximation for the problem in part (b).

(i) State the assumptions needed so that binomial distribution can be approximated by the normal distribution.

(ii) Assume that number of employees that suffers from job-related stress has a normal distribution with mean and standard deviation which are equal to the mean and standard deviation calculated in part (b)(iii). Hence, find the probability that at most 2 of these employees suffers from job-related stress.

(iii) Compare your answer in part (d)(ii) to your answer in part (b)(ii). Is the normal distribution a good approximation in this case? Explain. (Note that an error of 0.005 or less would indicate a good approximation.)

Answer 1

n=100 p=0.0122

(b)(i)

• Each trial (i.e. individual) has one of the two outcomes-success(i.e., suffering from job related stress) or failure(i.e., not suffering from job related stress).

• Trials are evidently independent.

• Success probability is same for each trial.

(ii)

X denotes number of employees who are suffering from job related stress. X follow Bin(n,p)

P(X<=2)= 0.8761799

(iii)

mean=100*.0122=1.22

standard deviation=(100*.0122*(1-.0122))^.5= 1.097778

(c)(i)

• n should be large
• p should be quite small
• n*p should be finite

(ii)

Y denotes number of employees who are suffering from job related stress. Y follow Poisson(1.22)

P(Y<=2)= 0.8751213

(iii)

Difference in probabilities=|0.8761799-0.8751213|= 0.0010586<0.005

Hence, it is, indeed, a good approximation.

(d)(i)

• n should be quite large
• p should be close to 0.5
• n*p>5

(ii)

Z denotes number of employees who are suffering from job related stress. Y follow Normal(1.22,1.097778)

P(Z<=2)=  0.761311

(iii)

Difference in probility=|0.761311-0.8751213|=0.1138103>>0.005

Hence it is not at all a good approximation.