Question

Agnes Hammer is a senior majoring in management science. She has interviewed with several companies for a job when she graduates, and she is curious about what starting salary offers she might receive. She asked 12 of her classmates at random what their annual staring salary offers were, and she received the following responses

$28500	$35500
$32600	$36000
$34000	$25700
$27500	$29000
$24600	$31500
$34500	$26800

1. Compute the sample mean and sample variance for these data. (Please write down the calculation process)
   

2. Suppose the starting salaries are normally distributed, the mean is the same as the sample mean and variance is the same as sample variance that you calculate from the previous question. What is the probability that Agnes will receive a salary offer of less than $27000? (For probability of normal distribution, please use the probability table posted on Blackboard)
   

3. Suppose the starting salaries are normally distributed, the mean is the same as the sample mean and variance is the same as sample variance that you calculate from the previous question. What is the probability that Agnes will receive a salary offer of between $27000 and $40000 (meaning 27000<= salary <= 40000)? (For probability of normal distribution, please use the probability table posted on Blackboard)

Answer 1

P(27000<X<40000) = P(X<40000) – P(X<27000)

First find P(X<40000)

Z = (X – mean) / SD

We have

Mean = 30517

SD = 3997

Z = (40000 - 30517)/3997

Z = 2.372529397

P(Z<2.372529397) = P(X<40000) = 0.991166621

(by using z-table or excel)

Now, find P(X<27000)

Z = (X – mean) / SD

We have

Mean = 30517

SD = 3997

Z = (27000 - 30517)/3997

Z =-0.87990993

P(Z<-0.87990993) = P(X<27000) = 0.189454052

(by using z-table)

P(27000<X<40000) = P(X<40000) – P(X<27000)

P(27000<X<40000) = 0.991166621 - 0.189454052

P(27000<X<40000) = 0.801712569

Required probability = 0.801712569