Question

2. The average starting salary for this year's graduates at a large university (LU) is $50,000 with a standard deviation of $10,000. Furthermore, it is known that the starting salaries are normally distributed.

a. What is the probability that a randomly selected LU graduate will have a starting salary of at least $45,000?

b. Individuals with starting salaries of less than $36,900 receive a low income tax break. What percentage of the graduates will receive the tax break? c. What are the minimum and the maximum starting salaries of the middle 95% of the LU graduates?

d. If 300 of the recent graduates have salaries of at least $65,000, how many students graduated this year from this university?

Answer 1

a)

for normal distribution z score =(X-μ)/σx
here mean=       μ=	50000
std deviation   =σ=	10000.0000

  probability that a randomly selected LU graduate will have a starting salary of at least $45,000:

probability =

P(X>45000)

=

P(Z>-0.5)=

1-P(Z<-0.5)=

1-0.3085=

0.6915

b)

percentage of the graduates will receive the tax break:

probability =

P(X<36900)

=

P(Z<-1.31)=

0.0951~9.51%

c)

for middle 95% values critical z=1.96

hence  minimum starting salaries =mean-z*std deviaiton =30400

and maximum starting salaries =mean+z*std deviaiton =69600

d)

probability of having salary at least 65000:

probability =

P(X>65000)

=

P(Z>1.5)=

1-P(Z<1.5)=

1-0.9332=

0.0668

as 0.0668 proportion corresponds to 300 graduates

therefore number of students graduated=300/0.0668 =4491