Question

Let us suppose that the first-year salaries of Hamilton college graduates follow a normal distribution with a mean of $50,000 and a standard deviation of $15,000.

a. Find the proportion of students who will earn less than $40,000?

b. What is the probability that a student will earn more than $80,000?

c. How many students will earn between 35,000 and 65,000?

d. What is the lowest salary for those in the top 15%? e. What is the maximum salary for those in the 1st quartile?

Answer #1

Answer)

As the data is normally distributed we can use standard normal z table to estimate the answers

Z = (x-mean)/s.d

Given mean = 50,000

S.d = 15000

A)

P(x<40,000)

Z = (40,000 - 50,000)/15000 = -0.67

From z table, P(z<-0.67) = 0.2514

B)

P(x>80,000)

Z = (80,000 - 50,000)/15000 = 2

From z table, P(z>2) = 0.0228

C)

P(35000<x<65000) = p(x<65000) - p(x<35000)

P(x<65000) = p(z<1) = 0.8413

P(x<35000) = p(z<-1) = 0.1587

Required proportion is 0.8413 - 0.1587 = 0.6826

D)

From z table, P(z>1.3) = 15%

1.3 = (x - 50,000)/15000

X = 69500

E)

We know that 25% lies below first quartile

From z table, P(z<-0.67) = 25%

-0.67 = (x - 50,000)/(15,000)

X = 39,950

SOLVE WITHOUT EXCEL
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