Question

The average starting salary for this year's graduates of a large community college is $30,000 with a standard deviation of $8,000. 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 $30,400? b. Individuals with starting salaries of less than $15,600 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 303 of the recent graduates have salaries of at least $43,120, how many students graduated this year from this university?

Answer 1

0.048 ; 0.03593 ; (14320, 45680) ; 6000

Explanation:

Given a normal distribution :

Mean salary (m) = $30000

Standard deviation (s) = $8000

A.) probability of having a starting salary of atleast 30400

P(x ≥ 30400)

Obtain the standardized value (Z) :

Z = (x - m) / s

Z = (30400 - 30000) / 8000

Z = 400 / 8000 = 0.05

P(Z ≥ 0.05) = 0.48006 (Z probability calculator)

b. Individuals with starting salaries of less than $15,600 receive a low income tax break. What percentage of the graduates will receive the tax break?

P(X < 15600)

Obtain the standardized value (Z) :

Z = (x - m) / s

Z = (15600 - 30000) / 8000

Z = - 14400 / 8000 = - 1.8

P(Z < - 1.8) = 0.03593

c. what are the minimum and the maximum starting salaries of the middle 95% of the LU graduates?

Zscore which corresponds to 0.95 is 1.96

Minimum:

-1.96 = (x - 30000) / 8000

-15680 = x - 30000

x = 14320

Maximum :

1.96 = (x - 30000) / 8000

15680 = x - 30000

x = 45680

d. If 303 of the recent graduates have salaries of at least $43,120, how many students graduated this year from this university?

P(X ≥ 43120)

Z = (43120 - 30000) / 8000

Z = 13120 / 8000

Z = 1.64

P(Z ≥ 1.64) = 0.050503

Hence,

0.050503 = 303

1 = x

x = 303 / 0.05050

= 6000