Question

A survey of the alumni of Mo University revealed that the starting annual salaries of Mo Graduates is normally distributed with a mean of AUD 60,000 and a standard deviation of AUD 15,000.

(a) Find the probability that a randomly selected Mo graduate earns less than AUD 45,000 in the first year of employment.

(b) Find the probability that a randomly selected Mo graduate earns more than AUD 80,000 in the first year of employment.

(c) Find the range for the top 15% all of earners amongst Monash graduates. [2 marks]

Answer 1

Solution :

(a)

P(x < 45000) = P[(x - ) / < (45000 - 60000) / 15000]

= P(z < -1)

= 0.1587

Probability = 0.1587

(b)

P(x > 80000) = 1 - P(x < 80000)

= 1 - P[(x - ) / < (80000 - 60000) / 15000)

= 1 - P(z < 1.33)

= 1 - 0.09082

= 0.0918

Probability = 0.9018

(c)

Using standard normal table,

P(Z > z) = 15%

1 - P(Z < z) = 0.15

P(Z < z) = 1 - 0.15 = 0.85

P(Z < 1.04) = 0.85

z = 1.04

Using z-score formula,

x = z * +

x = 1.04 * 15000 + 60000 = 75600

Range = 75600