Question

The Economic Policy Institute periodically issues reports on wages of entry-level workers. The institute reported that entry-level wages for male college graduates were $21.68 per hour and for female college graduates were $18.80 per hour in 2011 (Economic Policy Institute website, March 30, 2012). Assume the standard deviation for male graduates is $2.30, and for female graduates it is $2.05.

What is the probability that a sample of 50 male graduates will provide a sample mean within $.50 of the population mean, $21.68? (Round to four decimal places) Answer

What is the probability that a sample of 50 female graduates will provide a sample mean within $.50 of the population mean, $18.80? (Round to four decimal places) Answer

What is the probability that a sample of 120 female graduates will provide a sample mean more than $.30 below the population mean? (Round to four decimal places)

Answer 1

Solution:-

a) The probability that a sample of 50 male graduates will provide a sample mean within $0.50 of the population mean, $21.68 is 0.8758.

x₁ = 21.18

x₂ = 22.18

By applying normal distribution:-

$z = \frac{x-\mu }{\frac{\sigma }{\sqrt{n}}}$

z₁ = -1.537

z₂ = 1.537

P( -1.537 < z < 1.537) = P(z > -1.537) - P(z > 1.537)

P( -1.537 < z < 1.537) = 0.9379 - 0.0621

P( -1.537 < z < 1.537) = 0.8758

b) The probability that a sample of 50 female graduates will provide a sample mean within $0.50 of the population mean, $18.80 is 0.9154.

x₁ = 18.30

x₂ = 19.30

By applying normal distribution:-

$z = \frac{x-\mu }{\frac{\sigma }{\sqrt{n}}}$

z₁ = -1.725

z₂ = 1.725

P( -1.725 < z < 1.725) = P(z > -1.725) - P(z > 1.725)

P( -1.725 < z < 1.725) = 0.9577 - 0.0423

P( -1.725 < z < 1.725) = 0.9154

c) The probability that a sample of 120 female graduates will provide a sample mean more than $0.30 is 0.9455.

x = 18.50

By applying normal distribution:-

$z = \frac{x-\mu }{\frac{\sigma }{\sqrt{n}}}$

z = -1.603

P(z > -1.603) = 0.9455