Question

4. According to the Bureau of Labor Statistics, the average weekly pay for a U.S. production worker was $441.84. Assume that available data indicate that production worker wages were normally distributed with a standard deviation of $90. What is the probability that a worker earned between $400 and $500? (Round to four decimal places) How much did a production worker have to earn to be in the top 20% of wage earners? (Round to two decimal places) For a randomly selected production worker, what is the probability the worker earned less than $250 per week? (Round to four decimal places)

Answer 1

Solution:-

Mean = 441.84, S.D = 90

a) The probability that a worker earned between $400 and $500 is 0.4186.

x₁ = 400

x₂ = 500

By applying normal distribution:-

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

z₁ = - 0.465

z₂ = 0.6462

P( - 0.465 < z < 0.642) = P(z > -0.465) - P(z > 0.642)

P( - 0.465 < z < 0.642) = 0.679 - 0.2604

P( - 0.465 < z < 0.642) = 0.4186

b) The production worker have to earn to be in the top 20% of wage earners is 517.53.

p-value for the top 20% = 1- 0.20 = 0.80

z-score for the p-value = 0.841

By applying normal distribution:-

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

x = 517.53

c) The probability the worker earned less than $250 per week is 0.0165.

x = 250

By applying normal distribution:-

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

z = - 2.132

P(z < -2.132) = 0.0165