Question

The average hourly wage of workers at a fast food restaurant is $6.50/hr. assume the wages are normally distributed with a standard deviation of $0.45. if a worker at this fast food restaurant is selected at random, what is the probability that the worker earns more than $6.75

Answer 1

Salaries are normally distributed

 Average salary per hour: $6.50 ⇒ μ=$6.50

 σ=$0.45 (Standard deviation)

 Probability that the selected worker will earn $6.75:

 P(x>6.75)

 Now you must standardize the normal random variables:

 If x~N (μσ²) => Z=(x-μ)/σ
 Z=(6.75-6.50)/0.45
 Z=0.556 => P(Z>0.556)= ? (To find this probability, you must use the tables for Standard normal distribution).

 Then:

 P(Z>0.556)=0.2877

 The probability of selecting a worker who earns more than $ 6.75 is 28.7%