Question

The average starting salary for this year's graduates at a large university (LU) is $50,000 with a standard deviation of $8,000. Furthermore, it is known that the starting salaries are normally distributed. What is the probability that a randomly selected LU graduate will have a starting salary of at least $55,200?

Answer 1

0.74215

Explanation:

The Z score formula = z = (x-μ)/σ, where

x is the raw score,

μ is the population mean,

and σ is the population standard deviation.

In the question, we are given

x is the raw score = Salary of a randomly selected LU graduate will be a starting salary of at least= 55,200

μ is the population mean = average starting salary for this year's graduates at a large university (LU) = 50,000

σ is the population standard deviation = 8,000

z = (x-μ)/σ,

z = (55,200 -50,000) ÷ 8000

z = 5,200 ÷ 8000

z = 0.65

Therefore,

P(z < 0.65)

Using the z score table for probability,

z < 0.65 = 0.74215

Therefore, the probability that a randomly selected LU graduate will have a starting salary of at least $55,200 in decimal form is 0.74215