Question

A company pays its employees an average wage of $17.90 an hour with a standard deviation of $1.50. If the wages are approximately normally distributed and paid to the nearest cent, the highest 2.5% of the employees hourly wages is greater than what amount

Answer 1

Answer:

$20.84

Step-by-step explanation:

To solve the above question, we would be using the z score formula

The formula for calculating a z-score :

z = (x - μ)/σ,

where x is the raw score

μ is the population mean

σ is the population standard deviation

x = unknown

μ = $17.90

σ = $1.50

We were not given z score in the above question but this can be determined.

We are told in the question to find the amount that the highest 2.5% of the employees hourly wages is greater than.

Hence, our confidence interval = 100 - 2.5 = 97.5%

The z score for 97.5% = 1.96

Below are inequalities equations with more explanation.

P(X ≥ x) = 2.5% = 0.025

P(X ≤ x) = 1 - 0.025 = 0.975

P (X - μ)/σ ≤ (x - μ)/σ) = 0.975

z ≤ (x - μ)/σ = 0.975

z ≤ 1.96 = 0.975

z = (x - μ)/σ,

1.96 = x - 17.90/1.5

Cross Multiply

1.96 × 1.5 = x - 17.90

x = (1.96 × 1.5) + 17.90

x = 2.94 + 17.90

x = 20.84

Therefore, the amount that the highest 2.5% of the employees hourly wages is greater than is $20.84