Question

The distribution of annual profit at a chain of stores was approximately normal with mean μ = $66,000, standard deviation σ = $22,000. The stores with profits in the top 5 percent each had a reward party for the employees to celebrate.
What is closest to the minimum annual profit for a store that had a reward party? Round to the nearest thousand dollars.

Answer 1

Answer:

$102,000.

Step-by-step explanation:

We have been given that the distribution of annual profit at a chain of stores was approximately normal with mean μ = $66,000, standard deviation σ = $22,000. The stores with profits in the top 5 percent each had a reward party for the employees to celebrate.

We will use z-score formula and normal distribution table to solve our given problem.

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

$z=\frac{x-66,000}{22,000}$

Top 5% of data would be equal to 95% and more.

Let us find z-score corresponding to 95% or 0.95 using normal distribution table.

$1.65=\frac{x-66,000}{22,000}$

$1.65*22,000=\frac{x-66,000}{22,000}*22,000$

$36,300=x-66,000$

$36300+66000=x-66,000+66,000$

$x=102,300$

Upon rounding to nearest thousand dollars, we will get:

$x\approx 102,000$

Therefore, the closest minimum annual profit, for a store that had a reward party, would be $102,000.

Answer 2

Answer:

$102,000

Step-by-step explanation: