Respuesta :
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.
[tex]z=\frac{x-\mu}{\sigma}[/tex]
[tex]z=\frac{x-66,000}{22,000}[/tex]
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.
[tex]1.65=\frac{x-66,000}{22,000}[/tex]
[tex]1.65*22,000=\frac{x-66,000}{22,000}*22,000[/tex]
[tex]36,300=x-66,000[/tex]
[tex]36300+66000=x-66,000+66,000[/tex]
[tex]x=102,300[/tex]
Upon rounding to nearest thousand dollars, we will get:
[tex]x\approx 102,000[/tex]
Therefore, the closest minimum annual profit, for a store that had a reward party, would be $102,000.
