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Anna007 [38]
3 years ago
7

The distribution of annual profit at a chain of stores was approximately normal with mean μ = $66,000, standard deviation σ = $2

2,000. 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.
Mathematics
1 answer:
alexandr1967 [171]3 years ago
7 0

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.

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