The distribution of annual profit at a chain of stores was approximately normal with mean μ = $66,000, standard deviation σ = $2
1 answer:
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.
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.
Upon rounding to nearest thousand dollars, we will get:
Therefore, the closest minimum annual profit, for a store that had a reward party, would be $102,000.
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The first year, the amount is 40,000
the second year is 40000 + 4.2% of 40000, or 0.042 * 4000, so 40000+(0.042*4000)
common factoring that we get 40000(1 + 0.042), or just 40000(1.042)
in short, the starting amount is 40000, and to get the next term's value you'd use the "common ratio" of 1.042, namely the multiplier of 1.042.
for the third year it'll be 40000(1.042) + (0.042 *40000(1.042) ), again, common factoring that
40000(1.042)(1 + 0.042) or 40000(1.042)(1.042) or 40000(1.042)²
therefore,
the second year is 40000 + 4.2% of 40000, or 0.042 * 4000, so 40000+(0.042*4000)
common factoring that we get 40000(1 + 0.042), or just 40000(1.042)
in short, the starting amount is 40000, and to get the next term's value you'd use the "common ratio" of 1.042, namely the multiplier of 1.042.
for the third year it'll be 40000(1.042) + (0.042 *40000(1.042) ), again, common factoring that
40000(1.042)(1 + 0.042) or 40000(1.042)(1.042) or 40000(1.042)²
therefore,