Mathematics: The annual profit at a chain of stores is normally distributed with a mean of $61,000 and a standard deviation of $23,000. The stores in the top 5% of annual profit were rewarded with a celebration. What was the annual profit required to have a party?

60 minutes in an hour. 1/4 × 60 = 153 quarters of an hour is3 × 15 = 45 minutes...

The annual profit at a chain of stores is normally

Ответ:

josemartinez2025

05.02.2022

Mathematics

$98,835.

Step-by-step explanation:

Given: Annual profit at a chain of stores is normally distributed with a mean ( $\mu$ ) of $61,000 and a standard deviation( $\sigma$ ) of $23,000.

Let denoted the profit.

Let x_0 be the minimum profit required to have a party.

The stores in the top 5% (i.e. 0.05) of annual profit were rewarded with a celebration.

i.e. P(xx_0)=0.05

$\Rightarrow\ P(\dfrac{x-\mu}{\sigma}\dfrac{x_0-61000}{23000})=0.05\\\\\Rightarrow\ P(z\dfrac{x_0-61000}{23000})=0.05\ \ \ [z=\dfrac{x-\mu}{\sigma}]\\\\\Rightarrow\ \dfrac{x_0-61000}{23000}=1.645\ \ [\text{critical z-value for p-value 0.05=1.645}]\\\\\Rightarrow\ x_0-61000 =1.645\times 23000\\\\\Rightarrow\ x_0-61000 =37835\\\\\Rightarrow\ x_0=37835+61000\\\\\Rightarrow\ x_0=98835$

Hence, the annual profit required to have a party was $98,835.

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The annual profit at a chain of stores is normally distributed with a mean of $61,000 and a standard deviation of $23,000. The stores in the top 5% of annual profit were rewarded with a celebration. What
was the annual profit required to have a party?

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