Question

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?

Answer 1

Answer: $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 $x$ 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. $P(x>x_0)=0.05$

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

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