Question

The distribution of annual profit at a chain of stores was approximately normal with mean \mu = \$66{,}000μ=$66,000mu, equals, dollar sign, 66, comma, 000 and standard deviation \sigma = \$21{,}000σ=$21,000sigma, equals, dollar sign, 21, comma, 000. The executives conducted an audit of the stores with the lowest 20\%20%20, percent of profits. What is closest to the maximum annual profit at a store where the executives conducted an audit?

Answer 1

Answer:

The closest to the maximum profit is  $x = \ 83682$

Step-by-step explanation:

From the question we are told that

  The  mean is  $\mu = \ 66,000$

   The standard deviation is  $\sigma = \ 21000$

   The percentage of profit is  20%

Generally the closest to the maximum annual profit at a store where the executives conducted an audit is mathematically evaluated  as follows

     $P(X > x ) = 0.20$

=>  $P(X > x ) = P(\frac{X -x}{\sigma } > \frac{x -66000}{21000} ) = 0.20$

From the z-table  the z-score for  0.20  is  

    $z-score = 0.842$

So

     $\frac{x -66000}{21000} = 0.842$

=>   $x = \ 83682$