Question

Customers at a certain pharmacy spend an average of $54.40, with a standard deviation of $13.50.What are the largest and smallest amounts spent by the middle 50% of these customers?

Answer 1

Answer:

Largest is  $x_2 = \ 63.506$

Smallest is  $x_1 = \ 45.29$

Step-by-step explanation:

From the question we are told that

  The  mean is  $\mu = \ 54.40$

   The  standard deviation is  $\sigma = \ 13.50$

 

The largest and the smallest amount spent by 50% of the costumer is mathematically represented as

        $P(x_1 < X < x_2) = P(\frac{ x_1 -\mu}{\sigma } < \frac{X - \mu }{ \sigma } < \frac{ x_2 -\mu}{\sigma } ) = 0.50$

=>    $P(x_1 < X < x_2 ) = P(\frac{x_1 - 54.40}{ 13.50} < X < \frac{x_2 - 54.40}{ 13.50} ) = 0.50$

Given that this is normally distributed we have that

      $P(X <- \frac{x_1 - 54.40}{ 13.50} ) = 0.25$

and

    $P(X < \frac{x_2 - 54.40}{ 13.50} ) = 0.25$

From the normal distribution curve that z-score having a probability of 0.25 is  

0.6745

So

    $\frac{x_1 - 54.40}{ 13.50} = -0.6745$

=>  $x_1 = \ 45.29$

and  

   $\frac{x_2 - 54.40}{ 13.50} = 0.6745$

=>  $x_2 = \ 63.506$