Question

The salaries of pharmacy techs are normally distributed with a mean of $32,000 and a standard deviation of $3,000. If a pharmacy tech is selected at random, what is the probability they make more than $39,000?

Answer 1

Answer:

The probability is  $P( X > 39000) =0.01$

Step-by-step explanation:

From the question we are told that

   The mean is  $\mu = \ 32000$

    The standard deviation is $\sigma = \ 3000$

Generally the probability that they make more than $39,000 is mathematically represented as

         $P( X > 39000) = P( \frac{X - \mu }{\sigma } > \frac{39000 - 32000}{3000} )$

 $\frac{X -\mu}{\sigma } = Z (The \ standardized \ value\ of \ X )$

          $P( X > 39000) = P(Z > 2.33 )$

From the z table  the area under the normal curve to the right corresponding to  2.33  is      

         $P(Z > 2.33 ) = 0.01$

So

         $P( X > 39000) =0.01$