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

         $P( X > 39000) =0.01$

     

Answer 2

The probability they make more than $39,000 is 1%

Z score is used to determine by how many standard deviations the raw score is above or below the mean. It is given by:

$z=\frac{x-\mu}{\sigma} \\\\where\ x=raw\ score,\mu=mean,\sigma=standard\ deviation\\\\Given \ that:\\\\\mu=32000,\sigma=3000.For\ x>39000:\\\\z=\frac{39000-32000}{3000}=2.33$

From the normal distribution table:

P(x > 39000) = P(z > 2.33) = 1 - P(z < 2.33) = 1 - 0.9901 = 0.0099 = 1%

Therefore the probability they make more than $39,000 is 1%

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