Answer to Question #320027 in Statistics and Probability for Mirylle Anne

Question

A company pay its employees an average wage of $15.90 an hour with a standard deviation of $1.50 , if the wages are approximately normally distributed and paid to the nearest cent,

a) what percentage of the workers receive wages between $13.75 and $16.22 an hour inclusive.

b) The highest 5% of the employee hourly wages is greater than what amount?

Answer 1

a)

P(13.75<x<16.22)=P(z1<x<z2)

mean =$15.90

standard deviation =$ 1.5

z=(x-mean)/standard deviation

z1=(13.75-15.90)/1.5

z1=-1.43

z2=(16.22-15.90)/1.5

z2=0.21

P(z1<x<z2)=P(z2=0.21)-P(z1=-1.43)

P(z1<x<z2)=0.5832-0.0764

=0.5068

Converting to percentage

0.5068 *100=50.68%

50.68 %of the workers receive wages between $13.75 and $16.22 an hour inclusive.

b)

From the z table the z value of 1.645 correspond to 95 %

Thus we have;

z=1.645

mean =15.90

standard deviation =1.5

x=?

z=(x-mean)/standard deviation

1.645=(x-15.90)/1.5

x=1.645*1.5+15.90

x=18.3675

The highest 5% of the employee hourly wages is greater than $18.3675