The mean hourly wage for males is given to be $37.39 with a standard deviation of $4.60. Both values are representative of the population, but we are to determine the probability that the sample mean for90male graduates is between $36.39 and $38.39. That is,

P(36.39x38.39).

The normal distribution will be used to find this probability, so the probability statement must be converted to the standard normal random variable,z. Recall the formula forz, given below.

z=

x

The given mean will be used for, but the standard deviation will need to be modified since we are dealing with a sample mean, not a particular value ofx. Thus, the standard deviation will be modified using the formula for the standard error below.

_x=

n

Thus,zwill be found using the formula below which uses the sample mean and standard error.

z=

x

_x

=

x

n

Find thezvalue corresponding to

x= 36.39,

rounding the result to two decimal places.

z

=

x

n

=

36.3937.39

90

=

Find thezvalue corresponding to

x= 38.39,

rounding the result to two decimal places.

z

=

x

n

=

38.3937.39

=