The requested probability for a group of35tax preparation customers to be within $16 of the population mean, $273, was determined to beP(257x289).

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

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,zwill be found using the formula below which uses the sample meanx

and standard error_x.

z=x

_x

=x

n

The mean is= 273

and the standard deviation was given to be= 100.

The size of the group is35customers,son=

.

Find thezvalue corresponding tox= 257,

rounding the result to two decimal places.

z=x

n

=257273

100

=

Find thezvalue corresponding tox= 289,

rounding the result to two decimal places.

z=x

n

=289273

100

=

n 257 - 273 100 35 2.7 X Find the z value corresponding to x = 289, rounding the result to two decimal places. X - H n 289 - 273 100 35 270.27 X