Question

The overhead reach distances of adult females are normally distributed with a mean of 205 cm205 cm and a standard deviation of 8.6 cm8.6 cm.
a. find the probability that an individual distance is greater than 215.00215.00 cm.
b. find the probability that the mean for 1515 randomly selected distances is greater than 202.80 cm.202.80 cm.
c. why can the normal distribution be used in part​ (b), even though the sample size does not exceed​ 30?

Answer 1

A) the probability is 0.123 or 12.3%.
B) the probability is 0.8589 or 85.89%.
C) the normal distribution can be used even though the sample is small because the population the sample is taken from is normally distributed.

Explanation
A) We first find the z-score that corresponds with this value.
$z=\frac{X-\mu}{\sigma} \\ \\=\frac{215-205}{8.6}= 1.1628$

Using a z-table (http://www.z-table.com) we find 1.16; this corresponds to a probability of 0.8770.  However, This is the probability that a number will be lower than this; since we want the probability that the value is higher, we subtract from 1:

1-0.8770 = 0.123.

B) To find the probability of a sample mean, we use the formula
$z=\frac{X-\mu}{\frac{\sigma}{\sqrt{n}}} \\ \\=\frac{202.8-205}{\frac{8.6}{\sqrt{15}}}=-0.9908$

Looking in our z-table we see that this corresponds with 0.1611.  Again, however, this is the area left of this value, which is the probability below; we want the probability above, so we subtract from 1:

1-0.1611 = 0.8589