Answer:
a) 78.81% probability that a randomly selected woman has a foot length less than 10.0 in.
b) 78.74% probability that a randomly selected woman has a foot length between 8.0 in and 10.0 in.
c) 2.28% probability that 25 women have foot lengths with a mean greater than 9.8 in.
Step-by-step explanation:
The Central Limit Theorem estabilishes that, for a random variable X, with mean
and standard deviation
, a large sample size can be approximated to a normal distribution with mean
and standard deviation
.
Normal probability distribution
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean
and standard deviation
, the zscore of a measure X is given by:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
.
a. Find the probability that a randomly selected woman has a foot length less than 10.0 in
This probability is the pvalue of Z when
.
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![Z = \frac{10 - 9.6}{0.5}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B10%20-%209.6%7D%7B0.5%7D)
![Z = 0.8](https://tex.z-dn.net/?f=Z%20%3D%200.8)
has a pvalue of 0.7881.
So there is a 78.81% probability that a randomly selected woman has a foot length less than 10.0 in.
b. Find the probability that a randomly selected woman has a foot length between 8.0 in and 10.0 in.
This is the pvalue of Z when X = 10 subtracted by the pvalue of Z when X = 8.
When X = 10, Z has a pvalue of 0.7881.
For X = 8:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![Z = \frac{8 - 9.6}{0.5}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B8%20-%209.6%7D%7B0.5%7D)
![Z = -3.2](https://tex.z-dn.net/?f=Z%20%3D%20-3.2)
has a pvalue of 0.0007.
So there is a 0.7881 - 0.0007 = 0.7874 = 78.74% probability that a randomly selected woman has a foot length between 8.0 in and 10.0 in.
c. Find the probability that 25 women have foot lengths with a mean greater than 9.8 in.
Now we have
.
This probability is 1 subtracted by the pvalue of Z when
. So:
![Z = \frac{X - \mu}{s}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7Bs%7D)
![Z = \frac{9.8 - 9.6}{0.1}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B9.8%20-%209.6%7D%7B0.1%7D)
![Z = 2](https://tex.z-dn.net/?f=Z%20%3D%202)
has a pvalue of 0.9772.
There is a 1-0.9772 = 0.0228 = 2.28% probability that 25 women have foot lengths with a mean greater than 9.8 in.