Question

The distribution of heights of adult males has a mean of 69 inches and a standard deviation of 4 inches. A random sample of 36 adult males is selected. Find the probability that the average height will be more than 70 inches.
a. 0.668
b. 0.858
c. 0.908
d. 0.995

Answer 1

Answer:

$P(\bar X >70)= P(Z>\frac{70-69}{\frac{4}{\sqrt{36}}})= P(Z>1.5)$

And for this case we can use the complement rule and the normal standard distribution of excel and we got:

$P(Z>1.5)=1-P(Z,1.5) = 1-0.933=0.0668$

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".  

Solution to the problem

Let X the random variable that represent the heights of a population, and for this case we know the distribution for X is given by:

$X \sim N(69,4)$  

Where $\mu=69$ and $\sigma=4$

And we select a sample size of n =70

From the central limit theorem  (n>30)we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

And we want to find this probability:

$P(\bar X >70)$

And we can use the z score formula given by:

$z= \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And using this formula we got:

$P(\bar X >70)= P(Z>\frac{70-69}{\frac{4}{\sqrt{36}}})= P(Z>1.5)$

And for this case we can use the complement rule and the normal standard distribution of excel and we got:

$P(Z>1.5)=1-P(Z,1.5) = 1-0.933=0.0668$