Question

Assume that the heights of men are normally distributed with a mean of 70.270.2 inches and a standard deviation of 2.12.1 inches. If 3636 men are randomly​ selected, find the probability that they have a mean height greater than 71.271.2 inches.

Answer 1

Answer:

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:

$X \sim N(70.2,2.12)$  

Where $\mu=70.2$ and $\sigma=2.12$

Since the disgtribution for X is normal then the distribution for the sample mean $\bar X$ is given by:

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

We can find the probability of interest we got:

$P(\bar X >71.2)=P(Z>\frac{71.2-70.2}{\frac{2.12}{\sqrt{36}}}=2.83)$

And using the complement rule and a calculator, excel or the normal standard table we got this:

$P(Z>2.83)=1-P(Z$

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:

$X \sim N(70.2,2.12)$  

Where $\mu=70.2$ and $\sigma=2.12$

Since the disgtribution for X is normal then the distribution for the sample mean $\bar X$ is given by:

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

We can find the probability of interest we got:

$P(\bar X >71.2)=P(Z>\frac{71.2-70.2}{\frac{2.12}{\sqrt{36}}}=2.83)$

And using the complement rule and a calculator, excel or the normal standard table we got this:

$P(Z>2.83)=1-P(Z$