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The percent of fat calories of a person in America is normally distributed with a mean of about $36$and a standard deviation of about ten. Let role="math" localid="1648361713521" $\stackrel{¯}{X}=$average percent of fat calories.

According to the given information,

$\stackrel{¯}{X}=$average percent of fat calories.

Mean$36$

Standard deviation $=10$

The number of individuals$=16$

Therefore,

$\stackrel{¯}{X}~N\left(36,\frac{10}{\sqrt{16}}\right)$

$\stackrel{¯}{X}~N(36,2.5)$

The percent of fat calories of a person in America is normally distributed with a mean of about $36$ and a standard deviation of about ten. Let $\stackrel{¯}{X}=$average percent of fat calories.

We have to find the probability,

$P(\stackrel{¯}{X}>5)=1-P(\stackrel{¯}{X}<5)$

$=1-P\left(\frac{\stackrel{¯}{X}-36}{10}\sqrt{16}<\frac{5-36}{10}\sqrt{16}\right)$

$=1-P\left(\frac{\stackrel{¯}{X}-36}{10}\sqrt{16}<.12.4\right)$

$=1-\mathrm{Ï•}(-12.4)$

$=1-(1-\mathrm{Ï•}(12.4\left)\right)$

$=\mathrm{Ï•}(12.4)$

$=1$

There is a one-in-five chance that the average % of fat calories consumed is greater than five.

Therefore, the graph is shown below :

The percent of fat calories of a person in America is normally distributed with a mean of about $36$and a standard deviation of about ten. Let $\stackrel{¯}{X}=$average percent of fat calories.

In order to find the first quartile, we have to find $z$from the probability:

$P(\stackrel{¯}{X}<z)=0.25$

This is equivalent to

$P\left(\frac{\stackrel{¯}{X}-36}{10}4<\frac{z-36}{10}4\right)=0.25$

Since variable $\frac{\stackrel{¯}{X}-36}{10}4$has a standard normal distribution, we can check the table of standard normal distribution to get that

$P\left(\frac{\stackrel{¯}{X}-36}{10}4<-0.67\right)=0.25$

Now we can find $z$is

$\frac{z-36}{10}4=-0.67$

$z-36=-1.675$

$z=34.325$