6.90. A variable is normally distributed with mean $68$and standard deviation $10$. Find the percentage of all possible values of the variable that
a. lie between $73$and $80$.
b. are at least $75$.
c. are at most $90$ .

To find the percentage of all possible values of the variable that lie between $73$ and $80$.

Let, the mean is $\mu =68$.

And the standard deviation is $\sigma =10$.
Calculate the probability of the variable $x$being between $73$and $80$, or $P(73<X<80)$.
Determine the value as follows:
$P(73<X<80)=P(73-\mu <X-\mu <80-\mu )$
$=P(73-68<X-\mu <80-68)$
$=P\left(\frac{73-68}{10}<\frac{X-\mu }{\sigma }<\frac{80-68}{10}\right)$
$=P(0.5<Z<1.2)$
$=0.1935$

The percentage will be $0.1935\times 100\%=19.35\%$.

To the percentage of all possible values of the variable that are at least $75$.

Let, the mean is $\mu =68$.

And the standard deviation is $\sigma =10$.
Determine the probability of the variable $x$being greater than $75$, that is, $P(75>X)$.
Then determine the value as follows:
$P(X>75)=P(X-\mu >75-\mu )$
$=P(X-\mu >75-68)$
$=P\left(\frac{X-\mu }{\sigma }>\frac{75-68}{10}\right)$
$=P(Z>0.7)$
$=0.999$
The percentage will be $0.999\times 100\%=99.9\%$.
Conclusion:
As a result, the percentage of all possible values of the variable that are at least $75$is $99.9\%$.

To find the percentage of all possible values of the variable that are at most $90$.

Let, the mean is $\mu =68$.

And the standard deviation is $\sigma =10$.

Determine the probability of the variable $X$being less than $90$, that is, $P(X<90)$.
Then determining the value as follows:

$P(X<90)=P(X-\mu <90-\mu )$
$=P(X-\mu <90-68)$
$=P\left(\frac{X-\mu }{\sigma }<\frac{90-68}{10}\right)$
$=P(Z<2.2)$
$=P(0<Z<2.2)$
$=0.0047$
The percentage will be $0.0047\times 100\%=0.47\%$.
As a result, the percentage of all possible values of the variable that are at most $90$ is $0.47\%$.