A variable is normally distributed with a mean $68$and standard deviation $10$. Find the percentage of all possible values of the variable that

a. lie between $73$and $80$.

h. are at least $75$.

c. are at most $90$.

The given mean is $68$

Standard deviation is$10.$

The given data is

Mean, $\mu =68$

Standard deviation, $\sigma =10$

If X is a continuous random variable with mean $\mu$and standard deviation $\sigma$, then a normal curve with X has the following equation:

$f\left(x\right)=\frac{1}{\sigma \sqrt{2\pi }}{e}^{\frac{-(x-\mu {)}^2}{2{\sigma }^2}};-\infty <x<\infty ,-\infty <\mu <\infty ,\sigma >0.$

Moreover, the equation of a normal curve with random variable $Z$ is as follows:

$f\left(z\right)=\frac{1}{\sqrt{2\pi }}{e}^{\frac{-z^2}{2}}$

Using the values, Find$P(73<X<80)$

$P(73<X<80)=P(73-\mu <X-\mu <80-\mu )$

$=P(73-68<X-\mu <80-68)$

$=P\left(\frac{73-68}{10}<\frac{X-\mu }{\sigma }<\frac{80-68}{10}\right)$

$=P(0.5<Z<1.2)$

$=0.1935$

$=19.35\%.$

The given mean is $68$

Standard deviation is$10.$

If X is a continuous random variable with mean $\mu$and standard deviation $\sigma$, then a normal curve with X has the following equation:

$f\left(x\right)=\frac{1}{\sigma \sqrt{2\pi }}{e}^{\frac{-(x-\mu {)}^2}{2{\sigma }^2}};-\infty <x<\infty ,-\infty <\mu <\infty ,\sigma >0.$

Moreover, the equation of a normal curve with random variable $Z$ is as follows:

$f\left(z\right)=\frac{1}{\sqrt{2\pi }}{e}^{\frac{-z^2}{2}}$

then find,

$P(X>75)=P(X-\mu >75-\mu )$

$=P(X-\mu >75-68)$

$=P\left(\frac{X-\mu }{\sigma }>\frac{75-68}{10}\right)$

$=P(Z>0.7)$

$=0.999$

$=99.9\%.$

The given mean is $68$

Standard deviation is$10$

If X is a continuous random variable with mean $\mu$and standard deviation $\sigma$, then a normal curve with X has the following equation:

$f\left(x\right)=\frac{1}{\sigma \sqrt{2\pi }}{e}^{\frac{-(x-\mu {)}^2}{2{\sigma }^2}};-\infty <x<\infty ,-\infty <\mu <\infty ,\sigma >0.$

Moreover, the equation of a normal curve with random variable $Z$ is as follows:

$f\left(z\right)=\frac{1}{\sqrt{2\pi }}{e}^{\frac{-z^2}{2}}$

Then find,

$P(X<90)=P(X-\mu <90-\mu )$

$=P(X-\mu <90-68)$

$=P\left(\frac{X-\mu }{\sigma }<\frac{90-68}{10}\right)$

$=P(Z<2.2)$

$=P(0<Z<2.2)$

$=0.0047$

$=0.47\%.$