91Ó°ÊÓ

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

Let, the mean is $\mathrm{μ}=68$.

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

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

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

Let, the mean is $\mathrm{μ}=68$.

And the standard deviation is $\mathrm{σ}=10$.
Determine the probability of the variable $x$being greater than $75$, that is, $P(75>X)$.
Then determine the value as follows:
$P(X>75)=P(X-\mathrm{μ}>75-\mathrm{μ})$
$=P(X-\mathrm{μ}>75-68)$
$=P\left(\frac{X-\mathrm{μ}}{\mathrm{σ}}>\frac{75-68}{10}\right)$
$=P(Z>0.7)$
$=0.999$
The percentage will be $0.999×100\%=99.9\%$.
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 $\mathrm{μ}=68$.

And the standard deviation is $\mathrm{σ}=10$.

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

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