91Ó°ÊÓ

The variable is normally distributed with a mean $10$and a standard deviation $3$.

$\mathrm{μ}=10;\mathrm{σ}=3$

First, find the probability of the variable $X$such that it is between $6$and $7$that is, $P(6<X<7)$

Now calculating its value:

$P(6<X<7)=P(6-\mathrm{μ}<X-\mathrm{μ}<7-\mathrm{μ})$

$=P(6-10X-\mathrm{μ}<7-10)$

$=P\left(\frac{6-10}{3}<\frac{X-\mathrm{μ}}{\mathrm{σ}}<\frac{7-10}{3}\right)$

$=P(-1.33<Z<-1)$$=P(-1.33<Z<-1)$

$=P(1<Z<1.33)$

$=0.0006$

The percentage will be $0.0006×100\%=0.06\%$.

The variable is normally distributed with a mean $10$and a standard deviation$3.$

First, find the probability of the variable $X$such that it is more than $10$that is, $P(10<X).$

Now calculating its value:

$P(X>10)=P(X-\mathrm{μ}>10-\mathrm{μ})$

$=P(X-\mathrm{μ}>10-10)$

$=P\left(\frac{X-\mathrm{μ}}{\mathrm{σ}}>\frac{10-10}{3}\right)$

$=P(Z>0)$

$=0.9996$

The percentage will be$0.9996×100\%=99.96\%$ .

The variable is normally distributed with a mean of $10$ and a standard deviation of $3$.

First, find the probability of the variable $X$such that it is less than $17.5$that is, $P(X<17.5)$

Now calculating its value:

$P(X<17.5)=P(X-\mathrm{μ}<17.5-\mathrm{μ})$

$=P(X-\mathrm{μ}<17.5-10)$

$=P\left(\frac{X-\mathrm{μ}}{\mathrm{σ}}<\frac{17.5-10}{3}\right)$

$=P(Z<2.5)$

$=P(0<Z<2.5)$

$=0.0062$

The percentage will be $0.0062×100\%=0.62\%$.