91Ó°ÊÓ

Given in the question that ,

Population mean $\left(\mathrm{μ}\right)=21.1$$\left(\mathrm{μ}\right)=21.1$

Population standard deviation $\left(\mathrm{σ}\right)=5.1$$\left(\mathrm{σ}\right)=5.1$

we have to find that the probability that a single student randomly chosen from all those taking the test scores $23$or higher.

The formula to compute the $z-$Score is:

$z=\frac{x-\mathrm{μ}}{\mathrm{σ}}$

$x$is raw score

$\mathrm{μ}$Population mean

$\mathrm{σ}$population standard deviation

$X$be the random variable follows the normal distribution with mean$21.1$standard deviation$=5.1$

The probability that randomly selected student would score $23$or more can be computed as:

$P\left(X>23\right)=P\left(\frac{x-\mathrm{μ}}{\mathrm{σ}}>\frac{23-\mathrm{μ}}{\mathrm{σ}}\right)$

$=P\left(Z>\frac{23-21.1}{5.0}\right)$

$=P\left(Z>0.37\right)\left(Froms\tan dardnormaltable\right)$

$=0.3557$

The require probability is$0.3557$

Given in the question that, take an SRS of $50$students Who took the fest .we have to find the probability that the mean score $\overline{x}$of these students is $23$or higher.

The probability that mean score of $50$students is $23$or above is calculated as follows:

$P\left(\overline{x}>295\right)=P\left(\frac{{\displaystyle \frac{\overline{x}-\mathrm{μ}}{\mathrm{σ}}}}{\sqrt{n}}>\frac{{\displaystyle \frac{295-\mathrm{μ}}{\mathrm{σ}}}}{\sqrt{n}}\right)$

$=P\left(Z>\frac{{\displaystyle \frac{23-21.1}{5.1}}}{\sqrt{50}}\right)$

$=P(Z>0.26)(Froms\tan dardnormaltable)$

$=0.3974$

The probability is$0.3974$