Question

ACT scores The composite scores of individual students on the ACT college entrance examination in $2009$followed a Normal distribution with mean $21.1$and standard deviation$5.1$

(a) What is the probability that a single student randomly chosen from all those taking the test scores $23$or higher? Show your work.

(b) Now take an SRS of $50$students Who took the fest. What is the probability that the mean score $\overline{x}$of these students is $23$or higher? Show your work

Short answer

(a) The probability is $0.3557$

(b) The probability is$0.3974$

Step-by-step solution

Part (a) Step 1: Given Information

Given in the question that ,

Population mean $\left(\mu \right)=21.1$$\left(\mu \right)=21.1$

Population standard deviation $\left(\sigma \right)=5.1$$\left(\sigma \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.

Part (a) Step-2 Explanation

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

$z=\frac{x-\mu }{\sigma }$

$x$is raw score

$\mu$Population mean

$\sigma$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:

$P\left(X>23\right)=P\left(\frac{x-\mu }{\sigma }>\frac{23-\mu }{\sigma }\right)$

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

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

$=0.3557$

The require probability is$0.3557$

Part (b) Step 1: Given Information

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.

Part (b) Step 2:  Explanation

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

$P\left(\overline{x}>295\right)=P\left(\frac{{\displaystyle \frac{\overline{x}-\mu }{\sigma }}}{\sqrt{n}}>\frac{{\displaystyle \frac{295-\mu }{\sigma }}}{\sqrt{n}}\right)$

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

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

$=0.3974$

The probability is$0.3974$