Chapter EP, Problem 12.5.3EP

To determine

To find the percent that would be scored above $510$ .

Answer to Problem 12.5.3EP

  $50\%$

Explanation of Solution

Given:

The scores have a mean of $\mu =510$ and standard deviation $\sigma =80$ .

Calculation:

The probability that number of fittings that the score would be above $510$ is represented by $P(X>510)$ . Compute the corresponding z-value for $X=510$ as follow:

  $\begin{array}{l}z=\frac{x-\mu }{\sigma }\\ =\frac{510-510}{80}\\ =0\end{array}$

Hence, the required probability will be calculated below:

  $\begin{array}{l}P(X>510)=P(Z>0)\\ =0.5\\ =50\%\end{array}$

To determine

To find the percent of score would you expect between $430$ and $590$

Answer to Problem 12.5.3EP

  $68\%$

Explanation of Solution

Given:

The scores have a mean of $\mu =510$ and standard deviation $\sigma =80$ .

Calculation:

The probability that the score will be between $430$ and $590$ is represented by by $P(430<X<590)$ . Compute the corresponding z-value for $X=430$ as follow:

  $\begin{array}{l}z=\frac{x-\mu }{\sigma }\\ =\frac{430-510}{80}\\ =-1\end{array}$

Compute the corresponding z-value for $X=590$ as follow:

  $\begin{array}{l}z=\frac{x-\mu }{\sigma }\\ =\frac{590-510}{80}\\ =1\end{array}$

Hence, the required probability will be calculated below:

  $\begin{array}{l}P(430<X<590)=P(-1<z<1)\\ =0.68\\ =68\%\end{array}$

To determine

To find the probability that a student is chosen having the scores between $350$ and $670$

Answer to Problem 12.5.3EP

  $\text{0}\text{.95}$

Explanation of Solution

Given:

The scores have a mean of $\mu =510$ and standard deviation $\sigma =80$ .

Calculation:

The probability that student is chosen having the scores between $350$ and $670$ is represented by by $P(350<X<670)$ . Compute the corresponding z-value for $X=350$ as follow:

  $\begin{array}{l}z=\frac{x-\mu }{\sigma }\\ =\frac{350-510}{80}\\ =-2\end{array}$

Compute the corresponding z-value for $X=670$ as follow:

  $\begin{array}{l}z=\frac{x-\mu }{\sigma }\\ =\frac{670-510}{80}\\ =2\end{array}$

Hence, the required probability will be calculated below:

  $\begin{array}{l}P(350<X<670))=P(-2<Z<2)\\ =\text{0}\text{.95}\end{array}$