Question

Ms. Hall gave her class a $10$-question multiple-choice quiz. Let $X=$the number of questions that a randomly selected student in the class answered correctly. The computer output below gives information about the probability distribution of X. To determine each student’s grade on the quiz (out of localid="1649489099543" $100$), Ms. Hall will multiply his or her number of correct answers by 10. Let localid="1649489106434" $G=$the grade of a randomly chosen student in the class.

localid="1649489113566" $\begin{array}{llllllll}\mathrm{N}& \text{Mean}& \text{Median}& \text{StDev}& \text{Min}& \text{Max}& Q_1& Q_3\\ 30& 7.6& 8.5& 1.32& 4& 10& 8& 9\end{array}$

(a) Find the mean of localid="1649489121120" $G$. Show your method.

(b) Find the standard deviation of localid="1649489127059" $G$. Show your method.

(c) How do the variance of localid="1649489132289" $X$and the variance oflocalid="1649489138146" $G$compare? Justify your answer.

Short answer

(a) The mean of $G$is localid="1649489147240" ${\mu }_G=76$.

(b) The standard deviation of $G$is ${\sigma }_G=13.2$.

(c) The variance of $G$is $100$times the variance of $X$.

${\sigma }_G^2=100{\sigma }_X^2$

Step-by-step solution

Part (a) Step 1: Given Information 

In the output, the mean for the variable $X$is given:

${\mu }_X=7.6$

Part (a) Step 2: Explanation

Ms. Hall multiplies the number of correct answers $X$by $10$:

$G=10X$

Property mean:

${\mu }_{aX+b}=a{\mu }_X+b$

Then we can determine the mean for $G$:

localid="1649909794929" $$\begin{gathered} {\mu }_G={\mu }_{10X} \\ =10{\mu }_X \\ =10(7.6) \\ =76 \end{gathered}$$

Part (b) Step 1: Given Information 

In the output, the standard deviation "StDev" for the variable $X$is given:

${\sigma }_X=1.32$

Part (b) Step 2: Explanation 

Ms. Hall multiplies the number of correct answers $X$by $10$:

$G=10X$

Property standard deviation:

${\sigma }_{aX+b}=a{\sigma }_X$

Then we can determine the standard deviation for $G$:

localid="1649909831105" $$\begin{gathered} {\sigma }_G={\sigma }_{10X} \\ =10{\sigma }_X \\ =10(1.32) \\ =13.2 \end{gathered}$$

Part (c) Step 1: Given Information 

Given

$\begin{array}{llllllll}\mathrm{N}& \text{Mean}& \text{Median}& \text{StDev}& \text{Min}& \text{Max}& Q_1& Q_3\\ 30& 7.6& 8.5& 1.32& 4& 10& 8& 9\end{array}$

Part (c) Step 2: Explanation 

Ms. Hall multiplies the number of correct answers $X$by $10$:

$G=10X$

Property standard deviation:

${\sigma }_{aX+b}=a{\sigma }_X$

Then we can determine the standard deviation for $G$:

localid="1649909863743" $$\begin{gathered} {\sigma }_G={\sigma }_{10X} \\ =10{\sigma }_X \end{gathered}$$

The variance is the square of the standard deviation:

localid="1649909871172" $$\begin{gathered} {\sigma }_G^2={10}^2{\sigma }_X^2 \\ =100{\sigma }_X^2 \end{gathered}$$

Thus the variance of $G$is $100$times the variance of $X$.