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 $100$), Ms. Hall will multiply his or her number of correct answers by $10$. Let $G=$the grade of a randomly chosen student in the class.

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

(a) Find the median of $G$. Show your method.

(b) Find the IQR of $G$. Show your method.

(c) What shape would the probability distribution of $G$have? Justify your answer

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

${MEDIAN}_X=8.5$

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

$G=10X$

Property median (same as for the mean, because both are measures of center):

${MEDIAN}_{aX+b}=a{MEDIAN}_X+b$

Then we can determine the median for $G$:

localid="1649909519882" $$\begin{gathered} {MEDIAN}_G={\text{MEDIAN}}_{10X} \\ =10{\text{MEDIAN}}_X \\ =10(8.5) \\ =85 \end{gathered}$$

The interquartile range is the difference between the third and the first quartile:

${IQR}_X=Q_3-Q_1=9-8=1$

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

$G=10X$

Property IQR (same as for the standard deviation, because both are measures of spread):

${\mathrm{IQR}}_{aX+b}=a{\mathrm{IQR}}_X$

Then we can determine the interquartile range for $G$:

localid="1649909558629" $$\begin{gathered} {\mathrm{IQR}}_G={\mathrm{IQR}}_{10X} \\ =10{\mathrm{IQR}}_X \\ =10\left(1\right) \\ =10 \end{gathered}$$

Given

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

Result exercise $39a-40a-40b:$

$$\begin{gathered} MEA{N}_G=76 \\ MEDIA{N}_G=85 \\ IQ{R}_G=10 \end{gathered}$$

The fact that the mean is lower than the median suggests that the distribution is skewed to the left (since the mean is influenced by outliers, which have to be very small because the mean is less than the median).