A government report looked at the amount borrowed for college by students who graduated in $2000$and had taken out student loans.12 The mean amount was $x=\backslash (17,776$and the standard deviation was $sx=\backslash )12,034$. The median was$\backslash (15,532$and the quartiles were $Q1=\backslash )9900$and $Q3=\backslash (22,500.$

(a) Compare the mean and the median. Also compare the distances of Q1 and Q3 from the median. Explain why both comparisons suggest that the distribution is right-skewed.

(b) The right-skew pulls the standard deviation up. So a Normal distribution with the same mean and standard deviation would have a third quartile larger than the actual Q3.Find the third quartile of the Normal distribution with$\mu =\backslash )17,776$and $\sigma =\backslash (12,034$and compare it with$Q3=\backslash )22,500$.

Given in the question that, a government report looked at the amount borrowed for college by students who graduated in $2000$and had taken out student loans. The mean amount was $x=\$17,776$and the standard deviation was $sx=\$12,034$. The median was$\$15,532$and the quartiles were $Q1=\$9900$and $Q3=\$22,500$.

Given:

$\overline{x}=\$17,776$

$MEDIAN=\$15,532$

We note that the median and mean differ by more than $\$2,000$.

Moreover the mean is greater than the median, which indicates outliers to the right and thus the distribution appears to be right-skewed.

Given:

$MEDIAN=\$15,532$

$Q_3=\$22,500$

We note that the first quartile$Q_1$lies closer to the median than the third quartile$Q_3$, which indicates a tail to the right and thus the distribution appears to be right-skewed.

A government report looked at the amount borrowed for college by students who graduated in $2000$and had taken out student loans. The mean amount was$x=\$17,776$ and the standard deviation was $sx=\$12,034$. The median was $\$15,532$ and the quartiles were$Q1=\$9900$ and $Q3=\$22,500$.

Given:

$\sigma =\$12,034$

The third quantile corresponds with a probability of $0.75$in table A:

$z=0.67$

The corresponding value is the mean increased by the product of the z-score and the standard deviation:

localid="1650004878896" $$\begin{gathered} x=\mu +z\sigma \\ =17776+0.67\left(12034\right) \\ =25838.78 \end{gathered}$$

We note that the third quartile of the normal distribution is greater than the given $Q_3$.