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A government report looked at the amount borrowed for college by students who graduated in 2000and had taken out student loans.12 The mean amount was x=\(17,776and the standard deviation was sx=\)12,034. The median was\(15,532and the quartiles were Q1=\)9900and Q3=\(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=\)17,776and =\(12,034and compare it withQ3=\)22,500.

Short Answer

Expert verified

(a)The distribution appears to be right-skewed.

(b)The third quartile of the Normal distribution is$25,838.78

Step by step solution

01

Part (a) Step 1: Given information

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

02

Part(a)Step 2: Explanation

Given:

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

Q3=$22,500

We note that the first quartileQ1lies closer to the median than the third quartileQ3, which indicates a tail to the right and thus the distribution appears to be right-skewed.

03

Part (b) Step 1: Given information

A government report looked at the amount borrowed for college by students who graduated in 2000and had taken out student loans. The mean amount wasx=$17,776 and the standard deviation was sx=$12,034. The median was $15,532 and the quartiles wereQ1=$9900 and Q3=$22,500.

04

Part (b) Step 2: Explanation

Given:

=$12,034

The third quantile corresponds with a probability of 0.75in 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" x=+z=17776+0.67(12034)=25838.78

We note that the third quartile of the normal distribution is greater than the given Q3.

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