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 = $ 17 , 776 and the standard deviation was s = $12,034. The quartiles were Q 1 = $9900, M = $15,532, and Q 3 = $22,500.

Question

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 Q 3. Find the third quartile of the Normal distribution with μ = $17,776 and σ = $12,034 and compare it with Q 3 = $22,500.

i know the answer is Q3 would be $25,838.78. This is a lot larger than our value of Q3 ($22,500).. but how what do i do to get that answer? thanks

Answer 1

To find the third quartile (Q3) of the normal distribution, we would use the Z-score formula:

Z = (X - μ) / σ

where X is the value we're interested in (Q3 in this case), μ is the mean, and σ is the standard deviation. In this problem, we have μ = $17,776 and σ = $12,034.

For Q3 in a normal distribution, we know that 75% of the data falls below that value. We can use a Z-table or calculator to find the Z-score that corresponds to 75% of the data. This Z-score is approximately 0.67.

Now, we'll use the Z-score formula above to find the value of Q3:

0.67 = (Q3 - 17776) / 12034

Now, solve for Q3:

0.67 * 12034 = Q3 - 17776
8022.78 = Q3 - 17776
Q3 = 8022.78 + 17776
Q3 = $25,798.78

So, the third quartile of the normal distribution with μ = $17,776 and σ = $12,034 is approximately $25,798.78. This is larger than the given Q3 value of $22,500, which confirms what we were expecting.