Using the normal distribution, it is found that:
a. X ~ N(26000, 12300).
b. 0.1913 = 19.13% probability that the college graduate has between $24,900 and $30,950 in student loan debt.
c. Low: $22,888, High: $29,112.
<h3>Normal Probability Distribution</h3>
In a normal distribution with mean
and standard deviation
, the z-score of a measure X is given by:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
- It measures how many standard deviations the measure is from the mean.
- After finding the z-score, we look at the z-score table and find the p-value associated with this z-score, which is the percentile of X.
Item a:
- The mean is of $26,000, hence
.
- The standard deviation is of $12,300, hence
.
Then, the distribution is:
X ~ N(26000, 12300).
Item b:
The probability is the <u>p-value of Z when X = 30950 subtracted by the p-value of Z when X = 24900</u>, hence:
X = 30950:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![Z = \frac{30950 - 26000}{12300}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B30950%20-%2026000%7D%7B12300%7D)
![Z = 0.4](https://tex.z-dn.net/?f=Z%20%3D%200.4)
has a p-value of 0.6554.
X = 24900:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![Z = \frac{24900 - 26000}{12300}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B24900%20-%2026000%7D%7B12300%7D)
![Z = -0.09](https://tex.z-dn.net/?f=Z%20%3D%20-0.09)
has a p-value of 0.4641.
0.6554 - 0.4641 = 0.1913
0.1913 = 19.13% probability that the college graduate has between $24,900 and $30,950 in student loan debt.
Item c:
- Between the 40th percentile(low) and the 60th percentile(high).
- The 40th percentile is X when Z has a p-value of 0.4, so X when Z = -0.253.
- The 60th percentile is X when Z has a p-value of 0.6, so X when Z = 0.253.
Then, the <em>40th percentile</em> is found as follows.
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![-0.253 = \frac{X - 26000}{12300}](https://tex.z-dn.net/?f=-0.253%20%3D%20%5Cfrac%7BX%20-%2026000%7D%7B12300%7D)
![X - 26000 = -0.253(12300)](https://tex.z-dn.net/?f=X%20-%2026000%20%3D%20-0.253%2812300%29)
![X = 22888](https://tex.z-dn.net/?f=X%20%3D%2022888)
For the <em>60th percentile:</em>
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![0.253 = \frac{X - 26000}{12300}](https://tex.z-dn.net/?f=0.253%20%3D%20%5Cfrac%7BX%20-%2026000%7D%7B12300%7D)
![X - 26000 = 0.253(12300)](https://tex.z-dn.net/?f=X%20-%2026000%20%3D%200.253%2812300%29)
![X = 29112](https://tex.z-dn.net/?f=X%20%3D%2029112)
Hence:
Low: $22,888, High: $29,112.
You can learn more about the normal distribution at brainly.com/question/24663213