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:
- 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:
has a p-value of 0.6554.
X = 24900:
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.
For the <em>60th percentile:</em>
Hence:
Low: $22,888, High: $29,112.
You can learn more about the normal distribution at brainly.com/question/24663213