Question 1207315: 3. According to a PEW* Research Center survey, the mean student loan at graduation is $25,000. Suppose that student loans are normally distributed with a standard deviation of $5,000. A graduate with a student loan is selected at random. Find the following probabilities. Let random variable X denote the student loan. (2 points)
[* PEW Research Center is a nonpartisan American think tank based in Washington, D.C.]
a. The loan is greater than $30,000 (1 point)
b. The loan falls between #20,000 and $32,000. (1 point)
Answer by ikleyn(52776)   (Show Source):
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3. According to a PEW* Research Center survey, the mean student loan at graduation is $25,000.
Suppose that student loans are normally distributed with a standard deviation of $5,000.
A graduate with a student loan is selected at random. Find the following probabilities.
Let random variable X denote the student loan. (2 points)
[* PEW Research Center is a nonpartisan American think tank based in Washington, D.C.]
a. The loan is greater than $30,000 (1 point)
b. The loan falls between #20,000 and $32,000. (1 point)
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(a) To answer (a), use the Empirical rule of normal distribution.
It says that that 68% of the data falls within one standard deviation from the mean.
In your case (a), it means that half of it, i.e. 34% of data falls
into the interval from the mean of $25,000 to $25,000 plus one standard deviation of $5,000.
So, the sum is $25,000 + $5,000 = $30,000.
Thus the probability that the loan is greater than $30,000 is 0.5-0.34 = 0.16, or 16%.
It is your answer to (a).
(b) To answer (b), use your regular calculator like TI-83/84.
Use the function normcdf, which represents normal cumulative distribution function.
The formatting pattern is p = normcdf(z1, z2, mean, SD).
In your case, z1 = 20000, z2 = 32000, mean = 25000 and SD = 5000.
Then the sough probability is
p = normcdf(20000, 32000, 25000, 32000) = 0.7606.
Alternatively, you may use free of charge online calculator
https://onlinestatbook.com/2/calculators/normal_dist.html
The answer is the same, p = 0.7606, or 76.06%.
It is the area under the described normal curve between the raw marks 20000 and 32000.
Solved.
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