C) f(n) = 73

Mathematics

1 answer:

Mandarinka [93]3 years ago

6 0

Answer:

I think the correct anwser to this question is b

Step-by-step explanation:

glad I could help :)

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The average student loan debt for college graduates is $25,150.

Aleks04 [339]

Using the normal distribution, it is found that:

a) $X \approx N(25150, 12050)$

b) There is a 0.5859 = 58.59% probability that the college graduate has between $14,200 and $33,950 in student loan debt.

c) Low: $23,519.65, High: $26,580.35.

<h3>Normal Probability Distribution</h3>

The z-score of a measure X of a normally distributed variable with mean $\mu$ and standard deviation $\sigma$ is given by:

$Z = \frac{X - \mu}{\sigma}$

The z-score measures how many standard deviations the measure is above or below the mean.

Looking at the z-score table, the p-value associated with this z-score is found, which is the percentile of X.

The mean and the standard deviation are given, respectively, by:

$\mu = 25150, \sigma = 12050$ .

Hence the distribution of X can be described as follows:

$X \approx N(25150, 12050)$

The probability that the college graduate has between $14,200 and $33,950 in student loan debt is the <u>p-value of Z when X = 33950 subtracted by the p-value of Z when X = 14200</u>, hence:

X = 33950:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{33950 - 25150}{12050}$

Z = 0.73

Z = 0.73 has a p-value of 0.7673.

X = 14200:

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{14200 - 25150}{12050}$

Z = -0.91

Z = -0.91 has a p-value of 0.1814.

0.7673 - 0.1814 = 0.5859.

There is a 0.5859 = 58.59% probability that the college graduate has between $14,200 and $33,950 in student loan debt.

Considering the symmetry of the normal distribution, the middle 10% is between the 45th percentile(X when Z = -0.127) and the 55th percentile(X when Z = 0.127), hence:

$Z = \frac{X - \mu}{\sigma}$

$-0.127 = \frac{X - 25150}{12050}$