Respuesta :
Using the normal distribution, it is found that:
a) [tex]X \approx N(25150, 12050)[/tex]
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.
Normal Probability Distribution
The z-score of a measure X of a normally distributed variable with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex] is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
- 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:
[tex]\mu = 25150, \sigma = 12050[/tex].
Hence the distribution of X can be described as follows:
[tex]X \approx N(25150, 12050)[/tex]
The probability that the college graduate has between $14,200 and $33,950 in student loan debt is the p-value of Z when X = 33950 subtracted by the p-value of Z when X = 14200, hence:
X = 33950:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{33950 - 25150}{12050}[/tex]
Z = 0.73
Z = 0.73 has a p-value of 0.7673.
X = 14200:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{14200 - 25150}{12050}[/tex]
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:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]-0.127 = \frac{X - 25150}{12050}[/tex]
X - 25150 = -0.127(12050)
X = $23,519.65.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]0.127 = \frac{X - 25150}{12050}[/tex]
X - 25150 = 0.127(12050)
X = $26,580.35.
More can be learned about the normal distribution at https://brainly.com/question/4079902
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