Answer: [tex]X-N\text{ \lparen25200, 13500\rparen}[/tex]
The probability that a college student has between $31500 and $47550 is 0.2715
Explanation:
Given:
The average student loan = $25200
The standard deviation = $13500
The distribution is normal
To find:
a) the distribution of X
b) the probability that the college graduate has between $31500 and $47550 in student loan
a) X = the student loan debt of randomly selected college graduat
To write the value of X and N in the formwritten:
The 1st parenthesis = average student loan = $25200
The 2nd parenthesis = standard deviation = $13500
[tex]X-N\text{ \lparen25200, 13500\rparen}[/tex]
b) To get the probability that college student has between $31500 and $47550, we will apply the z-score formula:
[tex]\begin{gathered} \begin{equation*} z=\frac{X-μ}{σ} \end{equation*} \\ X=\text{ value/score} \\ \mu\text{ = mean} \\ \sigma\text{ = standard deviation} \end{gathered}[/tex][tex]\begin{gathered} when\text{ X = 31500} \\ z\text{ = }\frac{31500\text{ - 25200}}{13500} \\ z\text{ = 0.4667} \\ Using\text{ a standard normal table/calculator, the probability for z = 0.4667 is 0 67963} \end{gathered}[/tex][tex]\begin{gathered} when\text{ X = 47550} \\ z\text{ = }\frac{47550\text{ - 25200}}{13500} \\ z\text{ = 1.6556} \\ Using\text{ a standard normal table/calculator, the probability for z = 1.6556 is 0.95109} \end{gathered}[/tex]
The probability that a college student has between $31500 and $47550 will be the difference in probability
[tex]\begin{gathered} =\text{ 0.95109 - 0.67963 = 0.27146} \\ The\text{ probability }=\text{ 0.2715} \\ \end{gathered}[/tex]
