Respuesta :
Answer:
A. X ~ N(mean , variance) ~N( 25,250 , 156,250,000 )
B. P(20,000<X<30,000)=0.3108
C. Low: $20,433.75
High: $30,066.25
Step-by-step explanation:
The question is incomplete.
A. X ~ N( ____ , ____ )
B. Find the probability that the college graduate has between $20,000 and $30,000 in student loan debt.
C. The middle 30% of college graduates' loan debt lies between what two numbers?
Low: $
High: $
We have a normal distribution to represent the average student loan debt for college graduates, with mean $25,250 and standard deviation $12,500.
A) The mean of the distribution is given and is $25,250.
The other parameter is the variance and we can calculate as:
[tex]\sigma^2=(12,500)^2=156,250,000[/tex]
B) To calculate the probability that the debt is between $20,000 and $30,000, we use the z-score for both values:
[tex]z_1=\dfrac{X-\mu}{\sigma}=\dfrac{20000-25250}{12500}=\dfrac{-5250}{12500}=-0.42\\\\\\z_2=\dfrac{X-\mu}{\sigma}=\dfrac{30000-25250}{12500}=\dfrac{4750}{12500}=0.38\\\\\\ P(20000<X<30000)=P(-0.42<z<0.38)=P(z<0.38)-P(z<-0.42)\\\\P(20000<X<30000)=0.64803-0.33724=0.3108[/tex]
C) The middle 30% of college graduates' loan debt lies within the interval 15% below the mean and 15% over the mean.
The values can be looked up in a standard normal distribution to know the z-value for this.
The z-value for this interval is z=±0.3853.
Then, we can calculate the lower and upper bound of this interval for the debt's distribution as:
[tex]LL=\mu-z\cdot \sigma=25250-0.3853*12500=25250-4816.25=20433.75\\\\ UL=\mu+z\cdot \sigma=25250+0.3853*12500=25250+4816.25= 30066.25[/tex]
