Question

suppose that the outstanding credit card balance of a randomly selected young man has an unknown distribution with a mean of 680 dollars and a standard deviation of 119.9 dollars. let ( x ) be the outstanding credit card balance for a randomly selected young man and let ( overline{x} ) be the average outstanding credit card balance for a random sample of size 34. (for any parts that are not possible enter dne)

1. describe the probability distribution of ( x ) and state its parameters ( mu ) and ( sigma ):

( x sim ) select an answer ( (mu=square, sigma=square) )
and find the probability that the outstanding credit card balance for a randomly selected young man is between 580 and 804 dollars.
( square ) (round the answer to 4 decimal places)

1. use the central limit theorem

select an answer
to describe the probability distribution of ( overline{x} ) and state its parameters ( mu_{overline{x}} ) and ( sigma_{overline{x}} ): (round the answers to 1 decimal place)
( overline{x} sim ) select an answer ( (mu_{overline{x}}=square, sigma_{overline{x}}=square) )
and find the probability that the average outstanding credit card balance for a sample of 34 randomly selected young men is more than 736 dollars.
use the standard deviation you entered above in your calculations.
( square ) (round the answer to 4 decimal places)
question help: video written example

Explanation:

Step1: Distribution of \(X\)

Since the distribution of \(X\) (out - standing credit card balance of a single young man) is unknown, but we are given the mean \(\mu = 680\) and standard deviation \(\sigma=119.9\).
To find \(P(580\lt X\lt804)\), we first calculate the \(z\) - scores.
The \(z\) - score formula is \(z=\frac{x-\mu}{\sigma}\).
For \(x = 580\), \(z_1=\frac{580 - 680}{119.9}=\frac{- 100}{119.9}\approx - 0.83\)
For \(x = 804\), \(z_2=\frac{804 - 680}{119.9}=\frac{124}{119.9}\approx1.03\)
Then \(P(580\lt X\lt804)=P(-0.83\lt Z\lt1.03)\)
Using the standard normal table \(P(-0.83\lt Z\lt1.03)=\Phi(1.03)-\Phi(-0.83)\)
where \(\Phi(z)\) is the cumulative distribution function of the standard normal distribution. \(\Phi(1.03) = 0.8485\), \(\Phi(-0.83)=0.2033\)
\(P(-0.83\lt Z\lt1.03)=0.8485 - 0.2033=0.6452\)

Step2: Distribution of \(\overline{X}\) (Central Limit Theorem)

By the Central Limit Theorem, for a sample of size \(n = 34\), the sampling distribution of \(\overline{X}\) (the sample mean) has mean \(\mu_{\overline{X}}=\mu = 680\)
and standard deviation \(\sigma_{\overline{X}}=\frac{\sigma}{\sqrt{n}}=\frac{119.9}{\sqrt{34}}\approx\frac{119.9}{5.83}\approx20.6\)
To find \(P(\overline{X}\gt736)\), we calculate the \(z\) - score for \(\overline{X}\): \(z=\frac{\overline{x}-\mu_{\overline{X}}}{\sigma_{\overline{X}}}=\frac{736 - 680}{20.6}=\frac{56}{20.6}\approx2.72\)
\(P(\overline{X}\gt736)=1 - P(\overline{X}\leq736)=1-\Phi(2.72)\)
From the standard normal table, \(\Phi(2.72)=0.9967\)
\(P(\overline{X}\gt736)=1 - 0.9967=0.0033\)

Answer:

1. \(X\sim\) (unknown distribution), \(\mu = 680\), \(\sigma = 119.9\), probability \(=0.6452\)
2. \(\overline{X}\sim N(\mu_{\overline{X}} = 680.0,\sigma_{\overline{X}}=20.6)\) (by Central Limit Theorem, since \(n = 34\geq30\)), probability \(=0.0033\)