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 \\) unknown (\\( \mu=680, \sigma=119 \\))
and find the probability that the outstanding credit card balance for a randomly selected young man is between 580 and 804 dollars.
dne (round the answer to 4 decimal places)

1. use the central limit theorem

select an answer
select an answer
wers to 1
to describ the original population is normally distributed
decimal p the sample size is large (n>30) although the distribution of the original population is unknown
the distribution of the original population is unknown
and find the sample size is small (n<30) and the distribution of the original population is unknown mly selected young men is more than 736 dollars.
use the standard deviation you entered above in your calculations.
0.00 (round the answer to 4 decimal places)

Explanation:

Step1: Calculate z - scores for \(X = 580\) and \(X = 804\)

The formula for the z - score is \(z=\frac{X-\mu}{\sigma}\).
For \(X = 580\), \(z_1=\frac{580 - 680}{119.9}=\frac{- 100}{119.9}\approx - 0.834\)
For \(X = 804\), \(z_2=\frac{804 - 680}{119.9}=\frac{124}{119.9}\approx1.034\)

Step2: Find the probability \(P(580<X<804)\)

Since the distribution of \(X\) is unknown, we cannot use the standard normal distribution to find \(P(-0.834<Z<1.034)\) using the formula \(P(-0.834<Z<1.034)=\Phi(1.034)-\Phi(-0.834)\) (where \(\Phi(z)\) is the cumulative distribution function of the standard normal distribution). So the probability for part 1 is DNE.

Step3: Justify the use of the Central Limit Theorem for \(\bar{X}\)

The Central Limit Theorem states that if the sample size \(n\) is large (\(n>30\)), the sampling distribution of \(\bar{X}\) is approximately normal, regardless of the distribution of the population. Here \(n = 34>30\), so we use the reason “the sample size is large (\(n > 30\)) although the distribution of the original population is unknown”

Step4: Calculate the standard deviation of \(\bar{X}\)

The standard deviation of the sampling distribution of \(\bar{X}\) (also known as the standard error) is \(\sigma_{\bar{X}}=\frac{\sigma}{\sqrt{n}}\). Given \(\sigma = 119.9\) and \(n = 34\), \(\sigma_{\bar{X}}=\frac{119.9}{\sqrt{34}}\approx\frac{119.9}{5.831}\approx20.56\)

Step5: Calculate the z - score for \(\bar{X}=736\)

The formula for the z - score of \(\bar{X}\) is \(z=\frac{\bar{X}-\mu}{\sigma_{\bar{X}}}\). Substituting \(\bar{X} = 736\), \(\mu = 680\) and \(\sigma_{\bar{X}}\approx20.56\), we get \(z=\frac{736 - 680}{20.56}=\frac{56}{20.56}\approx2.72\)

Step6: Find \(P(\bar{X}>736)\)

\(P(\bar{X}>736)=1 - P(\bar{X}\leq736)\). Using the standard normal distribution, \(P(Z\leq2.72)\) can be found from the standard normal table. \(P(Z\leq2.72)\approx0.9967\). So \(P(\bar{X}>736)=1 - 0.9967 = 0.0033\)

Answer:

1. The probability is DNE.
2. The reason for using the Central Limit Theorem is “the sample size is large (\(n>30\)) although the distribution of the original population is unknown” and \(P(\bar{X}>736)\approx0.0033\)