Normal Distribution: Credit card balances, consisting of mean = $5000, standard deviation = $900.

Question

If credit card balances are normally distributed with a mean of $4200 and a standard deviation of $900, find the probability that a randomly selected credit card holder has a credit card balance of less than $5000. 

Answer 1

To find the probability that a randomly selected credit card holder has a credit card balance of less than $5000, you need to use the z-score formula and the standard normal distribution table. The z-score formula is:

\[Z = \frac{X - \mu}{\sigma}\]

Where:

- \(Z\) is the z-score.

- \(X\) is the value you want to find the probability for (in this case, $5000).

- \(\mu\) is the mean (average) credit card balance.

- \(\sigma\) is the standard deviation.

Given the mean (\(\mu\)) is $4200, the standard deviation (\(\sigma\)) is $900, and you want to find the probability for \(X = $5000\), you can calculate the z-score as follows:

\[Z = \frac{5000 - 4200}{900} = \frac{800}{900} = \frac{4}{4.5} = 0.8889\]

Now, you can find the probability associated with this z-score using a standard normal distribution table or calculator. In this case, you want to find the probability that a credit card holder has a balance less than $5000, so you're looking for \(P(Z < 0.8889)\).

Using a standard normal distribution table or calculator, you can find that \(P(Z < 0.8889)\) is approximately 0.7867 or 78.67%.

So, the probability that a randomly selected credit card holder has a credit card balance of less than $5000 is about 78.67%.