Question

1. Suppose that the average outstanding credit balance for young couples is $650 with a standard deviation of $420. In an SRS of 100 couples, what is the probability that the mean outstanding credit balance exceeds $700?

Answer 1

Answer:

11.70% probability that the mean outstanding credit balance exceeds $700

Step-by-step explanation:

To solve this question, we need to understand the normal probability distribution and the central limit theorem.

Normal probability distribution

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$, the sample means with size n of at least 30 can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 650, \sigma = 420, n = 100, s = \frac{420}{\sqrt{100}} = 42$

In an SRS of 100 couples, what is the probability that the mean outstanding credit balance exceeds $700?

This is 1 subtracted by the pvalue of Z when X = 700. So

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{700 - 650}{42}$

$Z = 1.19$

$Z = 1.19$ has a pvalue of 0.8830

1 - 0.8830 = 0.1170

11.70% probability that the mean outstanding credit balance exceeds $700