Question

The Dow Jones Industrial Average has had a mean gain of 432 pear year with a standard deviation of 722. A random sample of 40 years is selected. What is the probability that the mean gain for the sample was between 250 and 500?

Answer 1

Answer:

66.98% probability that the mean gain for the sample was between 250 and 500.

Step-by-step explanation:

To solve this problem, it is important to know the Normal probability distribution and the Central limit theorem.

Normal probability distribution

Problems of normally distributed samples can be 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 $\sigma$, a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $\frac{\sigma}{\sqrt{n}}$.

In this problem, we have that:

$\mu = 432, \sigma = 722, n = 40, s = \frac{722}{\sqrt{40}} = 114.16$

What is the probability that the mean gain for the sample was between 250 and 500?

This is the pvalue of Z when X = 500 subtracted by the pvalue of Z when X = 250.

So

X = 500

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

By the Central Limit Theorem

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

$Z = \frac{500 - 432}{114.16}$

$Z = 0.6$

$Z = 0.6$ has a pvalue of 0.7257.

X = 250

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

By the Central Limit Theorem

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

$Z = \frac{250 - 432}{114.16}$

$Z = -1.59$

$Z = -1.59$ has a pvalue of 0.0559.

So there is a 0.7257 - 0.0559 = 0.6698 = 66.98% probability that the mean gain for the sample was between 250 and 500.