Question

find the indicated probability and interpret the result. from 1975 through 2020, the mean annual gain of the dow jones industrial average was 653. a random sample of 32 years is selected from this population. what is the probability that the mean gain for the sample was between 400 and 800? assume $sigma = 1540$. the probability is
(round to four decimal places as needed.)

Explanation:

Step1: Calculate the standard error

The formula for the standard error of the mean is $\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}$, where $\sigma = 1540$ and $n = 32$. So, $\sigma_{\bar{x}}=\frac{1540}{\sqrt{32}}\approx272.23$.

Step2: Calculate the z - scores

The z - score formula is $z=\frac{\bar{x}-\mu}{\sigma_{\bar{x}}}$. For $\bar{x}_1 = 400$, $z_1=\frac{400 - 653}{272.23}\approx - 0.93$. For $\bar{x}_2 = 800$, $z_2=\frac{800 - 653}{272.23}\approx0.54$.

Step3: Find the probability

We want $P(-0.93

Answer:

$0.5292$