Question

find the indicated probability and interpret the result. from 1975 through 2020, the mean annual gain of the dow jones industrial average was 652. a random sample of 33 years is selected from this population. what is the probability that the mean gain for the sample was between 500 and 800? assume $sigma = 1539$. 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 $\sigma$ is the population standard - deviation and $n$ is the sample size. Given $\sigma = 1539$ and $n = 33$, we have $\sigma_{\bar{x}}=\frac{1539}{\sqrt{33}}\approx\frac{1539}{5.7446}\approx268$.

Step2: Calculate the z - scores

The z - score formula is $z=\frac{\bar{x}-\mu}{\sigma_{\bar{x}}}$, where $\bar{x}$ is the sample mean, $\mu$ is the population mean.
For $\bar{x}_1 = 500$, $z_1=\frac{500 - 652}{268}=\frac{- 152}{268}\approx - 0.57$.
For $\bar{x}_2 = 800$, $z_2=\frac{800 - 652}{268}=\frac{148}{268}\approx0.55$.

Step3: Find the probabilities corresponding to the z - scores

We use the standard normal distribution table. The probability corresponding to $z_1=-0.57$ is $P(Z < - 0.57)=0.2843$. The probability corresponding to $z_2 = 0.55$ is $P(Z < 0.55)=0.7088$.

Step4: Calculate the probability between the two z - scores

$P(-0.57

Answer:

$0.4245$