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 σ = 1539.
the probability is 0.4245 (round to four decimal places as needed.)
interpret the result. select the correct choice and fill in the answer box to complete your choice. (round to two decimal places as needed.)
○ a. about % of samples of 33 years will have an annual mean gain between 500 and 800.
○ b. about % of samples of 33 years will have an annual mean gain between 500 and 652.
○ c. about % of samples of 33 years will have an annual mean gain between 652 and 800.
○ d. about % of samples of 46 years will have an annual mean gain between 500 and 800.

Explanation:

Step1: Calculate z - scores

The formula for the z - score is $z=\frac{\bar{x}-\mu}{\frac{\sigma}{\sqrt{n}}}$, where $\mu = 652$, $\sigma = 1539$, $n = 33$.
For $\bar{x}=500$, $z_1=\frac{500 - 652}{\frac{1539}{\sqrt{33}}}=\frac{- 152}{\frac{1539}{5.745}}\approx\frac{-152}{267.9}\approx - 0.57$.
For $\bar{x}=800$, $z_2=\frac{800 - 652}{\frac{1539}{\sqrt{33}}}=\frac{148}{\frac{1539}{5.745}}\approx\frac{148}{267.9}\approx0.55$.

Step2: Find the probability

We want $P(500<\bar{X}<800)=P(-0.57 < Z < 0.55)$.
Using the standard normal table, $P(-0.57Since $\Phi(-z)=1 - \Phi(z)$, $\Phi(-0.57)=1-\Phi(0.57)$.
$\Phi(0.55)\approx0.7088$ and $\Phi(0.57)\approx0.7157$.
So $P(-0.57 < Z < 0.55)=0.7088-(1 - 0.7157)=0.7088 - 0.2843=0.4245$.
The interpretation of the probability $P(500<\bar{X}<800) = 0.4245$ is that about $42.45\%$ of samples of 33 years will have an annual mean gain between 500 and 800.

Answer:

A. About 42.45% of samples of 33 years will have an annual mean gain between 500 and 800.