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 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.)
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 653
b. about % of samples of 46 years will have an annual mean gain between 500 and 800.
c. about % of samples of 33 years will have an annual mean gain between 500 and 800.
d. about % of samples of 33 years will have an annual mean gain between 653 and 800.

Explanation:

Step1: Calculate the standard error

The formula for the standard error of the mean is $\sigma_{\bar{x}}=\frac{\sigma}{\sqrt{n}}$.
Given $\sigma = 1539$ and $n = 33$, then $\sigma_{\bar{x}}=\frac{1539}{\sqrt{33}}\approx269.77$.

Step2: Calculate the z - scores

The z - score formula is $z=\frac{\bar{x}-\mu}{\sigma_{\bar{x}}}$.
For $\bar{x}=500$, $z_1=\frac{500 - 653}{269.77}\approx - 0.57$.
For $\bar{x}=800$, $z_2=\frac{800 - 653}{269.77}\approx0.54$.

Step3: Find the probabilities using the standard normal distribution

We know that $P(Z\lt z_1)=\Phi(-0.57)$ and $P(Z\lt z_2)=\Phi(0.54)$.
From the standard normal table, $\Phi(-0.57)=0.2843$ and $\Phi(0.54)=0.7054$.

Step4: Calculate the probability between the two z - scores

$P(500\lt\bar{x}\lt800)=P(-0.57\lt Z\lt0.54)=\Phi(0.54)-\Phi(-0.57)$.
$P(-0.57\lt Z\lt0.54)=0.7054 - 0.2843=0.4211$.

Answer:

The probability is $0.4211$.
For the interpretation:
Since we are dealing with a sample of $n = 33$ years, and we found the probability that the sample mean is between $500$ and $800$.
The correct choice is C. About $42.11\%$ of samples of 33 years will have an annual mean gain between 500 and 800.