Answer:
The probability that the sample mean will be within 0.5 of the population mean is 0.3328.
Step-by-step explanation:
It is provided that a random variable <em>X</em> has mean, <em>μ</em> = 50 and<em> </em>standard deviation, <em>σ</em> = 7.
A random sample of size, <em>n</em> = 36 is selected.
According to the Central Limit Theorem if we have a population with mean μ and standard deviation σ and we take appropriately huge random samples (n ≥ 30) from the population with replacement, then the distribution of the sample mean will be approximately normally distributed.
Then, the mean of the distribution of sample mean is given by,
![\mu_{\bar x}=\mu=50](https://tex.z-dn.net/?f=%5Cmu_%7B%5Cbar%20x%7D%3D%5Cmu%3D50)
And the standard deviation of the distribution of sample mean is given by,
![\sigma_{\bar x}=\frac{\sigma}{\sqrt{n}}=\frac{7}{\sqrt{36}}=1.167](https://tex.z-dn.net/?f=%5Csigma_%7B%5Cbar%20x%7D%3D%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D%3D%5Cfrac%7B7%7D%7B%5Csqrt%7B36%7D%7D%3D1.167)
So, the distribution of the sample mean of <em>X</em> is N (50, 1.167²).
Compute the probability that the sample mean will be within 0.5 of the population mean as follows:
![P(|\bar X-\mu_{\bar x}|\leq 0.50)=P(-0.50](https://tex.z-dn.net/?f=P%28%7C%5Cbar%20X-%5Cmu_%7B%5Cbar%20x%7D%7C%5Cleq%200.50%29%3DP%28-0.50%3C%28%5Cbar%20X-%5Cmu_%7B%5Cbar%20x%7D%29%3C0.50%29)
![=P(\frac{-0.50}{1167}](https://tex.z-dn.net/?f=%3DP%28%5Cfrac%7B-0.50%7D%7B1167%7D%3C%5Cfrac%7B%5Cbar%20X-%5Cmu_%7B%5Cbar%20x%7D%7D%7B%5Csigma_%7B%5Cbar%20x%7D%7D%3C%5Cfrac%7B0.50%7D%7B1.167%7D%29%5C%5C%5C%5C%3DP%28-0.43%3CZ%3C0.43%29%5C%5C%5C%5C%3DP%28Z%3C0.43%29-P%28Z%3C-0.43%29%5C%5C%5C%5C%3D0.66640-0.33360%5C%5C%5C%5C%3D0.3328)
Thus, the probability that the sample mean will be within 0.5 of the population mean is 0.3328.