Question

Suppose that a random sample of size 64 is to be selected from a population with mean 40 and standard deviation 5 . a. What are the mean and standard deviation of the \(\bar{x}\) sampling distribution? Describe the shape of the \(\bar{x}\) sampling distribution. b. What is the approximate probability that \(\bar{x}\) will be within \(0.5\) of the population mean \(\mu\) ? c. What is the approximate probability that \(\bar{x}\) will differ from \(\mu\) by more than \(0.7\) ?

Short answer

a. The mean and standard deviation of the \(\bar{x}\) sampling distribution are 40 and 0.625 respectively. The shape of the \(\bar{x}\) sampling distribution is approximately normal. b. The approximate probability that \(\bar{x}\) will be within 0.5 of the population mean \(\mu\) will have to be calculated using the Z scores. c. The approximate probability that \(\bar{x}\) will differ from \(\mu\) by more than 0.7 will also be calculated using the Z scores.

Step-by-step solution

Finding Mean and Standard Deviation of Sampling Distribution

According to Central Limit Theorem, the mean of the sampling distribution, \(\mu_{\bar{x}}\), is equal to the mean of the population, \(\mu\), i.e., \(\mu_{\bar{x}} = \mu = 40\). The standard deviation of the sampling distribution, \(\sigma_{\bar{x}}\), is equal to the standard deviation of the population, \(\sigma\), divided by the square root of sample size n, i.e., \(\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{5}{\sqrt{64}} = 0.625\)

Describing the Shape of the Sampling Distribution

According to the Central Limit Theorem, if the sample size is large enough (n > 30), the shape of the sampling distribution will be approximately normal, regardless of the shape of the population distribution. So, the shape of the \(\bar{x}\) sampling distribution will be approximately normal.

a: Finding the Approximate Probability that \(\bar{x}\) is Within 0.5 of \(\mu\)

We need to convert the score of 0.5 into a Z score (standard score), as the Z score helps determine how many standard deviations an element is from the mean. The Z score is calculated as \(Z = \frac{\bar{x} - \mu}{\sigma_{\bar{x}}}\). So, we substitute \(\bar{x} = \mu + 0.5\) and \(\bar{x} = \mu - 0.5\) to find the Z scores, calculate the corresponding probabilities from the Standard Normal Distribution table, and then subtract to find the difference, which is the required probability.

b: Finding the Approximate Probability that \(\bar{x}\) Differs from \(\mu\) by more than 0.7

Similarly, we convert the scores of 0.7 into Z scores and find the corresponding probabilities. To find the probability that the sample mean differs by more than 0.7, we find the probabilities at \(Z1 = \frac{\mu + 0.7 - \mu}{\sigma_{\bar{x}}}\) and \(Z2 = \frac{\mu - 0.7 - \mu}{\sigma_{\bar{x}}}\). Subtract each probability from 1 (as these are 'more than' probabilities), and add to get the final probability.