91Ó°ÊÓ

To find the probability, we need to determine the z-score corresponding to the given number of standard errors from the mean. Let's call the given number of standard errors \(k\). The z-score is a measure of how far the value (in this case, the sample mean) is from the mean (µ), measured in terms of the standard deviation (σ). It is calculated as follows: \[z = \frac{(\underline{X} - \mu)}{(\frac{\sigma}{\sqrt{n}})}\] where: - \(\underline{X}\) is the sample mean, - \(\mu\) is the population mean, - \(\sigma\) is the population standard deviation, and - \(n\) is the sample size. a) For one standard error (k = 1), we are looking for the z-score corresponding to \(\underline{X} = \mu \pm 1\frac{\sigma}{\sqrt{n}}\). b) For two standard errors (k = 2), we are looking for the z-score corresponding to \(\underline{X} = \mu \pm 2\frac{\sigma}{\sqrt{n}}\).

Now that we have the z-scores, we can use the standard normal distribution table or an online calculator to find the corresponding probabilities. a) For one standard error (k = 1): We need to find the probability that the z-score is between -1 and 1. \[P(-1 \leq z \leq 1)\] Using the standard normal distribution table or an online calculator: \[P(-1 \leq z \leq 1) \approx 0.6827\] b) For two standard errors (k = 2): We need to find the probability that the z-score is between -2 and 2. \[P(-2 \leq z \leq 2)\] Using the standard normal distribution table or an online calculator: \[P(-2 \leq z \leq 2) \approx 0.9545\]

Now we can interpret the results and answer the exercise: a) The probability that a random sample mean lies within one standard error of the mean is approximately \(68.27\%\). b) The probability that a random sample mean lies within two standard errors of the mean is approximately \(95.45\%\).