91Ó°ÊÓ

We know the sample size n=100 and the population standard deviation \(\sigma=10\). We need to find the probability that the sample mean \(\bar{x}\) is within 2 of the population mean \(\mu\). Additionally, we need to determine the values for which \(\bar{x}\) will lie within \(\mu\) 95% of the time, and the value from which \(\bar{x}\) will be farther than \(\mu\) 0.3% of the time.

The standard error (SE) can be calculated using the formula SE = \(\sigma /\sqrt{n}\), where \(\sigma\) is the population standard deviation and \(n\) is the sample size. In this case, \(\sigma=10\) and \(n=100\). Thus, SE = \(10/\sqrt{100}=1\). This standard error means that on average, the sample mean should typically deviate from the population mean by 1.

The z-score represents the number of standard errors a data point is away from the mean. To find the probability that the sample mean will be within 2 of the population mean, calculate the z-score corresponding to 2 using the formula z = (X - \(\mu\))/SE. Since we want the sample mean to be within 2 of the population mean, X will be \(\mu\)+2 or \(\mu\)-2. The z-scores associated with these are (+2-0)/1 = 2 and (-2-0)/1 = -2. We want the probability that the sample mean is between these z-scores. From the standard normal table, the value corresponding to a z-score of 2 is 0.9772 and for -2 is 0.0228. So, P(-2 < Z < 2) = 0.9772 - 0.0228 = 0.9544. This means there is approximately a 95.44% chance that the sample mean will be within 2 of the population mean.

To complete the statements, we first rewind to the z-scores. For 95% of the time (or 0.9500 probability), the z-scores from the standard normal table are approximately -1.96 and 1.96. Therefore, \(\bar{x}\) will be within +- \(1.96*SE =1.96*1 = 1.96\) of \(\mu\). Next, for 0.3% of the time (or 0.0030 probability), the z-scores are approximately -2.75 and 2.75 (since probability for z-scores -2.75 and 2.75 is 0.003). Hence, \(\bar{x}\) will be farther than \(2.75*1 = 2.75\) from \(\mu\).