- The sampling distribution of the mean has a standard error of \(\sigma_{\bar{X}} = \frac{460}{\sqrt{27}} \approx 88.5270\).
- The z-score for \(\bar{X} = 3920\) is \(-1.63\), yielding a probability \(P(\bar{X} < 3920) = \boxed{0.0516}\).
- The z-score for \(\bar{X} = 4110\) is \(0.52\), yielding a probability \(P(\bar{X} > 4110) = \boxed{0.3015}\).
Explanation
- Understand the sampling distribution of the mean
We are given that the yearly Medicare Hospital Insurance benefits are normally distributed with a mean \(\mu = \$4064\) and a standard deviation \(\sigma = \$460\). We are taking a random sample of \(n = 27\) patients. We want to find the probability that the sample mean \(\bar{X}\) is less than \(\$3920\) and greater than \(\$4110\). To do this, we use the property of the sampling distribution of the sample mean, which is also normally distributed with:
- Mean \(\mu_{\bar{X}} = \mu = 4064\)
- Standard error \(\sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}} = \frac{460}{\sqrt{27}} \approx 88.5270\).
- Calculate the probability for part (a)
For part (a), we want to find \(P(\bar{X} < 3920)\). We calculate the z-score:
\[z = \frac{\bar{X} - \mu_{\bar{X}}}{\sigma_{\bar{X}}} = \frac{3920 - 4064}{88.5270} \approx -1.6266\]
Rounding to \(2\) decimal places as requested, we get \(z = -1.63\).
Using the standard normal distribution table for \(z = -1.63\), we find the cumulative probability:
\(P(Z < -1.63) \approx 0.0516\).
- Calculate the probability for part (b)
For part (b), we want to find \(P(\bar{X} > 4110)\). We calculate the z-score:
\[z = \frac{\bar{X} - \mu_{\bar{X}}}{\sigma_{\bar{X}}} = \frac{4110 - 4064}{88.5270} \approx 0.5196\]
Rounding to \(2\) decimal places, we get \(z = 0.52\).
Using the standard normal distribution table, we find the cumulative probability for \(z = 0.52\) is approximately \(0.6985\).
The probability of being greater than \(4110\) is:
\(P(Z > 0.52) = 1 - P(Z < 0.52) = 1 - 0.6985 = 0.3015\).
Examples
This methodology is vital in fields like healthcare administration or economics, where professionals analyze average costs across patient groups to plan budgets. By understanding how sample means behave, they can predict how likely it is for an average benefit to fall within or outside a expected range, helping in financial forecasting and policy adjustments. For instance, calculating these probabilities ensures institutions are prepared for deviations from the expected average beneficiary cost.