- The z-score for \(\$47,000\) is \(2.27\), and the corresponding probability is \(P(X > 47,000) = \boxed{0.0116}\).
- The standard error for a sample of \(50\) is \(\approx 721.2489\), the z-score for \(\$35,746\) is \(0.42\), and the corresponding probability is \(P(\bar{X} > 35,746) = \boxed{0.3372}\).
Explanation
- Problem analysis and methodology
We are analyzing the distribution of teacher salaries, which is normally distributed with a mean \(\mu = \$35,441\) and a standard deviation \(\sigma = \$5,100\). To solve the given problems, we will use the concept of the standard normal distribution and z-scores, which measure how many standard deviations a value is from the mean. The formula for the z-score is \(z = \frac{X - \mu}{\sigma}\). We will round all intermediate z-scores to 2 decimal places and the final probabilities to 4 decimal places.
- Part (a): Probability for a single teacher's salary
To find the probability that a randomly selected teacher's salary is greater than \(\$47,000\), we first calculate the corresponding z-score:
\[z = \frac{47,000 - 35,441}{5,100} = \frac{11,559}{5,100} \approx 2.2665\]
Rounding the z-score to 2 decimal places gives \(z = 2.27\).
Next, we find the probability \(P(Z > 2.27)\) using the standard normal distribution table.
The area to the left of \(z = 2.27\) is approximately \(0.9884\).
Thus, the area to the right (the probability of a salary being greater than \(\$47,000\)) is:
\[P(X > 47,000) = P(Z > 2.27) = 1 - 0.9884 = 0.0116\]
- Part (b): Probability for the sample mean
For a sample of \(n = 50\) teachers, the sampling distribution of the sample mean \(\bar{X}\) follows a normal distribution with the same mean \(\mu = \$35,441\) and a standard error \(\sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}}\).
Calculating the standard error:
\[\sigma_{\bar{X}} = \frac{5,100}{\sqrt{50}} \approx \frac{5,100}{7.0711} \approx 721.2489\]
Now, we calculate the z-score for the sample mean \(\bar{X} = 35,746\):
\[z = \frac{\bar{X} - \mu}{\sigma_{\bar{X}}} = \frac{35,746 - 35,441}{721.2489} = \frac{305}{721.2489} \approx 0.4229\]
Rounding the z-score to 2 decimal places gives \(z = 0.42\).
Using the standard normal distribution table, the area to the left of \(z = 0.42\) is approximately \(0.6628\).
Thus, the probability of the sample mean being greater than \(\$35,746\) is:
\[P(\bar{X} > 35,746) = P(Z > 0.42) = 1 - 0.6628 = 0.3372\]
Examples
Understanding the distribution of salaries is vital for researchers and policy makers. For instance, if you are planning a research study, this method helps you determine how likely it is to observe certain sample averages compared to the population, allowing you to assess if a specific group of teachers has significantly different compensation patterns than the national average. By calculating \(\sigma_{\bar{X}} = \frac{\sigma}{\sqrt{n}}\), you effectively account for sample size when estimating the precision of your results.