To solve these questions, we need to use the Central Limit Theorem (CLT), which states that for a large enough sample size, the sampling distribution of the sample mean will be approximately normal. Given that the sample size here is 12 (n = 12), we will assume normality for practical purposes.

Given Information:

• Population Mean ($\mu$) = 53,600
• Population Standard Deviation ($\sigma$) = 3,200
• Sample Size (n) = 12

Standard Error of the Mean (SEM)

The standard error of the mean, $\sigma_{\bar{x}}$, for a sample of size $n$ is given by: $\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}}$ Plugging in the values: $\sigma_{\bar{x}} = \frac{3200}{\sqrt{12}} \approx 923.38$

Now, we can calculate the probabilities for each part by using the Z-score formula for sample means: $Z = \frac{\bar{x} - \mu}{\sigma_{\bar{x}}}$

i) Probability that the average salary is less than $50,000

For this part, we calculate the Z-score when $\bar{x} = 50,000$: $Z = \frac{50,000 - 53,600}{923.38} \approx -3.90$ Using standard normal tables, a Z-score of -3.90 corresponds to a very low probability, close to $0.00005$. Thus, the probability that the average salary is less than $50,000 is approximately 0.00005 or 0.005%.

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ii) Probability that the average salary is between $52,000 and $54,000

1. Calculate the Z-score for $52,000: $Z = \frac{52,000 - 53,600}{923.38} \approx -1.73$ From standard normal tables, $P(Z < -1.73) \approx 0.0418$.

2. Calculate the Z-score for $54,000: $Z = \frac{54,000 - 53,600}{923.38} \approx 0.43$ From standard normal tables, $P(Z < 0.43) \approx 0.6664$.

3. Calculate the probability for the range: $P(52,000 < \bar{x} < 54,000) = P(Z < 0.43) - P(Z < -1.73)$ $\approx 0.6664 - 0.0418 = 0.6246$ So, the probability that the average salary is between $52,000 and $54,000 is approximately 0.6246 or 62.46%.

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iii) Probability that the average salary exceeds $55,000

For this part, we calculate the Z-score when $\bar{x} = 55,000$: $Z = \frac{55,000 - 53,600}{923.38} \approx 1.52$ Using standard normal tables, $P(Z < 1.52) \approx 0.9357$.

To find the probability of exceeding $55,000: $P(\bar{x} > 55,000) = 1 - P(Z < 1.52)$ $= 1 - 0.9357 = 0.0643$ So, the probability that the average salary exceeds $55,000 is approximately 0.0643 or 6.43%.

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Summary of Results

1. Probability less than $50,000: 0.00005 (0.005%)
2. Probability between $52,000 and $54,000: 0.6246 (62.46%)
3. Probability exceeding $55,000: 0.0643 (6.43%)

Would you like additional details on any of these steps, or do you have any questions?

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Related Questions

1. How would the probabilities change if the sample size were increased to 30?
2. What effect does a higher standard deviation have on the probabilities calculated above?
3. How is the Central Limit Theorem applicable here, and what conditions make it suitable for this calculation?
4. If we wanted to find the probability that the average salary is more than $51,000, how would we approach this?
5. Can these results help in decision-making for salary expectations in different regions or industries?

Tip

When using the Central Limit Theorem, remember that larger sample sizes provide more accurate approximations for the normal distribution, especially when the population distribution is not normal.