To solve this problem, we will apply the Central Limit Theorem to determine the probability that the sample mean salary offered to these 85 students is $64,250 or less.

Given Information:

• Population mean ($\mu$) = $64,245
• Population standard deviation ($\sigma$) = $3678
• Sample size ($n$) = 85
• Desired sample mean ($\bar{X}$) = $64,250

Steps:

1. Calculate the Standard Error of the Mean (SEM): $\text{SEM} = \frac{\sigma}{\sqrt{n}}$ Substituting the values: $\text{SEM} = \frac{3678}{\sqrt{85}} \approx \frac{3678}{9.2195} \approx 398.6541$

2. Calculate the Z-score: The Z-score for the sample mean is calculated as: $Z = \frac{\bar{X} - \mu}{\text{SEM}}$ Substituting the values: $Z = \frac{64250 - 64245}{398.6541} \approx \frac{5}{398.6541} \approx 0.0125$

3. Find the Probability: Using the Z-score table, we find the probability that $Z \leq 0.0125$.

   The Z-score of 0.0125 corresponds to a probability very close to 0.5050.

Answer:

The probability that the mean starting salary offered to these 85 students was $64,250 or less is approximately 0.505.

Would you like further details or have any questions?

Here are five related questions to expand your understanding:

1. What if the sample size was larger, say 200 students? How would the probability change?
2. How does the Central Limit Theorem help in approximating probabilities for sample means?
3. If the sample mean desired was $64,000 instead, how would that affect the Z-score and probability?
4. What would happen to the probability if the standard deviation were higher, say $5,000?
5. Can we still apply the Central Limit Theorem if the sample size were only 20?

Tip: Always remember that as the sample size increases, the sampling distribution of the sample mean becomes closer to a normal distribution, even if the population distribution is not normal.