The mean starting salary offered to graduating students with a certain major in a recent year was $64,260, with a standard deviation of $3,712.

Question

The mean starting salary offered to graduating students with a certain major in a recent year was $64,260, with a standard deviation of $3,712. A random sample of 80 of that year's graduating students with the major has been selected.What is the probability that the mean starting salary offered to these 80 students was $65,000 or more?

Answer 1

Given the data, the mean starting salary offered to graduating students with a specific major in a recent year was $64,260, with a standard deviation of $3,712. We are interested in finding the probability that the mean starting salary for a random sample of 80 students from that year's graduates will be $65,000 or more.

To calculate this probability, we'll use the Central Limit Theorem, which states that for a large enough sample size, the sampling distribution of the sample means will be approximately normally distributed, regardless of the shape of the original population distribution.

First, we find the standard error of the mean (SE) using the formula:

SE = σ / √n

where σ is the standard deviation of the population, and n is the sample size.

In our case, SE = 3712 / √80 ≈ 415.910.

Next, we standardize the sample mean using the Z-score formula:

Z = (x̄ - μ) / SE

where x̄ is the sample mean, μ is the population mean, and SE is the standard error.

For the given scenario, we have:

Z = (65000 - 64260) / 415.910 ≈ 17.78.

Now, to find the probability corresponding to this Z-score, we consult a standard normal distribution table or use a calculator. However, it's essential to note that the Z-score is exceptionally high, suggesting an extreme value. Consequently, the probability of obtaining a mean starting salary of $65,000 or more, given the population mean and standard deviation, is practically negligible.

In real-world terms, it would be highly improbable to have a sample mean starting salary of $65,000 or above when the population mean is $64,260, and the standard deviation is $3,712.

In conclusion, the probability of the mean starting salary offered to the 80 students being $65,000 or more is almost zero, making it practically impossible in this context.

Answer 2

use the Central Limit Theorem, which states that the distribution of sample means from a large enough sample will be approximately normally distributed, regardless of the shape of the original population.

Given:

Population mean (μ) = $64,260

Population standard deviation (σ) = $3,712

Sample size (n) = 80

First, we need to find the standard error of the mean (SE), which is the standard deviation of the sampling distribution of the mean:

SE = σ / √n

SE = $3,712 / √80 ≈ $415.7559 (rounded to 4 decimal places)

Next, we calculate the Z-score for the mean starting salary of $65,000:

Z = (X - μ) / SE

Z = ($65,000 - $64,260) / $415.7559 ≈ 1.7784 (rounded to 4 decimal places)

Now, we need to find the probability that the Z-score is greater than 1.7784 using a standard normal distribution table or calculator:

P(Z > 1.7784) ≈ 0.0379 (rounded to 4 decimal places)

Therefore, the probability that the mean starting salary offered to these 80 students was $65,000 or more is approximately 0.0379 or 3.79%.