The mean starting salary offered to graduating students with a certain major in a recent year was $\$ 64,215$, with a standard deviation of $\$ 3712$. A random sample of 85 of that year's graduating students with the major has been selected.
What is the probability that the mean starting salary offered to these 85 students was $\$ 63,750$ or more?
Carry your intermediate computations to at least four decimal places. Round your answer to at least three decimal places.
Answer
Final Answer: The probability that the mean starting salary offered to these 85 students was \$63,750 or more is approximately \(\boxed{0.876}\).
Step 1 ：We are given the population mean (\(\mu = \$64,215\)), the population standard deviation (\(\sigma = \$3,712\)), and the sample size (\(n = 85\)). We are asked to find the probability that the sample mean (\(\bar{x}\)) is \$63,750 or more.
Step 2 ：The first step is to calculate the standard error of the mean (SE), which is the standard deviation of the sampling distribution of the mean. The formula for the standard error of the mean is \(\sigma/\sqrt{n}\).
Step 3 ：Next, we need to calculate the z-score, which measures how many standard errors away from the mean our sample mean is. The formula for the z-score is \((\bar{x} - \mu)/SE\).
Step 4 ：Finally, we can use the z-score to find the probability that the sample mean is \$63,750 or more. This is equivalent to finding the area under the standard normal curve to the right of the z-score. We can use the standard normal distribution table to find this probability.
Step 5 ：Using the given values, we find that the standard error (SE) is approximately 402.6229, the z-score is approximately -1.1549, and the probability (p) is approximately 0.8759.
Step 6 ：Final Answer: The probability that the mean starting salary offered to these 85 students was \$63,750 or more is approximately \(\boxed{0.876}\).
