Assume the population distribution of FEV is normal with standard deviation 400 mLi.

(a) find P(E) is n=15 and population mean is 2800 mLi

(B) find P(E) is n=15 and the population mean is 2600mLi

(C) How does P(E) depend on the population mean?

To find P(E) given a normal distribution with a known standard deviation, you can use the standard normal distribution table or calculator to convert the given values into z-scores and then find the corresponding probabilities.

(a) To find P(E) when n=15 and the population mean is 2800 mLi:
1. Calculate the standard error of the mean (SE) by dividing the standard deviation by the square root of the sample size:
SE = 400 / √15 ≈ 103.28 mLi

2. Calculate the z-score using the formula:
z = (x - μ) / SE
where x is the population mean and μ is the sample mean. In this case, x = 2800 mLi and μ = 2800 mLi, so the z-score will be 0.

3. Use the z-score to find the corresponding probability using the standard normal distribution table or calculator. The probability corresponding to a z-score of 0 is 0.5000, or 50%.

Therefore, P(E) is 50% when n=15 and the population mean is 2800 mLi.

(b) To find P(E) when n=15 and the population mean is 2600 mLi:
1. Calculate the standard error of the mean (SE) as done previously:
SE = 400 / √15 ≈ 103.28 mLi

2. Calculate the z-score using the formula:
z = (x - μ) / SE
where x is the population mean and μ is the sample mean. In this case, x = 2800 mLi and μ = 2600 mLi. Plugging in these values, we get:
z = (2600 - 2800) / 103.28 = -1.9342 (rounded to four decimal places)

3. Use the z-score to find the corresponding probability using the standard normal distribution table or calculator. The probability corresponding to a z-score of -1.9342 is approximately 0.0276, or 2.76%.

Therefore, P(E) is approximately 2.76% when n=15 and the population mean is 2600 mLi.

(c) P(E) depends on the population mean because the z-score is computed using the difference between the population mean and the sample mean. As the population mean becomes smaller in relation to the sample mean, the z-score decreases, resulting in a smaller probability P(E). Conversely, if the population mean becomes larger in relation to the sample mean, the z-score increases, resulting in a larger probability P(E). In other words, the probability P(E) is influenced by how much the sample mean deviates from the population mean.