Question

Suppose we are interested in the Annual Salary of Electronics Associated, Inc. (EAI) managers and we find that for a sample of n = 30 managers, there was 0.5034 probability of obtaining a sample mean within ± $500 of the population mean (see the figure below). The title of the diagram is "Sampling Distribution of x". A bell-shaped curve divided into three areas is above a horizontal axis labeled x. The text sigma sub x bar = 730.30 is on the figure. The horizontal axis has three tick marks. In the order they appear, from the left side of the figure to the right, they are: 51,300, 51,800, and 52,300. The label 51,800 is below the maximum point on the curve and in the center of the horizontal axis. The first area under the curve is to the left of 51,300, is shaded, and is labeled P(x < 51,300). The second area under the curve is between 51,300 and 52,300, is shaded, and is labeled P(51,300 ≤ x ≤ 52,300). The third area under the curve is to the right of 52,300 and is shaded. There is no label. For parts (a) and (b), let the population mean be 51,800 and = 4,000. (Round your answers to four decimal places.)

(a) What is the probability that x is within ±$500 of the population mean if a sample of size 50 is used?
(b) Answer part (a) for a sample of size 100.

Answer 1

(a) The probability that x is within ±$500 of the population mean for a sample of size 50 is approximately 0.9562.

(b) The probability that x is within ±$500 of the population mean for a sample of size 100 is approximately 0.9974.

Step-by-step explanation:

(a) For a sample size of 50, the standard error (SE) is calculated using the formula SE = σ / √n, where σ is the population standard deviation and n is the sample size. In this case, SE = 4000 / √50 ≈ 565.6854.

The margin of error (ME) is then $500, so the z-score for ±$500 is found using the formula z = ME / SE, resulting in z ≈ 0.8842. Using a standard normal distribution table, the probability P(-0.8842 < Z < 0.8842) is approximately 0.9562.

(b) For a sample size of 100, the SE is now SE = 4000 / √100 = 400. The z-score for ±$500 is calculated as z ≈ 500 / 400 = 1.25. The probability P(-1.25 < Z < 1.25) from the standard normal distribution table is approximately 0.7940.

To find the probability within ±$500, we subtract the tails from 1: 1 - 2(0.7940) ≈ 0.9974. Therefore, the probability is approximately 0.9974 for a sample size of 100.

Answer 2

The question asks for the probability of a sample mean being within ±$500 of the population mean for two different sample sizes. By using the Central Limit Theorem and calculating Z-scores, this probability can be determined for both sample sizes of 50 and 100.

Step-by-step explanation:

The problem concerns calculating the probability that the sample mean salary of Electronics Associated, Inc. (EAI) managers is within ±$500 of the population mean for different sample sizes. To solve this, we utilize the Central Limit Theorem which states that the sampling distribution of the sample mean will be approximately normally distributed if the sample size is large enough.

For the sample size of 50, we can calculate the standard error of the mean by dividing the given population standard deviation (σ = 4,000) by the square root of the sample size (n). The probability can be found using the standard normal distribution and Z-scores.

(a) The new standard error for a sample size of 50 is σ/√n = 4,000/√50 = 4,000/7.071 = 565.685. We then calculate the Z-score for ±$500 as Z = 500/565.685 = 0.884. The probability that x is within ±$500 of the mean can be obtained from the standard normal distribution table or using a statistical software. Assuming a normal distribution, we double the one-tailed probability for Z = 0.884 to get the two-tailed probability.

(b) Similarly, for a sample size of 100, we calculate the standard error as σ/√n = 4,000/√100 = 4,000/10 = 400. Then a Z-score for ±$500 is Z = 500/400 = 1.25. Again, find the two-tailed probability for this Z-score using the same method as above.