Final answer:
A) The probability that the sample mean will be greater than 50 is approximately 0.0228. B) The probability that the sample mean will be less than 56 is approximately 0.2119. C) The probability that the sample mean will be within 5 points of the population mean is approximately 0.6826.
Step-by-step explanation:
A) To find the probability that the sample mean will be greater than 50, we need to calculate the z-score. The formula for the z-score is (X - μ) / (σ / sqrt(n)), where X is the sample mean, μ is the population mean, σ is the population standard deviation, and n is the sample size. Plugging in the given values, we have (50 - 60) / (20 / sqrt(16)) = -2. The probability that the sample mean will be greater than 50 can be found by looking up the corresponding z-score in the standard normal distribution table. For a z-score of -2, the probability is approximately 0.0228.
B) To find the probability that the sample mean will be less than 56, we use the same formula and calculate the z-score: (56 - 60) / (20 / sqrt(16)) = -0.8. Looking up the z-score in the standard normal distribution table, we find that the probability is approximately 0.2119.
C) To find the probability that the sample mean will be within 5 points of the population mean, we need to find the z-scores for both 55 and 65 using the same formula. For 55: (55 - 60) / (20 / sqrt(16)) = -1. The z-score for 65 is (65 - 60) / (20 / sqrt(16)) = 1. The probability of the sample mean being within 5 points of the population mean is the difference between the cumulative probabilities for the two z-scores: p(-1 < Z < 1) = p(Z < 1) - p(Z < -1). Using the standard normal distribution table, we find that p(Z < 1) is approximately 0.8413 and p(Z < -1) is approximately 0.1587. Therefore, the probability is 0.8413 - 0.1587 = 0.6826.