Answer :
The probability that a random sample of size 63 has a mean value greater than 73.9 given a population with a normal distribution, mean of 65 and standard deviation of 59.1, is 0.1151 or 11.51%.
The subject of this question is Statistics, specifically it's to do with Probability in a Normal Distribution. You are being asked to find the probability that a random sample of size 63 has a mean greater than 73.9 when the population mean (μ) is 65 and the standard deviation (σ) is 59.1.
First, the standard error (SE) needs to be calculated, which is σ/√n, where n is the sample size. In this case, SE = 59.1 / √63 = 7.45 (rounded to two decimal places).
Next, let's calculate the Z-score, which is (X - μ) / SE, where X is the sample mean. Here, Z = (73.9 - 65) / 7.45 = 1.2 (rounded to one decimal place).
Finally, we use the Standard Normal Distribution table (also known as the z-table) to find the probability that Z is greater than 1.2. However, these tables generally only have values for P(Z < x). So, in this specific case, to get P(Z > 1.2), we subtract the probability shown on the table for Z = 1.2 from 1. If you check a standard z-table, you'll see that for Z=1.2, the value is 0.8849. Hence, P(Z > 1.2) = 1 - 0.8849 = 0.1151 (rounded to four decimal places).
Therefore, the probability that a sample of size 63 has a mean greater than 73.9 is 0.1151 or 11.51%.
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