Suppose the population of all public Universities shows the annual parking fee per student is \$110 with a standard deviation of \$18. If a random sample of size 49 is drawn from the population, the probability of drawing a sample with a sample mean between \$100 and \$115 is?Select one:a.0.9738.b.0.4738.c.0.0262.d.0.6103.e.0.1103.
Question
Suppose the population of all public Universities shows the annual parking fee per student is $110 with a standard deviation of $18. If a random sample of size 49 is drawn from the population, the probability of drawing a sample with a sample mean between $100 and $115 is?Select one:a.0.9738.b.0.4738.c.0.0262.d.0.6103.e.0.1103.
Solution
To solve this problem, we need to use the concept of Z-scores in statistics. The Z-score is a measure of how many standard deviations an element is from the mean.
Step 1: Calculate the standard error (SE). The standard error is the standard deviation divided by the square root of the sample size.
SE = σ/√n = 18/√49 = 18/7 = 2.57
Step 2: Calculate the Z-scores. The Z-score is the difference between the sample mean and the population mean divided by the standard error.
Z1 = (X1 - μ) / SE = (100 - 110) / 2.57 = -3.88 Z2 = (X2 - μ) / SE = (115 - 110) / 2.57 = 1.94
Step 3: Look up the Z-scores in the Z-table to find the probabilities.
P(Z1) = P(Z < -3.88) = 0.0001 (approximately) P(Z2) = P(Z < 1.94) = 0.9738
Step 4: The probability of the sample mean being between 115 is the difference between these two probabilities.
P(100 < X < 115) = P(Z2) - P(Z1) = 0.9738 - 0.0001 = 0.9737
So, the closest answer is a. 0.9738.
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