Question

A population of values has a normal distribution with μ=174.3 and σ=17.6. A random sample of size n=192 is drawn.

A) Find the probability that a single randomly selected value is greater than 172.5
B) Find the probability that a sample of size n=192 is randomly selected with a mean greater than 172.5

Answer 1

A) The probability that a single randomly selected value is greater than 172.5, given a normal distribution with μ=174.3 and σ=17.6, can be calculated using the z-score formula. Substituting the given values into the z-score formula, the probability is approximately 0.6451, or 64.51%.

B) To find the probability that a sample of size n=192 has a mean greater than 172.5, we use the Central Limit Theorem (CLT) for sample means. First, calculate the standard error of the mean (SE) using σ/√n. Then, compute the z-score for the sample mean and find the corresponding probability. After calculations, the probability is approximately 0.9452, or 94.52%.

Step-by-step explanation:

A) To find the probability that a single randomly selected value from the distribution is greater than 172.5, we use the z-score formula:
`\(z = ((X - \mu))/(\sigma)\)`, where (X) is the value (172.5), (mu) is the mean (174.3), and (sigma) is the standard deviation (17.6). Substituting these values,
`\(z = ((172.5 - 174.3))/(17.6)`= -0.1023\). Using a standard normal distribution table or a calculator, we find the area to the right of this z-score, which is approximately 0.6451 or 64.51%.

B) For a sample mean with n=192, the Central Limit Theorem states that the sample mean will be approximately normally distributed regardless of the population's distribution. First, calculate the standard error of the mean (SE) using
`\(SE = (\sigma)/(√(n)) = (17.6)/(√(192))`≈ 1.2723). Then, find the z-score for the sample mean using
`(z = ((X - \mu))/(SE) = ((172.5 - 174.3))/(1.2723)` ≈ -1.4107). Using a standard normal distribution table or a calculator, find the area to the right of this z-score, which is approximately 0.9452 or 94.52%.