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Answer :
Final answer:
To find the requested probabilities, apply the Central Limit Theorem, calculate the sample standard deviation, and apply z-scores. P(ẍ > 90.9) is nearly 1, and P(95.8<ẍ<97.3) is approximately 0.3448.
Explanation:
To determine the probabilities asked in the question, we first need to understand that when the sample size is large (typically n > 30), the Central Limit Theorem (CLT) suggests that the distribution of sample means approximates a normal distribution, regardless if the source population is skewed or not. In this case, it is stated that the distribution of the runtimes of all movies is skewed to the left, however our sizable sample (n=34) allows us to apply the CLT.
The CLT also tells us that the mean (µ) of this distribution is equal to the population mean, which is 96 minutes, and the standard deviation (σ) is the population standard deviation (9.22 minutes) divided by the square root of the sample size (√n). Therefore, the standard deviation for the sample distribution is 9.22/√34 ≈ 1.582.
Now, we calculate z-scores, which measures how many standard deviations an element is from the mean. For P(ẍ > 90.9), calculate the z-score with z = (90.9 - 96)/1.582 ≈ -3.219. The probability that a z-score being greater than -3.219 is very high, close to 1. Therefore, P(ẍ > 90.9) ≈ 1.
For P(95.8<ẍ<97.3), we need to find two z-scores: z1 = (95.8 - 96)/1.582 ≈ -0.126, and z2 = (97.3 - 96)/1.582 ≈ 0.822. Using a standard normal distribution table, find the corresponding probabilities: P(z <-0.126) = 0.4495, and P(z < 0.822) = 0.7943. Subtract these two to get the probabilities between z1 and z2, which is
P(95.8<ẍ<97.3) ≈ 0.7943 - 0.4495 = 0.3448.
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