Answer :
To solve this problem, we use the Central Limit Theorem and Z-scores. After calculating the standard error and the Z-score, we find the probability associated with our Z-score. The final probability that the sample mean is less than 39.4 months is approximately 0.3085.
This problem can be solved using the Central Limit Theorem and Z-scores. The Central Limit Theorem tells us that the sampling distribution of the mean of any independent random variable will be normal or near-normal if the sample size is large enough. In this case, 127 is sufficiently large.
First, we calculate the standard deviation of the mean. The standard deviation of the mean, also known as the standard error, is the standard deviation of the population divided by the square root of the sample size. Since the variance is given as 81, the population standard deviation would be √81 = 9. Therefore, the standard error would be 9/√127 = 0.7984 months.
Next, we calculate the Z-score which is the number of standard errors a particular score is from the mean. The Z-score for 39.4 months would be (39.4-39)/0.7984 = 0.5003.
To find the probability that the sample mean is less than 39.4 months, we look at a standard Z-table or use a calculator to find the probability associated with a Z-score of 0.5003. This gives us a probability of 0.6915, however this figure represents the probability of getting a value up to 39.4 months. Since we need the probability of getting less than 39.4 months, we subtract this value from 1, giving us 1-0.6915 = 0.3085.
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