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The mean life of a television set is 121 months with a variance of 256. If a sample of 95 televisions is randomly selected, what is the probability that the sample mean would differ from the true mean by greater than 2.2 months?

User Ianpojman
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Final answer:

To find the probability that the sample mean would differ from the true mean by greater than 2.2 months, we can use the Central Limit Theorem. First, calculate the standard deviation of the sampling distribution of the sample mean. Next, convert the difference of 2.2 months into a z-score. Finally, find the probability of obtaining a z-score greater than the calculated value.

Step-by-step explanation:

To find the probability that the sample mean would differ from the true mean by greater than 2.2 months, we can use the Central Limit Theorem. Since the sample size is large (n = 95) and the population variance is known (variance = 256), the sampling distribution of the sample mean will be approximately normally distributed.

First, we calculate the standard deviation of the sampling distribution of the sample mean, which is equal to the population standard deviation divided by the square root of the sample size. In this case, the standard deviation of the sampling distribution is square root of variance = 16.

Next, we convert the difference of 2.2 months into a z-score using the formula z = (x - μ) / σ, where x is the difference, μ is the true mean, and σ is the standard deviation of the sampling distribution. In this case, z = (2.2 - 0) / 16 = 0.1375.

Finally, we find the probability of obtaining a z-score greater than 0.1375 using a standard normal distribution table or a calculator. In this case, the probability is approximately 0.4447, or 44.47%. Therefore, the probability that the sample mean would differ from the true mean by greater than 2.2 months is 44.47%.

User Sanchezcl
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