Answer:
The probability that the mean of the sample would differ from the population mean by less than 0.88 months is approximately 0.7286. Please check the explanation below to help you remember the steps.
Explanation:
The problem is asking for the probability that the sample mean differs from the population mean by less than 0.88 months. This is a problem of sampling distributions, and we can use the Central Limit Theorem to solve it.
First, we need to calculate the standard deviation of the sampling distribution (also known as the standard error). The standard error (SE) is given by the formula:
SE = a/n
Where o is the standard deviation of the population and n is the sample size. The standard deviation of the population is the square root of the variance, so o = /64 = 8. The sample size n is 100. So,
SE = 8/100 = 0.8
Next, we need to calculate the z-score for the difference of 0.88 months. The z-score is given by the formula:
z = x-u / SE
where x is the sample mean, u is the population mean, and SE is the standard error. In this case, we want the probability that the sample mean differs from the population mean by less than 0.88 months, so
x - u = +0.88/0.8 = +1.1
The probability that a z-score is less than 1.1 is 0.8643 and the probability that a z-score is less than -1.1 is 0.1357 (these values are found in a standard z-table). The probability that the z-score is between -1.1 and 1.1 is the difference of these two probabilities:
P(-1.1 ≤ z ≤ 1.1) = P(z ≤ 1.1) - P(z ≤ -1.1) = 0.8643 - 0.1357
So, the probability that the mean of the sample would differ from the population mean by less than 0.88 months is approximately 0.7286.