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I'm trying to do the following problem:

The mean of a sample size n = 25 is used to estimate the mean of a very large population, consisting of the periods of time during which a person over 65 can concentrate, which has a standard deviation equal to 2 ,4 minutes. What can we say about the probability of the error being less than 1.2 minutes, if we apply:

b) the central limit theorem.

Here is my attempt:

The Central Limit Theorem states that the sampling distribution of the sample mean will be approximately normally distributed if the sample size is sufficiently large.

We have that,

Population standard deviation, $\sigma = 2.4$

Sample size, $n = 25$

We calculate the standard error (SE) of the mean:

$SE = \frac{\sigma}{\sqrt{n}} = \frac{2.4}{\sqrt{25}} = 0.48$

Now, we want to find the probability that the sample mean is within 1.2 minutes of the population mean. We convert this to a Z-score:

$Z = \frac{1.2}{0.48} = 2.5$

Looking at the table for $Z = 2.5$, I have $0.9938$.

However I should get 0.9876 as answer. Can someone help me?

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This is a 2 tailed problem. It is asking for the error to be less than 1.2 minutes, that is in the positive and negative direction. ie: P(23.8 < x < 26.2)

What you did was find P(x < 26.2)

Since the normal distribution is symmetric, you can take 1 - 0.9938 to find P(23.8 < x)

So you will get: P(23.8 < x < 26.2) = (0.9938 - (1 - 0.9938)) = 0.9876

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