A normally distributed population with 200 elements has a mean of 60 and a standard deviation of 10. The probability that the mean of a sample of 25 elements taken from this population will be smaller than 56 is:

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To find the probability that the mean of a sample of 25 elements taken from a normally distributed population will be smaller than 56, we can use the Central Limit Theorem.

The Central Limit Theorem states that the sampling distribution of the sample mean approaches a normal distribution as the sample size becomes large, regardless of the shape of the population distribution.

In this case, since the population is normally distributed, we can assume that the sampling distribution of the sample mean will also be normally distributed. The mean of the sampling distribution of the sample mean will be the same as the population mean, which is 60. The standard deviation of the sampling distribution of the sample mean, also known as the standard error, can be calculated as the population standard deviation divided by the square root of the sample size:

Standard Error (SE) = population standard deviation / sqrt(sample size)
SE = 10 / sqrt(25)
SE = 10 / 5
SE = 2

Now, we want to find the probability that the sample mean is smaller than 56. We can standardize this value by subtracting the population mean and dividing by the standard error:

Z = (sample mean - population mean) / standard error
Z = (56 - 60) / 2
Z = -4 / 2
Z = -2

Using a Z-table or a statistical software, we can find the probability associated with a Z-score of -2. According to the standard normal distribution, the probability of obtaining a Z-score less than -2 is approximately 0.0228.

Therefore, the probability that the mean of a sample of 25 elements taken from this normally distributed population will be smaller than 56 is approximately 0.0228, or 2.28%.