If all possible samples of size 16 are drawn from a normal population with mean equal to 50 and standard deviation equal to 5, what is the probability that a sample mean X¯ will fall in the interval from μX¯ −1.9σX¯ to μX¯ −0.4σX¯ ? Assume that the sample means can be measured to any degree of accuracy.
Question
If all possible samples of size 16 are drawn from a normal population with mean equal to 50 and standard deviation equal to 5, what is the probability that a sample mean X¯ will fall in the interval from μX¯ −1.9σX¯ to μX¯ −0.4σX¯ ? Assume that the sample means can be measured to any degree of accuracy.
Solution
To solve this problem, we need to understand the concept of sampling distribution of the mean. The sampling distribution of the mean is a probability distribution of all possible sample means of a given sample size from the same population.
The mean of the sampling distribution (μX¯) is equal to the population mean (μ), which is 50 in this case. The standard deviation of the sampling distribution (σX¯), also known as the standard error, is equal to the population standard deviation (σ) divided by the square root of the sample size (n), which is 5/√16 = 1.25 in this case.
The problem asks for the probability that a sample mean (X¯) will fall in the interval from μX¯ −1.9σX¯ to μX¯ −0.4σX¯. This is equivalent to asking for the probability that a Z score will fall in the interval from -1.9 to -0.4, because Z = (X¯ - μX¯) / σX¯.
We can find these probabilities using a standard normal distribution table or a Z-score calculator. The probability that Z is less than -1.9 is approximately 0.0287, and the probability that Z is less than -0.4 is approximately 0.3446.
Therefore, the probability that a sample mean will fall in the given interval is 0.3446 - 0.0287 = 0.3159, or 31.59%.
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