Given that X is normally distributed with population mean 50

Given that X is normally distributed with population mean 50 and population standard deviation 4. Compute the following for n = 25. Mean and variance of X^bar (i.e., sample mean). P(X^bar lessthanorequalto 49) P(49 lessthanorequalto x^bar lessthanorequalto 51.5)

Solution

a) mean of sample mean = mean = 50

variance of sample mean = variance / n

= 4²/25 = 0.64

b) For sample mean less than 49.

z-score = (49-50)/std = -1/sqrt(0.64) = -1.25

You look into the z-score to find the area under the distribution curve that is to the left of a z-score of -1.25 which is 0.1056.

You can find z-table to do this anywhere. I used the table here: http://www.stat.ufl.edu/~athienit/Tables/Ztable.pdf

Hence, the probability is 10.56%

c) For sample mean less than 49.

z-score = (49-50)/std = -1/sqrt(0.64) = -1.25

You look into the z-score to find the area under the distribution curve that is to the left of a z-score of -1.25 which is 0.1056.

For sample mean less than 51.5.

z-score = (51.5-50)/std = 1.5/sqrt(0.64) = 1.875

Since, z-table has values with a precision of 2 decimal places, we take it as 1.87.

You look into the z-score to find the area under the distribution curve that is to the left of a z-score of 1.87 which is 0.9693.

But we need to subtract the area under the curve to the left of 49. Therefore, 0.9693 - 0.1056 = 0.8637.

Hence, the probability is 86.37%.