Question 1031436: It is given that population mean is 200 with a standard deviation of 40. A sample of size 50 is taken with mean 60 then what is the probability of observing a mean value of upto 60(assume that the underlying population is not normal).
Norm.dist(60,200,40/sqrt(50), true)
1-Norm.dist(60,200,40,true)
Norm.dist(60,200,40,true)
None of these
I think the answer is (C). IS that correct?
Answer by ewatrrr(24785)   (Show Source):
You can put this solution on YOUR website! •If the population is not normally distributed, but the sample size is sufficiently large, then the sample means will have an approximately normal distribution. Some books define sufficiently large as at least 30 and others as at least 31.
The formula for a z-score when working with the sample means is: z = 
P(x-bar <60) = P(z < ) =Norm.dist(60,200,40/sqrt(50), true)
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