Question 1161705: Let x be a continuous random variable that follows a distribution skewed to the left with mu=93, and standard deviation=16. Assuming n/N is less than/equal to 0.05, find the probability that the sample mean, x bar, for a random sample of 60 taken from this population will be:
a) less than 85 (round to 4 decimal places)
b)greater than 88.9 (round to 4 decimal places)
I'm not sure if skewed to the left changes anything but I can't find it in my textbook or notes. So mean should be 93, and standard deviation I got as 16/square root of 60, to get 2.0656. To find the z score for mean of 85, I did (85-93)/2.0656. I got a z score of -3.87, which is not in the table of probabilities, so now I'm stuck. Otherwise I would find the probability corresponding to the z scores for 85 and 88.9, and since it wants greater than 88.9, I would do 1 minus the probability for 88.9. I can't seem to understand what I'm doing wrong because this z score isn't in the table. Help please!!! Thank you very much
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! Your work is correct and z=-3.87
on a calculator you get a probability of z <-3.87 as 0.000054 or 0.0001 rounded to four places.
For the other one, z>-4.1/16/sqrt(60). This is z>-4.1*sqrt(60)/16, or z>-1.98, which has probability of 0.9764.
I interpret that as a new problem, the mean of the sample >88.9
The left skew doesn't change anything with a sample size of 60, where the central limit theorem will apply.
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