SOLUTION: For the households in Ankara, we know that the average amount of money spent on electricity each month is normally distributed with a mean of 52.30 TRY and a standard deviation of
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Question 1158549: For the households in Ankara, we know that the average amount of money spent on electricity each month is normally distributed with a mean of 52.30 TRY and a standard deviation of 18.23 TRY. Assume that these values represent all households in Ankara. For a sample of 25 households, what is the probability that the average amount spent exceeds 60 TRY? Answer by Boreal(15235) (Show Source):
You can put this solution on YOUR website! This is z> (x bar-mean)/sigma/sqrt(n)
>(60-52.30)/18.23/sqrt(25)
>7.70*5/18.23
>2.11
That probability of z>2.11 is 0.0174
