Suppose that battery lives are normally distributed with a mean of 11.8 hours and a standard deviation of 1.9 hours. What is the minimum sample size that would be required so that the probability of obtaining a sample mean below 11.2 is less than 1%?
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Mean = 11.8 hours
Stdev = 1.9 hours
P(X<11.2) < .01
P( Z < (11.2-11.8)/(1.9/sqrt(n)) < 0.01
Taking the NORMSINV formula we have a Z of -2.33
So, (11.2-11.8)/(1.9/sqrt(n)) = -2.33
((11.2-11.8)/ 2.33)*((11.2-11.8)/ 2.33) = 1.9*1.9/n
n = (1.9*1.9)/(((11.2-11.8)/ 2.33)*((11.2-11.8)/ 2.33)) = 54.44 or 55
The minimum sample size is 55
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