SOLUTION: In a normal distribution with a mean of 330 and a standard deviation of 60, find the upper and lower bounds that define the middle 80% of the distribution.

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Question 853759: In a normal distribution with a mean of 330 and a standard deviation of 60, find the upper and lower bounds that define the middle 80% of the distribution.
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
 
Hi,
mean of 330 and a standard deviation of 60,
80%, (1-.80)/2 = .10, z = NORMSINV(.10) = -1.2816 , z = NORMSINV(.90) = 1.2816
-1.2816 = (X -330)/60
60(-1.2816) + 330 = 253.104 or 254 to next whole number
60(1.2816) + 330 = 406.896 0r 407 rounded Up
254 & 407 define the upper and lower bounds of the middle 80% of the distribution