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.

Algebra.Com
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)   (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
RELATED QUESTIONS

In a normal distribution with a mean of 330 and a standard deviation of 60 what are the... (answered by ewatrrr)
consider that the historical return of an investment follows a normal distribution with a (answered by stanbon)
If random samples of size 9 is taken from a normal distribution with mean 50 and standard (answered by stanbon)
Suppose X is subject to normal distribution with the population mean 7, and the standard... (answered by stanbon)
For a normal distribution with mean of 100 and standard deviation of 20, find the... (answered by stanbon)
In a National Achievement Test, the mean was found to be 75 and the standard deviation... (answered by ikleyn)
Define a standard normal distribution by identifying its shape and the numeric values for (answered by solver91311)
suppose in a normal distribution mean = 75 and standard deviation = 10 . find the data... (answered by ewatrrr)
Suppose you have a normal distribution of values with a mean of 70 and a standard... (answered by mathmate)