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 ->  Probability-and-statistics -> 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.      Log On


   

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