Which is the correct excel formula for the 80th percentile of a distribution that is n(475, 33)? =norm. dist(80,475,33,1) =norm. inv(0.80,475,33) =norm. s.inv((80-475)/33)?

To find the 80th percentile, you have to use one of the inverse CDF functions. NORM.DIST is not one of them; it returns a probability. The 80th percentile the value of the random variable for which 80% of the distributions falls below that value.

NORM.INV(0.80, 475, 33) returns what you want.

NORM.S.INV could also work, but not with the given arguments. NORM.S.INV(0.80) alone will give you the 80th percentile for the standard normal distribution (mean 0, standard deviation 1), approximately 0.8416. From here, you would convert this value to the given distribution using

so that

NORM.INV gives this value directly.

it has changed from the width and the height.

step-by-step explanation: