The mean daily production of a herd of cows is assumed to be normally distributed with a mean of 31 liters, and standard deviation of 7.6 liters.

Question

A) What is the probability that daily production is less than 39.8 liters?

Answer= (Round your answer to 4 decimal places.)

B) What is the probability that daily production is more than 38.3 liters?

Answer= (Round your answer to 4 decimal places.)

Answer 1

You can use TI-84 calculator, Excel, or by-hand to solve this. Here's how you'd do it each way:

TI-84:

2nd -> Vars -> normalCDF

normalCDF(-10^99, 39.8, 31, 7.6) = 0.8765 for part A

normalCDF(38.3, 10^99, 31, 7.6) = 0.1684 for part B

Excel:

= NORM.DIST(39.8, 31, 7.6) = 0.8765 for part A

= 1 - NORM.DIST(38.3, 31, 7.6) = 0.1684 for part B

By-hand:

First, we find z-score by z = (x - mu) / sigma and then we look up the probability on the z-table.

Part A:

z = (39.8 - 31) / 7.6 = 1.16 and probability on the z-table is 0.8770

Part B:

z = (38.3 - 31) / 7.6 = 0.96 and probability on the z-table is 0.8315, but we have to subtract from 1 since it's an right hand probability. So, probability is 1 - 0.8315 = 0.1685