probability statistics

Question

The mean daily production of a herd of cows is assumed to be normally distributed with a mean of 40 liters, and standard deviation of 11.6 liters.

A) What is the probability that daily production is less than 50 liters? Use technology (not tables) to get your probability.

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

B) What is the probability that daily production is more than 69.6 liters? Use technology (not tables) to get your probability.

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

Warning: Do not use the Z Normal Tables...they may not be accurate enough since WAMAP may look for more accuracy than comes from the table. 

Answer 1

Pr(<z) = P(<0.862068965) = .8053 = 80.53%

Pr(>69.6L) = Pr(>z) = Pr(> 2.55172) = 0.0053596 = 0.5360%

using z calculators on line,

z= (50-40)/11.6 = .862068965 for part A

z =(69.6- 40)/11.6 = 29.6/11.6 = 2.55172 for part B

take the number of liters, subtract the mean, then divide by the standard deviation to get the z score

plug the z score into the online z calculator and it gives the areas under the normal curve

even if you're using calculators, it helps to check by using the empirical rule to see if you're in the right range

the empirical rule is 68- 95- 99.7 for percentages corresponding to 1,2 and 3 standard deviations

to the right of z=1 is .5-.64/2 = .5-.34 = .16

to the left of z=1 is 1-.16 = .84% which is > 80.53%

to the right of z=2 is .5- .95/2 = .5- .475 = .025

to the right of z=3 is .5-.997/2 = .5 - .4985 = .0015

.0015 < .005 < .025

to the right of z=2.55 is .005

the online calculator answers look like they're in the right range, using estimates from the empirical rule