Question

The mean daily production of a herd of cows is assumed to be normally distributed with a mean of 37 liters, and standard deviation of 12.3 liters. A) What is the probability that daily production is between 14.9 and 60.4 liters? Do not round until you get your your final answer. 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

Solution:

Given:

The mean daily production of a herd of cows is assumed to be normally distributed with a mean of 37 liters, and standard deviation of 12.3 liters.

Thus mean = and Standard deviation =

We have to find  the probability that daily production is between 14.9 and 60.4 liters.

That is:

P( 14.9 < X < 60.4 ) =...........?

Use following Excel command:

=NORM.DIST( upper x value , mean , std_dev , TRUE) - NORM.DIST( Lower x value , mean , std_dev , TRUE)

=NORM.DIST(60.4,37,12.3,TRUE)-NORM.DIST(14.9,37,12.3,TRUE)

=0.93525529

=0.9353

Thus

P( 14.9 < X < 60.4 ) = 0.9353

We can use TI 84 plus calculator to find the probability:

Press 2ND and VARS

Select normalcdf(

Enter numbers

Click on Paste and press Enter two times.

Thus we get:

P( 14.9 < X < 60.4 ) = 0.9352554085

P( 14.9 < X < 60.4 ) = 0.9353