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 3.5 liters. A) What is the probability that daily production is less than 48 liters? Answer= (Round your answer to 4 decimal places.) B) What is the probability that daily production is more than 47.7 liters? 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

Let X = the mean daily production of a herd of cows

So X ~ Normal ( )

A) What is the probability that daily production is less than 48 liters? Answer= (Round your answer to 4 decimal places.)

Here we want to find P(X < 48)

Let's use excel:

P(X < 48) = "=NORMDIST(48,40,3.5,1)" = 0.9889

B) What is the probability that daily production is more than 47.7 liters?

Here we want to find P (X > 47.5) = 1 - P(X < 47.5) ...( 1 )

P(X < 47.5) = "=NORMDIST(47.5,40,3.5,1)" = 0.9839

Plug this value in equation ( 1 ), we get

P (X > 47.5) = 1 - 0.9839 = 0.0161

u = 40,0 = 3.5

Let's use TI-84 Plus calculator:

Command

2ND>>>VARS>>>DISTR>>>2:normalcdf(

Input the required information

Then click on OK

so we get the following output

B)

Here we want to find P(X > 47.5)

So lower : 47.5

upper: 1111111 ( very large value)

Mean : 40

SD : 3.5

Then highlight Paste and enter then we get the following answer