Question

The mean daily production of a herd of cows is assumed to be normally distributed with a mean of 30 liters, and standard deviation of 3.3 liters. A) What is the probability that daily production is less than 36.4 liters? Answer= (Round your answer to 4 decimal places.) B) What is the probability that daily production is more than 27.6 liters? Answer= (Round your answer to 4 decimal places.)

Answer 1

a) 0.9738 = 97.38% probability that daily production is less than 36.4 liters.

b) 0.7673 = 76.73% probability that daily production is more than 27.6 liters

Explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the zscore of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

`\mu = 30, \sigma = 3.3`

A) What is the probability that daily production is less than 36.4 liters?

This is the pvalue of Z when X = 36.4. So

`Z = (X - \mu)/(\sigma)`

`Z = (36.4 - 30)/(3.3)`

`Z = 1.94`

`Z = 1.94` has a pvalue of 0.9738.

0.9738 = 97.38% probability that daily production is less than 36.4 liters.

B) What is the probability that daily production is more than 27.6 liters?

This is 1 subtracted by the pvalue of Z when X = 27.6. So

`Z = (X - \mu)/(\sigma)`

`Z = (27.6 - 30)/(3.3)`

`Z = -0.73`

`Z = -0.73` has a pvalue of 0.2327.

1 - 0.2327 = 0.7673

0.7673 = 76.73% probability that daily production is more than 27.6 liters