Question

2. The Hereford Cattle Society says that the mean weight of a one-year-old Hereford bull is 1135 pounds, with a standard deviation of 97 pounds. Suppose 40 bulls are randomly selected and loaded on a train car. Find the probability their combined weight exceeds 46000 pounds. (Hint: The combined weight exceeds 46000 pounds if the average weight exceeds 46000 40 = 1150 pounds.)

Answer 1

16.35% probability their combined weight exceeds 46000 pounds.

Explanation:

To solve this question, we have to understand the normal probability distribution and the central limit theorem.

Normal probability distribution:

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.

Central limit theorem:

The Central Limit Theorem estabilishes that, for a random variable X, with mean
`\mu` and standard deviation
`\sigma`, the sample means with size n of at least 30 can be approximated to a normal distribution with mean
`\mu` and standard deviation
`s = (\sigma)/(√(n))`

In this problem, we have that:

`\mu = 1135, \sigma = 97, n = 40, s = (97)/(√(40)) = 15.34`

Find the probability their combined weight exceeds 46000 pounds.

This is 1 subtracted by the pvalue of Z when
`X = (46400)/(40) = 1150`. So

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

By the Central Limit Theorem

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

`Z = (1150 - 1135)/(15.34)`

`Z = 0.98`

`Z = 0.98` has a pvalue of 0.8365

1 - 0.8365 = 0.1635

16.35% probability their combined weight exceeds 46000 pounds.