Question

It is known that Ayrshire cattle produce a mean of 49 pounds of milk per day, with a standard deviation of 5 pounds. A dairy farmer has a herd of 80 Ayrshire cattle. Answer questions 21-23 based on this information and assuming that this is a random sample of Ayrshire cattle. For this herd of 80 cattle, what is the probability that the mean daily production exceeds 50 pounds?

Answer 1

Answer:

3.67% probability that the mean daily production exceeds 50 pounds.

Step-by-step explanation:

To solve this problem, we need 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:

$Z = \frac{X - \mu}{\sigma}$

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 $\sigma$, a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 49, \sigma = 5, n = 80, s = \frac{5}{\sqrt{80}} = 0.559$

For this herd of 80 cattle, what is the probability that the mean daily production exceeds 50 pounds?

This probability is 1 subtracted by the value of Z when X = 50. So

$Z = \frac{X - \mu}{\sigma}$

By the Central Limit Theorem

$Z = \frac{X - \mu}{s}$

$Z = \frac{50 - 49}{0.559}$

$Z = 1.79$

$Z = 1.79$ has a pvalue of 0.9633

1 - 0.9633 = 0.0367

3.67% probability that the mean daily production exceeds 50 pounds.