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
and standard deviation
, the zscore of a measure X is given by:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
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
and standard deviation
, a large sample size can be approximated to a normal distribution with mean
and standard deviation ![s = \frac{\sigma}{\sqrt{n}}](https://tex.z-dn.net/?f=s%20%3D%20%5Cfrac%7B%5Csigma%7D%7B%5Csqrt%7Bn%7D%7D)
In this problem, we have that:
![\mu = 49, \sigma = 5, n = 80, s = \frac{5}{\sqrt{80}} = 0.559](https://tex.z-dn.net/?f=%5Cmu%20%3D%2049%2C%20%5Csigma%20%3D%205%2C%20n%20%3D%2080%2C%20s%20%3D%20%5Cfrac%7B5%7D%7B%5Csqrt%7B80%7D%7D%20%3D%200.559)
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}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
By the Central Limit Theorem
![Z = \frac{X - \mu}{s}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7Bs%7D)
![Z = \frac{50 - 49}{0.559}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B50%20-%2049%7D%7B0.559%7D)
![Z = 1.79](https://tex.z-dn.net/?f=Z%20%3D%201.79)
has a pvalue of 0.9633
1 - 0.9633 = 0.0367
3.67% probability that the mean daily production exceeds 50 pounds.