Mathematics, 04.04.2020 10:56 elenagarcia123

Suppose cattle in a large herd have a mean weight of 1217lbs1217 lbs and a variance of 10,40410,404. What is the probability that the mean weight of the sample of cows would differ from the population mean by more than 11lbs11 lbs if 116116 cows are sampled at random from the herd? Round your answer to four decimal places.

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Answer from: lukelara

0.2460 = 24.60% probability that the mean weight of the sample of cows would differ from the population mean by more than 11 lbs if 116 cows are sampled at random from the herd.

Step-by-step explanation:

To solve this question, 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(which is the square root of the variance) $\sigma$ , 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 normally distributed random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$ .

For a skewed variable, the Central Limit Theorem can also be applied, as long as n is at least 30.

In this problem, we have that:

$\mu = 1217, \sigma = \sqrt{10414} = 102, n = 116, s = \frac{102}{\sqrt{116}} = 9.475$

What is the probability that the mean weight of the sample of cows would differ from the population mean by more than 11 lbs if 116 cows are sampled at random from the herd?

This is 2 multiplied by the pvalue of Z when X = 1217 - 11 = 1206. So

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

By the Central Limit Theorem

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

$Z = \frac{1206 - 1217}{9.475}$

Z = -1.16

Z = -1.16 has a pvalue of 0.1230

2*0.1230 = 0.2460

0.2460 = 24.60% probability that the mean weight of the sample of cows would differ from the population mean by more than 11 lbs if 116 cows are sampled at random from the herd.

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