Mathematics, 24.06.2020 22:01 sgslayerkingminecraf

Suppose cattle in a large herd have a mean weight of 3181lbs and a standard deviation of 119lbs. What is the probability that the mean weight of the sample of cows would differ from the population mean by greater than 11lbs if 49 cows are sampled at random from the herd

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51.56% probability that the mean weight of the sample of cows would differ from the population mean by greater than 11lbs if 49 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

When the distribution is normal, we use 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 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 question, we have that:

$\mu = 3181, \sigma = 119, n = 49, s = \frac{119}{\sqrt{49}} = 17$

What is the probability that the mean weight of the sample of cows would differ from the population mean by greater than 11lbs if 49 cows are sampled at random from the herd

Lower than 3181 - 11 = 3170 lbs or greater than 3181 + 11 = 3192 lbs. Since the normal distribution is symmetric, these probabilities are equal. So i will find one of them, and multiply by 2.

Probability of mean weight lower than 3170 lbs:

This is 1 subtracted by the pvalue of Z when X = 3170. So

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

By the Central Limit Theorem

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

$Z = \frac{3170 - 3181}{17}$

Z = -0.65

Z = -0.65 has a pvalue of 0.2578

2*0.2578 = 0.5156

51.56% probability that the mean weight of the sample of cows would differ from the population mean by greater than 11lbs if 49 cows are sampled at random from the herd

Answer from: Quest

step-by-step explanation:

the multiplicative inverse of a fraction a/b is b/a. for the multiplicative inverse of a real number, divide 1 by the number. for example, the reciprocal of 5 is one fifth 1/5 or 0.2, and the reciprocal of 0.25 is 1 divided by 0.25, or 4.

Answer from: Quest

i believe it is 0.1m because the measurement of 30.4m limits how precise you can be

you can only be as precise as the least precise measurement

What is the level of precision for the solution to the addition problem below? 6.339 m 0.170 m + 30

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Suppose cattle in a large herd have a mean weight of 3181lbs and a standard deviation of 119lbs. Wha...

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