Mathematics, 30.07.2021 03:10 colemason223

The mean output of a certain type of amplifier is 102102 watts with a standard deviation of 1212 watts. If 6363 amplifiers are sampled, what is the probability that the mean of the sample would differ from the population mean by greater than 3.43.4 watts

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Answer from: kkeith121p6ujlt

0.0244 = 2.44% probability that the mean of the sample would differ from the population mean by greater than 3.4 watts.

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 normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score 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 p-value, we get the probability that the value of the measure is greater than X.

Central Limit Theorem

The Central Limit Theorem establishes 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.

Mean of 102, standard deviation of 12:

This means that $\mu = 102, \sigma = 12$

Sample of 63:

This means that $n = 63, s = \frac{12}{\sqrt{63}}$

What is the probability that the mean of the sample would differ from the population mean by greater than 3.4 watts?

Below 102 - 3.4 = 98.6 or above 102 + 3.4 = 105.4. Since the normal distribution is symmetric, these probabilities are equal, and thus, we find one of them and multiply by two.

Probability the mean is below 98.6.

p-value of Z when X = 98.6. So

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

By the Central Limit Theorem

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

$Z = \frac{98.6 - 102}{\frac{12}{\sqrt{63}}}$

Z = -2.25

Z = -2.25 has a p-value of 0.0122.

2*0.0122 = 0.0244

0.0244 = 2.44% probability that the mean of the sample would differ from the population mean by greater than 3.4 watts.

Answer from: Quest

step-by-step explanation:

Answer from: Quest

company a

step-by-step explanation:

using the linear regression models , it is found that company a pays about $14 more. but using the same regression (linear) company b pays about $10 more.

so company a pays more than b

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The mean output of a certain type of amplifier is 102102 watts with a standard deviation of 1212 wat...