Mathematics, 24.03.2020 18:03 lizzyyyvv

The mean output of a certain type of amplifier is 183 watts with a variance of 121. If 78 amplifiers are sampled, what is the probability that the mean of the sample would differ from the population mean by less than 1.7 watts? Round your answer to four decimal places.

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Answer from: bdgray7759

0.8278 = 82.78% probability that the mean of the sample would differ from the population mean by less than 1.7 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 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:

$\mu = 183, \sigma = \sqrt{121} = 11, n = 78, s = \frac{11}{\sqrt{78}} = 1.2455$

Probability that the mean of the sample would differ from the population mean by less than 1.7 watts?

This is the pvalue of Z when X = 183+1.7 =184.7 subtracted by the pvalue of Z when X = 183 - 1.7 = 181.3.

X = 184.7

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

By the Central Limit Theorem

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

$Z = \frac{184.7 - 183}{1.2455}$

Z = 1.365

Z = 1.365 has a pvalue of 0.9139

X = 181.3

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

$Z = \frac{181.3 - 183}{1.2455}$

Z = -1.365

Z = -1.365 has a pvalue of 0.0861

0.9139 - 0.0861 = 0.8278

0.8278 = 82.78% probability that the mean of the sample would differ from the population mean by less than 1.7 watts

Answer from: Quest

step-by-step explanation:

my butt

Answer from: Quest

it would more likely be bald eagle because its has more in the box so about 75%.

step-by-step explanation:

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The mean output of a certain type of amplifier is 183 watts with a variance of 121. If 78 amplifiers...