Mathematics, 21.03.2020 02:17 isabelperez063

An operation manager at an electronics company wants to test their amplifiers. The design engineer claims they have a mean output of 113 watts with a variance of 100. What is the probability that the mean amplifier output would be greater than 112.5 watts in a sample of 43 amplifiers if the claim is true

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Answer from: whitneyb13

62.93% probability that the mean amplifier output would be greater than 112.5 watts in a sample of 43 amplifiers

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:

$\mu = 113, \sigma = \sqrt{100} = 10, n = 43, s = \frac{10}{\sqrt{43}} = 1.5250$

What is the probability that the mean amplifier output would be greater than 112.5 watts in a sample of 43 amplifiers if the claim is true

This is 1 subtracted by the pvalue of Z when X = 112.5. So

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

By the Central Limit Theorem

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

$Z = \frac{112.5-113}{1.5250}$

Z = -0.33

Z = -0.33 has a pvalue of 0.3703

1 - 0.3707 = 0.6293

62.93% probability that the mean amplifier output would be greater than 112.5 watts in a sample of 43 amplifiers

Answer from: Quest

for this case we have by definition, if two lines are perpendicular, then the product of their slopes is -1.

$m_ {1} * m_ {2} = - 1$

then, given the following line:

$6x-3y = 18\\6x-18 = 3y$

$\frac {6x} {3} - \frac {18} {3} = y\\y = \frac {6x} {3} - \frac {18} {3}\\y = 2x-6$

so, $m_ {1} = 2$

we are looking for m_ {2}:

$m_{2}=\frac{-1}{m_{1}}\\m_{2}=\frac{-1}{2}\\m_{2}=-\frac{1}{2}$

then, the line is given by:

$y = - \frac {1} {2} x + b$

we find "b" replacing the given point:

$8 = - \frac {1} {2} (0) + b\\8 = b$

finally, the equation is:

$y = - \frac {1} {2} x + 8$

option c

Answer from: Quest

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step-by-step explanation:

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