Mathematics, 18.05.2021 19:10, jdjdjdjdjjffi7273

A quality control expert at LIFE batteries wants to test their new batteries. The design engineer claims they have a standard deviation of 86 minutes with a mean life of 505 minutes. Of the claim is true, in a sample of 120 batteries, what is the probability that the mean battery life would be greater than 523.8 minutes

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answered: pennygillbert

0.0082 = 0.82% probability that the mean battery life would be greater than 523.8 minutes

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 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.

Standard deviation of 86 minutes with a mean life of 505 minutes.

This means that $\sigma = 86, \mu = 505$

Sample of 120:

This means that $n = 120, s = \frac{86}{\sqrt{120}}$

What is the probability that the mean battery life would be greater than 523.8 minutes?

This is 1 subtracted by the p-value of Z when X = 523.8. So

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

By the Central Limit Theorem

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

$Z = \frac{523.8 - 505}{\frac{86}{\sqrt{120}}}$

Z = 2.39

Z = 2.39 has a p-value of 0.9918.

1 - 0.9918 = 0.0082

0.0082 = 0.82% probability that the mean battery life would be greater than 523.8 minutes

answered: Guest

maybe showing the graph would

step-by-step explanation:

-387129 iq

answered: Guest

x =±sqrt(1/3)

step-by-step explanation:

x+4/6x=1/x

multiply each side by 6x to get rid of the fractions

6x*(x+4/6x)=1/x*6x

distribute

6x^2 +4 = 6

subtract 4 from each side

6x^2 +4-4 = 6-4

6x^2 =2

divide by 6

6x^2/6 = 2/6

x^2 = 1/3

take the square root of each side

sqrt(x^2) = ±sqrt(1/3)

x =±sqrt(1/3)

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A quality control expert at LIFE batteries wants to test their new batteries. The design engineer cl...