Mathematics, 27.07.2021 01:10 dramaqueenactr2040

A quality control expert at Glotech computers wants to test their new monitors. The production manager claims they have a mean life of 83 months with a variance of 81. If the claim is true, what is the probability that the mean monitor life would be greater than 81.2 months in a sample of 146 monitors? Round your answer to four decimal places.

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Answer from: alyxkellar06

0.9922 = 99.22% probability that the mean monitor life would be greater than 81.2 months in a sample of 146 monitors.

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.

The production manager claims they have a mean life of 83 months with a variance of 81.

This means that $\mu = 83, \sigma = \sqrt{81} = 9$

Sample of 146:

This means that $n = 146, s = \frac{9}{\sqrt{146}}$

What is the probability that the mean monitor life would be greater than 81.2 months in a sample of 146 monitors?

This is 1 subtracted by the p-value of Z when X = 81.2. So

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

By the Central Limit Theorem

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

$Z = \frac{81.2 - 83}{\frac{9}{\sqrt{146}}}$

Z = -2.42

Z = -2.42 has a p-value of 0.0078.

1 - 0.0078 = 0.9922.

0.9922 = 99.22% probability that the mean monitor life would be greater than 81.2 months in a sample of 146 monitors.

Answer from: Quest

more detail

step-by-step explanation:

Answer from: Quest

in mathematics, a constraint is a condition of an optimization problem that the solution must satisfy. there are several types of constraints—primarily equality constraints, inequality constraints, and integer constraints. the set of candidate solutions that satisfy all constraints is called the feasible set.

hoped i'm right lol