Mathematics, 08.07.2021 23:20 shayneseaton

Thompson and Thompson is a steel bolts manufacturing company. Their current steel bolts have a mean diameter of 144 millimeters, and a variance of 49 . If a random sample of 46 steel bolts is selected, what is the probability that the sample mean would differ from the population mean by more than 2 millimeters? Round your answer to four decimal places.

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Answer from: rebecabosca

0.0524 = 5.24% probability that the sample mean would differ from the population mean by more than 2 millimeters.

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 diameter of 144 millimeters, and a variance of 49.

This means that $\mu = 144, \sigma = \sqrt{49} = 7$

Sample of 46:

This means that $n = 46, s = \frac{7}{\sqrt{46}}$

Wat is the probability that the sample mean would differ from the population mean by more than 2 millimeters?

Above 144 + 2 = 146 or below 144 - 2 = 142. Since the normal distribution is symmetric, these probabilities are equal, which means that we find one of them and multiply by two.

Probability the sample mean is below 142:

p-value of Z when X = 142, so:

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

By the Central Limit Theorem

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

$Z = \frac{142 - 144}{\frac{7}{\sqrt{46}}}$

Z = -1.94

Z = -1.94 has a p-value of 0.0262

2*0.0262 = 0.0524

0.0524 = 5.24% probability that the sample mean would differ from the population mean by more than 2 millimeters.

Answer from: Quest

it would be 6xy and -16xy.

Answer from: Quest

answer: i'm not sure

step-by-step explanation:

Which transformations, when performed together, would carry graph a onto graph b? choose all that a

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