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swat32
3 years ago
9

PLEASE HELP ME I NEED THE ANS IN 5 mins

Mathematics
1 answer:
AlexFokin [52]3 years ago
5 0

Answer:

A

Step-by-step explanation:.................

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Answer:

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Step-by-step explanation:

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The fracture strength of tempered glass averages 14 (measured in thousands of pounds per square inch) and has standard deviation
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Answer:

0.1587 = 15.87% probability that the average fracture strength of 100 randomly selected pieces of this glass exceeds 14.2.

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 fracture strength of tempered glass averages 14 (measured in thousands of pounds per square inch) and has standard deviation 2.

This means that \mu = 14, \sigma = 2

Sample of 100:

This means that n = 100, s = \frac{2}{\sqrt{100}} = 0.2

What is the probability that the average fracture strength of 100 randomly selected pieces of this glass exceeds 14.2?

This is 1 subtracted by the p-value of Z when X = 14.2. So

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

By the Central Limit Theorem

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

Z = \frac{14.2 - 14}{0.2}

Z = 1

Z = 1 has a p-value of 0.8413.

1 - 0.8413 = 0.1587

0.1587 = 15.87% probability that the average fracture strength of 100 randomly selected pieces of this glass exceeds 14.2.

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