Mathematics, 11.03.2020 22:54 HaseelM

The amount of warpage in a type of wafer used in the manufacture of integrated circuits has mean 1.3 mm and standard deviation 0.13 mm. A random sample of 200 wafers is drawn. What is the probability that the sample mean warpage exceeds 1.305 mm?

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Answer from: sierravick123owr441

29.46% probability that the sample mean warpage exceeds 1.305 mm

Step-by-step explanation:

To solve this question, we have 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 $\sigma$ , 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 random variable X, with mean $\mu$ and standard deviation $\sigma$ , the sample means with size n of at least 30 can be approximated to a normal distribution with mean $\mu$ and standard deviation $s = \frac{\sigma}{\sqrt{n}}$

In this problem, we have that:

$\mu = 1.3, \sigma = 0.13, n = 200, s = \frac{0.13}{\sqrt{200}} = 0.0092$

A random sample of 200 wafers is drawn. What is the probability that the sample mean warpage exceeds 1.305 mm

This is 1 subtracted by the pvalue of Z when X = 1.305. So

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

By the Central Limit Theorem

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

$Z = \frac{1.305 - 1.3}{0.0092}$

Z = 0.54

Z = 0.54 has a pvalue of 0.7054.

1 - 0.7054 = 0.2946

29.46% probability that the sample mean warpage exceeds 1.305 mm

Answer from: Quest

use an equation of ratios to determine the amount of gas needed to fill the tank.

0.2 0.8

288.75 x

231 gal

so, 0.2x = (0.8)(288.75 gal), and x = = 1155 gal.

0.2

it would take 1155 gal. to fill the tank.

the cost would be

1155 gal $3

* = $3465 (answer)

1 1 gal

hope this you

Answer from: Quest

my answer got deleted but it's a

step-by-step explanation:

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The amount of warpage in a type of wafer used in the manufacture of integrated circuits has mean 1.3...