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 ra

Question

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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Kennedy 3 weeks 2021-09-20T13:41:35+00:00 1 Answer 0

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    2021-09-20T13:42:50+00:00

    Answer:

    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

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