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?

Answer 1

29.46% probability that the sample mean warpage exceeds 1.305 mm

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 = (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 = (\sigma)/(√(n))`

In this problem, we have that:

`\mu = 1.3, \sigma = 0.13, n = 200, s = (0.13)/(√(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 = (X - \mu)/(\sigma)`

By the Central Limit Theorem

`Z = (X - \mu)/(s)`

`Z = (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