Question

Your company manufactures LCD panels. Let the probability that a panel has a dead pixel be p = 0.03. Assume different panels get such defects independently. In one week the company makes 2000 of these LCD panels. Using the CLT, what is the approximate probability that in this week more than 80 panels have dead pixels?

Answer 1

0.44% approximate probability that in this week more than 80 panels have dead pixels.

Explanation:

For each LCD panel, there are only two possible outcomes. Either it has a dead pixel, or it does not. So we use the binomial probability distribution to solve this problem.

However, we are working with samples that are considerably big. So i am going to aproximate this binomial distribution to the normal, by the Central Limit Theorem(CLT).

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

`E(X) = np`

The standard deviation of the binomial distribution is:

`√(V(X)) = √(np(1-p))`

Normal probability distribution

Problems of normally distributed samples can be 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.

When we are approximating a binomial distribution to a normal one, we have that
`\mu = E(X)`,
`\sigma = √(V(X))`.

In this problem, we have that:

`p = 0.03, n = 2000`

So

`\mu = np = 2000*0.03 = 60`

`\sigma = √(np(1-p)) = √(2000*0.03*0.97) = 7.63`

Using the CLT, what is the approximate probability that in this week more than 80 panels have dead pixels?

This is 1 subtracted by the pvalue of Z when X = 80. So

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

`Z = (80 - 60)/(7.63)`

`Z = 2.62`

`Z = 2.62` has a pvalue of 0.9956.

So there is a 1-0.9956 = 0.0044 = 0.44% approximate probability that in this week more than 80 panels have dead pixels.