Question

A simple random sample of 90 analog circuits is obtained at random from an ongoing production process in which 25% of all circuits produced are defective.

Let X be a binomial random variable corresponding to the number of defective circuits in the sample.
Use the normal approximation to the binomial distribution to compute P(19 ≤ X ≤ 28), the probability that between 19 and 28 circuits in the sample are defective. Report your answer to two decimal places of precision.

Answer 1

76.10% probability that between 19 and 28 circuits in the sample are defective

Explanation:

I am going to use the binomial approximation to the normal to solve this question.

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:

`n = 90, p = 0.25`

So

`\mu = E(X) = np = 90*0.25 = 22.5`

`\sigma = √(V(X)) = √(np(1-p)) = √(90*0.25*0.75) = 4.11`

P(19 ≤ X ≤ 28)

Using continuity correction, this is
`P(19-0.5 \leq X \leq 28+0.5) = P(18.5 \leq X \leq 28.5)`, which is the pvalue of Z when X = 28.5 subtracted by the pvalue of Z when X = 18.5. So

X = 28.5

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

`Z = (28.5 - 22.5)/(4.11)`

`Z = 1.46`

`Z = 1.46` has a pvalue of 0.9279

X = 18.5

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

`Z = (18.5 - 22.5)/(4.11)`

`Z = -0.97`

`Z = -0.97` has a pvalue of 0.1669

0.9279 - 0.1669 = 0.7610

76.10% probability that between 19 and 28 circuits in the sample are defective