Question

A simple random sample of 110 analog circuits is obtained at random from an ongoing production process in which 20% 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 ( 17 ≤ X ≤ 25 ) , the probability that between 17 and 25 circuits in the sample are defective. Report your answer to two decimal places of precision.

Answer 1

64.56% probability that between 17 and 25 circuits in the sample are defective.

Explanation:

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 = 110, p = 0.2`

So

`\mu = E(X) = np = 110*0.2 = 22`

`\sigma = √(V(X)) = √(np(1-p)) = √(110*0.2*0.8) = 4.1952`

Probability that between 17 and 25 circuits in the sample are defective.

This is the pvalue of Z when X = 25 subtrated by the pvalue of Z when X = 17. So

X = 25

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

`Z = (25 - 22)/(4.1952)`

`Z = 0.715`

`Z = 0.715` has a pvalue of 0.7626.

X = 17

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

`Z = (17 - 22)/(4.1952)`

`Z = -1.19`

`Z = -1.19` has a pvalue of 0.1170.

0.7626 - 0.1170 = 0.6456

64.56% probability that between 17 and 25 circuits in the sample are defective.