Question

A simple random sample of 85 analog circuits is obtained at random from an ongoing production process in which 21% 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 ( 14 ≤ X ≤ 20 ) , the probability that between 14 and 20 circuits in the sample are defective. Report your answer to two decimal places of precision.

Answer 1

63.81% probability that between 14 and 20 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 = 85, p = 0.21`.

So

`E(X) = np = 85*0.21 = 17.85`

`√(V(X)) = √(np(1-p)) = √(85*0.21*0.79) = 3.7552`

P ( 14 ≤ X ≤ 20 )

Using continuity correction, this is
`P(14 - 0.5 \leq X \leq 20 + 0.5) = P(13.5 \leq X \leq 20.5)`, which is the pvalue of Z when X = 20.5 subtracted by the pvalue of Z when X = 13.5. So

X = 20.5

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

`Z = (20.5 - 17.85)/(3.7552)`

`Z = 0.71`

`Z = 0.71` has a pvalue of 0.7611

X = 13.5

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

`Z = (13.5 - 17.85)/(3.7552)`

`Z = -1.16`

`Z = -1.16` has a pvalue of 0.1230

0.7611 - 0.1230 = 0.6381

63.81% probability that between 14 and 20 circuits in the sample are defective.