Question

Suppose that a random sample of 20 items is selected from the machine. If the machine produces 5% defectives, find the probability that the sample will contain at least three defectives, by using the following methods. (a) the normal approximation to the binomial (Round your answer to four decimal places.) (b) the exact binomial tables (Round your answer to three decimal places.)

Answer 1

a) 0.0618 = 6.18% probability that the sample will contain at least three defectives.

b) 0.076 = 7.6% probability that the sample will contain at least three defectives

Explanation:

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

`P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)`

In which
`C_(n,x)` is the number of different combinations of x objects from a set of n elements, given by the following formula.

`C_(n,x) = (n!)/(x!(n-x)!)`

And p is the probability of X happening.

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 distributions 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))`.

Sample of 20 items is selected from the machine.

This means that
`n = 20`

5% defectives

This means that
`p = 0.05`

(a) the normal approximation to the binomial

The mean is:

`\mu = E(X) = np = 20*0.05 = 1`

The standard deviation is:

`\sigma = √(V(X)) = √(np(1-p)) = √(20*0.05*0.95) = 0.9747`

The probability is, using continuity correction,
`P(X \geq 3 - 2.5) = P(X \geq 2.5)` , which is 1 subtracted by the pvalue of Z when X = 2.5. So

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

`Z = (2.5 - 1)/(0.9747)`

`Z = 1.54`

`Z = 1.54` has a pvalue of 0.9382

1 - 0.9382 = 0.0618

0.0618 = 6.18% probability that the sample will contain at least three defectives.

(b) the exact binomial tables

This is:

`P(X \geq 3) = 1 - P(X < 3)`

In which

`P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2)`

In which

`P(X = x) = C_(n,x).p^(x).(1-p)^(n-x)`

`P(X = 0) = C_(20,0).(0.05)^(0).(0.95)^(20) = 0.358`

`P(X = 1) = C_(20,1).(0.05)^(1).(0.95)^(19) = 0.377`

`P(X = 2) = C_(20,2).(0.05)^(2).(0.95)^(18) = 0.189`

`P(X < 3) = P(X = 0) + P(X = 1) + P(X = 2) = 0.358 + 0.377 + 0.189 = 0.924`

`P(X \geq 3) = 1 - P(X < 3) = 1 - 0.924 = 0.076`

0.076 = 7.6% probability that the sample will contain at least three defectives