Question

(Normal Approximation) A process yields 7% defective items. Suppose 2400 items are randomly selected from the process. Use the normal curve approximation (with half-unit correction) to find the probability that the number of defectives exceeds 191? Be sure to: • Define your random variable. • Check the success / failure condition. • Provide your z-score to 2 decimals. • Provide your final answer to 4 decimals.

Answer 1

`z = (191-168)/(12.50)= 1.84`

`P(X> 191)= P(Z>1.84)=1-P(Z<1.84)`

And we can use the normal standard distribution table or excel and we got:

`P(X> 191)= P(Z>1.84)=1-P(Z<1.84)=1-0.9671=0.0329`

Explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Solution to the problem

Let X the random variable of interest, on this case we now that:

`X \sim Binom(n=2400, p=0.07)`

The probability mass function for the Binomial distribution is given as:

`P(X)=(nCx)(p)^x (1-p)^(n-x)`

Where (nCx) means combinatory and it's given by this formula:

`nCx=(n!)/((n-x)! x!)`

The expected value is given by this formula:

`E(X) = np=2400*0.07=168`

And the standard deviation for the random variable is given by:

`sd(X)=√(np(1-p))=√(2400*0.07*(1-0.07))=12.50`

We need to check if we can use the normal approximation , the conditions are:

`np=2400*0.07=168>10` and
`n(1-p)=2400*(1-0.07)=2232>10`

So then we can apply the normal approximation to the binomial distribution in our case:

`X \sim N(\mu=168,\sigma=12.50)`

And for this case we cant this probability:

`P(X> 191)`

And we can use the z score given by:

`z = (x- \mu)/(\sigma)`

And replacing we got:

`z = (191-168)/(12.50)= 1.84`

`P(X> 191)= P(Z>1.84)=1-P(Z<1.84)`

And we can use the normal standard distribution table or excel and we got:

`P(X> 191)= P(Z>1.84)=1-P(Z<1.84)=1-0.9671=0.0329`