Question

A process produces batches of a chemical whose impurity concentrations follow a normal distribution with a variance of175. A random sample of 20 of these batches is chosen. Find the probability that the sample variance exceeds 3.10.

Answer 1

`P(s^2 >3.10) =P(((n-1)s^2)/(\sigma^2)>(19*3.10)/(1.75))`

`P(chi^2_(19)>33.657)=1-P(\chi^2_{19]<33.657)=0.0202`

Explanation:

Previous concepts

The Chi Square distribution is the distribution of the sum of squared standard normal deviates .

Data given and notation

For this case we can use the fact that the estimator of the population variance
`\sigma` is the sample variance
`s^2`, because
`E(s^2)=\sigma^2`

The proof is this one:

Since
`E(\chi^2) = n-1` and

`\chi^2 =((n-1) s^2)/(\sigma^2)`

When we take the expected value we got:

`E[((n-1) s^2)/(\sigma^2)]= n-1`

`E[s^2]=(n-1)/(n-1)\sigma^2`

`E[s^2]=\sigma^2`

We have the distribution on this case given chi square.

Solution to the problem

The degrees of freddom on this case are given by

`df=n-1=20-1=19`

On this case we want this probability:

`P(s^2 >3.10) =P(((n-1)s^2)/(\sigma^2)>(19*3.10)/(1.75))`

`P(chi^2_(19)>33.657)=1-P(\chi^2_(19)<33.657)=0.0202`

And we can use excel to find the probability with the following code:"=1-CHISQ.DIST(33.657,19,TRUE) "

Answer 2

the probability that the sample variance exceeds 3.10 is 0.02020 ( 2,02%)

Explanation:

since the variance S² of the batch follows a normal distribution , then for a sample n of 20 distributions , then the random variable Z:

Z= S²*(n-1)/σ²

follows a χ² ( chi-squared) distribution with (n-1) degrees of freedom

since

S² > 3.10 , σ²= 1.75 , n= 20

thus

Z > 33.65

then from χ² distribution tables:

P(Z > 33.65) = 0.02020

therefore the probability that the sample variance exceeds 3.10 is 0.02020 ( 2,02%)