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

Answer:

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

Step-by-step 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%)

Answer 2

Answer:

$P(s^2 >3.10) =P(\frac{(n-1)s^2}{\sigma^2}>\frac{19*3.10}{1.75})$

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

Step-by-step 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

$\chi^2 =\frac{(n-1) s^2}{\sigma^2}$

When we take the expected value we got:

$E[\frac{(n-1) s^2}{\sigma^2}]= n-1$

$E[s^2]=\frac{n-1}{n-1}\sigma^2$

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

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:

$P(s^2 >3.10) =P(\frac{(n-1)s^2}{\sigma^2}>\frac{19*3.10}{1.75})$

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

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