Question

A process produces batches of a chemical whose impurity concentrations follow a normal distribution with a variance of 1.75. A random sample of 20 of these batches is chosen. Find the probability that the sample variance exceeds 3.10.

Answer 1

Answer:

Probability that the sample variance exceeds 3.10 is less than 0.05%.

Step-by-step explanation:

We are given that a process produces batches of a chemical whose impurity concentrations follow a normal distribution with a variance of 1.75.

Also, a random sample of 20 of these batches is chosen.

The probability distribution that we will used here is of Chi-square distribution, i.e.;

                    $\frac{(n-1)s^{2} }{\sigma^{2} }$ ~ $\chi^{2} __n_-_1$

where, $s^{2}$ = sample variance

            $\sigma^{2}$ = population variance = 1.75

            n = sample of batches = 20

So, probability that the sample variance exceeds 3.10 is given by = P($s^{2}$ > 3.10)

        P($s^{2}$ > 3.10) = P( $\frac{(n-1)s^{2} }{\sigma^{2} }$ > $\frac{(20-1) \times 3.10^{2} }{1.75^{2} }$ )

                           = P( $\chi^{2} __2_0_-_1$ > 59.62) = P( $\chi^{2} __1_9$ > 59.62) = Less than 0.05%

From chi-square table we can observe that Probability that $\chi^{2} __1_9$ is greater than is less than 0.05%.                      

So, the probability that the sample variance exceeds 3.10 is less than 0.05%.