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.

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.;

                    ~

where, = sample variance

            = population variance = 1.75

            n = sample of batches = 20

So, probability that the sample variance exceeds 3.10 is given by = P( > 3.10)

        P( > 3.10) = P( > )

                           = P( > 59.62) = P( > 59.62) = Less than 0.05%

From chi-square table we can observe that Probability that is greater than is less than 0.05%.                      

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

answer: can and shows the work

step-by-step explanation:

(x3 + 6x2 + 2x - 6) • (x - 1)

step-by-step explanation:

polynomial long division :

3.2     polynomial long division

dividing :   x4+5x3-4x2-8x+6

                              ("dividend")

by         :     x-1     ("divisor")

dividend     x4   +   5x3   -   4x2   -   8x   +   6

- divisor   * x3     x4   -   x3            

remainder         6x3   -   4x2   -   8x   +   6

- divisor   * 6x2         6x3   -   6x2        

remainder             2x2   -   8x   +   6

- divisor   * 2x1             2x2   -   2x    

remainder               -   6x   +   6

- divisor   * -6x0               -   6x   +   6

remainder                     0

quotient :   x3+6x2+2x-6   remainder:   0