The weights of 83 randomly selected windshields were found to have a variance of 1.88. Construct the 95% confidence interval for the population variance of the weights of all windshields in this factory. Round your answers to two decimal places.

95% confidence interval for the population variance = (1.42 , 2.62).

Step-by-step explanation:

We are given that the weights of 83 randomly selected windshields were found to have a variance of 1.88.

So, firstly the pivotal quantity for 95% confidence interval for the population variance is given by;

        P.Q. = ~

where, = sample variance = 1.88

            = population variance

            n = sample of windshields = 83

So, 95% confidence interval for population variance, is;

P(58.85 < < 108.9) = 0.95 {As the table of at 82 degree of freedom

                                              gives critical values of 58.85 & 108.9}

P(58.85 < < 108.9) = 0.95

P( < < ) = 0.95

P( < < ) = 0.95

95% confidence interval for = ( , )

                                                  = ( , )

                                                  = (1.42 , 2.62)

Therefore, 95% confidence interval for the population variance of the weights of all windshields in this factory is (1.42 , 2.62).

5/20 so 25%

step-by-step explanation:

sorry i got nothing

step-by-step explanation: