Question

A camera manufacturer purchases glass to be ground into lenses. From experience it is known that the variance in the refractive index of the glass is 1.25 × 10-4. The company will reject any shipment of glass if the sample variance of 30 lenses measured at random exceeds 2.10 × 10 4, what is the probability that such a batch of glass will be rejected even though it is actually within the normal variance for refractive index?

Answer 1

Given data Refractive index of glass = 1.25 times 10^-4 sample variance = 20 lenses measured at random exceeds = 2.10 times 10^-4 we know that x^2_2 (v) = V.Sx^2/sigma^2 where V = degrees of freedom Sx^2 = sample variance sigma = population variance V = N-1 = 20 - 1 = 19 Xa^2 (V) = V.S x^2/sigma = 19 times (2.10 times 10^-4)/1.25 times 10^-4 = 31.92 probability of x^2 at 19 df = CHISQ.DIST. RT (31.92, 19) = 0.0319 = 3.19 % Probability will rejected even it is actually within normal variance for refractor index.