Question

[6.4] Beer cans are sampled to determine if the number of cans with little liquid is very frequent Suppose "p" is the proportion of cans that have less liquid than the nominal value (10 oz). If "p" is too large, the filling mechanism is adjusted. Every 4 hours a sample of 200 cans is selected and the volume of each is measured. Assume that X is the number of cans in the sample that have less liquid than the nominal value. Assume that the rule of the department of quality control is established: make an adjustment of the filling mechanism when X greater than or equal to 7 a) Calculate the probability of making the adjustment when the pump produces 3% of defective. b) Calculate the probability of not making the adjustment when the pump produces 5% of defective

Answer 1

a)

For p = 3% defective, standard error of proportion = = 0.01206

For X = 7, sample proportion = 7 / 200 = 0.035

Probability of making the adjustment when the pump produces 3% of defective = P(X 7)

= P(p 0.035)

= P(Z (0.035 - 0.03) / 0.01206)

= P(Z 0.4146)

= 0.3392

b)

For p = 5% defective, standard error of proportion = = 0.0154

Probability of not making the adjustment when the pump produces 5% of defective = P(X < 7)

= P(p < 0.035)

= P(Z < (0.035 - 0.05)/0.0154)

= P(Z < -0.9740)

= 0.1650