Question 18 2 marks A cylindrical grinder produces bushings

Question 18 [2 marks] A cylindrical grinder produces bushings with a mean Inside diameter of 12.7mm and an assumed standard deviation of 0.05mm. A bushing produced by this grinder is selected at random. Assuming that the inside diameter of bushings are normally distributed, what is the probability that its inside diameter exceeds 12.78mm? Round any z-scores to 2 decimal places for use on the Fawcett and Kent tables. Blackboard will mark your answer to 3 decimal places. Question 19 (3 marks] What is the probability that the inside diameter of the randomly chosen bushing from Question 18 is between 12.63mm and 12.73mm? Round any z-scores to 2 decimal places for use on the Fawcett and Kent tables. Blackboard will mark your answer to 3 decimal places.

Solution

for normal distribution

P(X =k) = P(Z = k - mean/std)

18)

P(X > 12.78) = P(Z > 12.78 - 12.7 /0.05)

= P( Z > 1.6)

= 1- P( Z < 1.6)

= 1 - 0.9452

= 0.0548 = 0.055

19)

P( 12.63 < X < 12.73)

= P ( X < 12.73) - P ( X < 12.63)

= P ( Z < 12.73 - 12.7/0.05) - P ( Z < 12.63 - 12.7/0.05)

= P( Z < 0.6) - P ( Z < -1.4)

= 0.7257 - 0.0808

= 0.6449

= 0.645