Question

PA 9-4 Consider again the company making tires for...

Consider again the company making tires for bikes is concerned about the exact width of their cyclocross tires. The company has a lower specification limit of 22.8 mm and an upper specification limit of 23.4 mm. The standard deviation is 0.19 mm and the mean is 23.1 mm.

(Round your answer to 4 decimal places.)

a.What is the probability that a tire will be too narrow?

(Round your answer to 4 decimal places.)

b.What is the probability that a tire will be too wide?

(Round your answer to 3 decimal places.)

c.What is the probability that a tire will be defective?

Answer 1

A. Probability{narrow} = NORM.DIST(LSL, mean, ?, 1)

= NORM.DIST(22.8, 23.1, 0.19, 1)

= 0.0572

B. Probability{wide} = NORM.DIST(USL, mean, ?, 1)

= NORM.DIST(23.4, 23.1, 0.19, 1)

= 0.0572

C. Probability{defective} = 0.0572 + 0.0572

= 0.1144

Answer 2

Cinen, Mean Mean (4) 23.1 6 = 0.19 It is normally distributed Specifications are :- . • UCL = 23.4 • LCL = 22.8 a) Now, PCXCLCL):- (CL-u = 22.8-23.) z - - 1.58 0:19 p= 0.0571 > from 2-table for z=-1158 P ( too narrow ) = b) Now, P(XJUCL) :- >) 0.0571 2= UCL-u 1.58 23.4-23.1 0.19 → from 2 table for z=1:58 P=o. 9429 > Pl too wide): 1-0.9429 = 0.0571 () > pl defective) defective ) - Pl for narrow) + Pl too wide) 0.0571 +0.0571 = 0.1142