Question

Please, I need help ONLY with number 5 and 6. Thank you!!!

ESI-3213 HW-01 FIU /10 pts. Problem No. 7.6 A cylindrical hole is drilled in a block, and a cylindrical piston is placed in the hole. The clearance is equal to one-half the difference between the diameters of the hole and the piston. The diameter of the hole is normally distributed with mean 14. cm and standard deviation 0.02 cm, and the diameter of the piston is normally distributed with mean 13.88 cm and standard deviation 0.012 cm. 1) Find the mean clearance 2) Find the standard deviation of the clearance 3) What is the probability that the clearance is less than 0.07 cm? 4) Find the 40.-th percentile of the clearance 5) Specifications call for the clearance to be between 0.07 and 0.09 cm. What is the probability that the clearance meets the specification? 6) It was possible to adjust the mean hole diameter. To what value should it be adjusted so as to maximize the probability that the clearance will be between 0.07 and 0.09 cm? Show your explanations. Displaying only the final answer is not enough to get credit. Note: round calculated numerical values to the fourth decimal place where applicable. Solution (Show all intermediate steps, formulas, calculations, explanations and comments below this line. Don't write above this line)

Answer 1

151 Clearance to be between 0.07 and oogcm What is probability that Clearance specification bi meet Z score is the value decreased by the mean, divided by standard deviation xall 07-0.06 - 0.8576 0.01166 z = X-l o og -O.( - 2.5728 Oo1166 - 2.6 Corresponding probabity P Co.07 <x co.og) = P C©.8576<2<26 2.5728- P (z<2.5728)- P(Z <0.8576) 0.9949 0.8023 P CO.07<x<0.09 ) - 0.1926 [ This all values are taken from normal probalily table at corresponding probalitg at 2.57 and 0.85 Solution = 0.1926 [pCooz<x<0.og) S

6- ux=13.94cm should be adjusted to get maximum probability