Question

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.

Answer 1

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Sol?? t is the let diametes of hole y is the diameter of piston X-N (14,0.02) YNN (13.88, 0.012²) Now the I defined as Clearance (-y) Now the mean Clearance is u lu -Clearance 4(x - 3) = 1 / Ma - 1 My ܢ (14) -(13-88) 14-13-88 0.06 Cm 2 clearance is a) The Standard deviation of the Clearance (x-4) (14)*6+W9* (0-02)+6) (0-012) = 0.01166 cm

(3) Here the diameters of hole & Piston are normally distributed, the is Aleo normally doebeiberteel clearance Now let C is the clearance (~N CO.06, 0.011662) low for is Z-Scores of C = 0.07 Ze=0.07 0.07- 0.06 oool166 0-86 Le 0.86 NOW from the Z-table f then the probability is 0.8051 Z table the closest area 1 (4 now from the to 0.4000 -0.253 Now -0.253 = x-0-06 0.0166 x = 0.05705 0.057 cm 0.057 Cm Clearance is 40th percentile of

(5) let compute 2-scores of C= 0.07 & C=0:09 0.07 -0.06 22:0.07 0.01166 0.86 2e=0.09 = 0-09-0-06 2.57 0.01166 Now Plo-07 30 30.09) - P (0-8657 32.57) = 0.9949-6-8051 0-1898 that Clearance 6) How to maximize the probability blu 0-07 & 0.09 I Me I My Mclearance á (2-13-88) 0-07 +0-09 а. x=14.04 & we must adjust mean hole diametes to 14-04cm Hole clearance - =(0.07 "ooft oog 2

Now the probability will be blo 00740.09 P(0.07 s Clearance <0.09) 0.09) - P 0.07 -0-08 cis 0-09-0-08 0.01166 0-01166 < P (-0.857 17.50.857) = 0.80428- 0-19572 = 1.22284