Question

The tensile strength of a metal part is normally distributed with mean 35 pounds and standard deviation 5 pounds. Suppose 40,000 parts are produced and specifications on the part have been established as 35.0 ± 4.2 pounds.

Find the percentage of parts that will fail to meet specification

Find the number of parts that will fail to meet specifications

Find the tensile strength at which 10% of the parts exceed the upper specification limit

Answer 1

onscoes! Let Mean - dl= 35 Standard deviation=6 - 5 UCL = 35.0+4,2 LEL=35.0-41.2 - 39.2 - 30.8 a) Picolul fail to meet specification) - P1x230.8) + P(x>39,2) - P(* 305-39) + p1*! 39.2-35) = P(Z2-41,2) + P(10737413) = P(Z2-0.50) + Plz>0-84) -0.20045+(1-0.79955) -0.0009 -) 40.09x,

b) Let No.of parts that coill fail to meet specifications *** : 40,000 loucoq'). 40,000 (0.4009). 16036 colere. P(x>0) = { 17.1-10}* P(x>U)= 7,11%, P(x>0) = 0.0711 Orien PLX20) = 1-0.0711 bet t es o re . . 0.99 M . P1x4 U PLC 0-39) -0.9289 = 0,9289 Z valle: 1.47 at 0.9289 0735=1,47 U-35 + (5x1147) U= 42,35