Question

a particular manufacturing design requires a shaft with a diameter between 23.92 and 24.018 mm. The manufacturing process yields shafts with diameters normally distributed, with a mean of 24.003 and standard deviation of .006. a) for this process what is the proportion of shafts with a diameter between of 23.92 and 24.00 mm b) The probability that the shaft is acceptable is _ c) The diameter that will be exceeded by only.5% of shafts is -

Answer 1

H = 24.003, 0 = 0.006 X-u 23.92 – 24.003 P(23.92 < X < 24.00) = P( 0.006 = P(-13.83<Z<-0.5) 24 – 24.003 0.006

= P(Z < -0.5) - P(Z < -13.83 = 0.3085 – 0.0000 = 0.3085

b.)

- 23.92 – 24.003 P(23.92 < X < 24.018) = PC 0.006 = P(-13.83 < Z < 2.5) 24.018 – 24.003 0.006 = P(Z < 2.5) - P(Z < -13.83) = 0.9938 - 0.0000 = 0.9938

c.)

P(X>x)=0.005

=>1-P(X<x)=0.005

=>P(X<x)=1-0.005

=>P(X<x)=0.995

= P(Z < 2 – 24.003 -) = 0.995 0.006 I – 24.003 0.006 = 2.58 => x = 24.003 + 0.01548 => x = 24.01848