Question

Assume that the diameter of 1000 brass plugs taken consecutively from a machine form a normal distribution with mean 0.7515 cm and standard deviation 0.0020 cm. How many brass plugs are likely to be approved if the acceptable diameter is 0.752 + 0.004 and 0.752 – 0.004. (Given : Area at z = 2.25 is 0.4878 and Area at z = 1.75 is 0.4599.)

Answer 1

Answer:

948 brass plugs are likely to be approved

Explanation:

z = (x - μ) / σ

x = LSL  & USL

LSL  = 0.752 - 0.004

USL = 0.752 + 0.004

μ = Mean = 0.7515

σ = standard deviation = 0.0020

z usl = (0.752 + 0.004 - 0.7515) /0.0020  = 0.0045/0.0020 = 2.25

z lsl =  (0.752 - 0.004 - 0.7515) /0.0020  = -0.0035/0.0020 = -1.75

Area at z = 2.25 is 0.4878

Area at z = 1.75 is 0.4599

Total Area = 0.4878 + 0.4599 = 0.9477

Brass plugs likely to be approved = (0.9477/1) * 1000

= 947.7

≈ 948

Answer 2

Explanation:

the no of rejected lamps have to calculated