Question

The width of a slot of a duralumin forging is (in inches) normally distributed with μ= 0.9000 and σ = 0.0030. The specification limits were given as 0.9000 ±0.0050. Show your calculations. Draw normal curves wherever appropriate.
a) What percentage of forgings will be defective?
b) What is the maximum allowable value of σ that will permit no more than 1 in 100 defectives when the widths are normally distributed with μ = 0.9000 and σ (revised value)?

Answer 1

Answer:

a) 9.56%

b) 0.0019

Step-by-step explanation:

a) Find the z-scores.

z = (x − μ) / σ

z₁ = (-0.0050) / 0.0030

z₁ = -1.67

z₂ = (0.0050) / 0.0030

z₂ = 1.67

Find the probability using a chart or calculator.

P(Z < -1.67 or Z > 1.67) = 2 P(Z < -1.67)

P(Z < -1.67 or Z > 1.67) = 2 (0.0478)

P(Z < -1.67 or Z > 1.67) = 0.0956

b) Use a chart or calculator to find the z-score.

P(Z < -z or Z > z) = 0.01

P(Z < -z) = 0.005

z = 2.576

Find the standard deviation.

z = (x − μ) / σ

2.576 = (0.0050) / σ

σ = 0.0019