Question

2. The defect length of a corrosion defect in a pressurized steel pipe is normally distributed with mean value 30mm and standard deviation 7.8 mm.

a. What is the probability that defect length is at most 20 mm?

b. What is the probability that defect length is less than 20 mm?

c. What is the 75th percentile of the defect length distribution?

d. What is the 15th percentile of defect length distribution?

e. What values separate the middle 80% of the defect length distribution from the smallest 10% and largest 10%?

Answer 1

Solution :

Given that,

mean = = 30

standard deviation = = 7.8

a ) P( x 20 )

P ( x - / ) ( 20 - 30 / 7.8)

P ( z    -10 / 7.8 )

P ( z - 1.28 )

= 0.1003

Probability = 0.1003

b ) P( x < 20 )

P ( x - / ) < ( 20 - 30 / 7.8)

P ( z < -10 / 7.8 )

P ( z < - 1.28 )

= 0.1003

Probability = 0.1003

c ) P( Z < z) = 75%

P(Z < z) = 0.75

z = 0.67

Using z-score formula,

x = z * +

x = 0.67 *7.8 + 30

= 35.226

P₇₅ = 35.22

d ) P( Z < z) = 15%

P(Z < z) = 0.15

z = -1.04

Using z-score formula,

x = z * +

x = -1.04 *7.8 + 30

= 21.888

P₁₅ = 21.89

e ) P(-z < Z < z) = 80%
P(Z < z) - P(Z < z) = 0.80
2P(Z < z) - 1 = 0.80
2P(Z < z ) = 1 + 0.80
2P(Z < z) = 1.80
P(Z < z) = 1.80 / 2
P(Z < z) = 0.90
z = 1.28 and z = - 1.28

x = z * +

x = - 1.28 * 7.8 + 30

= 20.016

x = 20.02

x = 1.28 * 7.8 + 30

= 39.98

x = 39.98