Question

The defect length of a corrosion defect in a pressurized steel pipe is normally distributed with mean value 28 mm and standard deviation 7.3 mm.

(a) What is the probability that defect length is at most 20 mm? Less than 20 mm? (Round your answers to four decimal places.)

at most 20mm .=

less than 20mm =

(b) What is the 75th percentile of the defect length distribution—that is, the value that separates the smallest 75% of all lengths from the largest 25%? (Round your answer to three decimal places.)

=

(c) What is the 15th percentile of the defect length distribution? (Round your answer to three decimal places.)

(d) What values separate the middle 80% of the defect length distribution from the smallest 10% and the largest 10%? (Round your answers to three decimal places.) smallest 10%=

mm largest 10%=

You may need to use the appropriate table in the Appendix of Tables to answer this question.

Answer 1

Solution :

Given that,

mean = = 28

standard deviation = = 7.3

a ) P( x 20 )

P ( x - / ) ( 20 - 28 / 7.3)

P ( z - 8 / 7.3 )

P ( z -1.09)

= 0.1379

Probability = 0.1379

P( x < 20 )

P ( x - / ) < ( 20 - 28 / 7.3)

P ( z < - 8 / 7.3 )

P ( z < -1.09)

= 0.1379

Probability = 0.1379

b ) P( Z < z) = 75%
P(Z < z) = 0.75

z = 0.67

Using z-score formula,

x = z * +

x = 0.67 * 7.3 + 28

x = 32.891

c ) P( Z < z) =15%
P(Z < z) = 0.15

z = -1.04

Using z-score formula,

x = z * +

x = -1.04 * 7.3 + 28

x = 20.408

Using standard normal table,

d ) 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

Using z-score formula,

x = z * +

x = -1.28 * 7.3 + 28

x = 18.656

Smallest Value =18.656

Using z-score formula,

x = z * +

x = 1.28 * 7.3 + 28

x = 37.344

Largest value = 37.344