The defect length of a corrosion defect in a pressurized steel pipe is normally distributed with mean value 32 mm and standard deviation 7.4 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.) mm (c) What is the 15th percentile of the defect length distribution? (Round your answer to three decimal places.) mm (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% mm

a) Probability of at most 20 mm = P(x ≤ 20) = 0.0526

Probability of less than 20 mm = P(x < 20) ≈ 0.0526

b) The 75th percentile of the defect length distribution = 36.988 mm

c) The 15th percentile of the defect length distribution = 24.334 mm

d) The values that separate the middle 80% of the defect length distribution from the smallest 10% and the largest 10%

= 22.513 mm and 41.487 mm respectively.

Step-by-step explanation:

This is a normal distribution problem with

Mean = μ = 32 mm

Standard deviation = σ = 7.4 mm

(a) The probability that defect length is at most 20 mm and Less than 20 mm

The required probabilities are P(x ≤ 20) and P(x < 20)

To find these, we first standardize/normalize 20 mm

The standardized score for any value is the value minus the mean then divided by the standard deviation.

z = (x - μ)/σ = (20 - 32)/7.4 = - 1.62

The required probabilities

We'll use data from the normal probability table for these probabilities

P(x ≤ 20) = P(z ≤ -1.62) = 0.05262

P(x < 20) = P(z < -1.62) ≈ P(z ≤ -1.62) = 0.05262

(b) The 75th percentile of the defect length distribution—that is, the value that separates the smallest 75% of all lengths from the largest 25%

Let the defect length be x'

P(x ≤ x') = 0.75

The z-score of this defect length = z'

P(z ≤ z') = 0.75

Using the normal distribution tables,

z' = 0.674

Then taking the defect length from standardized score to real length,

z' = (x' - μ)/σ

0.674 = (x' - 32)/7.4

x' = 36.9876 mm = 36.988 mm to 3 d.p

(c) The 15th percentile of the defect length distribution

Let the defect length be x'

P(x ≤ x') = 0.15

The z-score of this defect length = z'

P(z ≤ z') = 0.15

Using the normal distribution tables,

z' = -1.036

Then taking the defect length from standardized score to real length,

z' = (x' - μ)/σ

-1.036 = (x' - 32)/7.4

x' = 24.3336 mm = 24.334 mm to 3 d.p

(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% mm

Let the two of them be x' and x" and their z-scores be z' and z"

P(x' ≤ x ≤ x") = 0.8

But it is explained that the 0.8 is the middle one, with 10% of the smallest values in the lowest range of defect lengths and another 10% of topmost values in the upper range of defect length l

P(x ≤ x') = 0.10

P(x ≤ x") = 0.10 + 0.80 = 0.90

Taking it one at a time

P(x ≤ x') = 0.10

The z-score of this defect length = z'

P(z ≤ z') = 0.10

Using the normal distribution tables,

z' = - 1.282

Then taking the defect length from standardized score to real length,

z' = (x' - μ)/σ

-1.282 = (x' - 32)/7.4

x' = 22.5132 mm = 22.513 mm to 3 d.p

P(x ≤ x") = 0.90

The z-score of this defect length = z"

P(z ≤ z") = 0.90

Using the normal distribution tables,

z' = 1.282

Then taking the defect length from standardized score to real length,

z' = (x' - μ)/σ

1.282 = (x' - 32)/7.4

x' = 41.4868 mm = 41.487 mm to 3 d.p

Hope this Helps!!!

2 over 5 and 4 over 15

don’t know last one sorry

step-by-step explanation: