Question

The defect length of a corrosion defect in a pressurized steel pipe is normally distributed with mean value 28 mm and standard deviation 7.4 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.)
(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.)

Answer 1

The probability that the defect length is at most 20mm is approximately 0.1401, and the probability that it is less than 20mm is approximately 0.8599. The 75th percentile of the defect length distribution is approximately 33mm. The 15th percentile of the defect length distribution is approximately 20.358mm. The values that separate the middle 80% of the defect length distribution from the smallest 10% and the largest 10% are approximately 17.481mm and 38.519mm respectively.

Step-by-step explanation:

(a) We can calculate the probability that the defect length is at most 20mm by finding the area under the normal distribution curve to the left of 20mm. To do this, we calculate the z-score for 20mm using the formula:

z = (x - mean) / standard deviation

where x is the value we want to find the probability for, mean is the mean value of the distribution, and standard deviation is the standard deviation of the distribution. Plugging in the values, we get:

z = (20 - 28) / 7.4 = -1.08

We then look up the z-score in the standard normal distribution table or use a calculator to find the corresponding probability. In this case, the probability is approximately 0.1401.

To find the probability that the defect length is less than 20mm, we can subtract the probability that it is at most 20mm from 1 (since the total probability under the distribution curve is 1). Therefore, the probability that the defect length is less than 20mm is approximately 1 - 0.1401 = 0.8599.

(b) The 75th percentile of the defect length distribution represents the value that is larger than 75% of all defect lengths. To find this value, we can use the z-table and find the z-score that corresponds to a cumulative probability of 0.75. In this case, the z-score is approximately 0.6745. We can then use the z-score formula to find the corresponding defect length:

x = mean + (z * standard deviation) = 28 + (0.6745 * 7.4) = 32.9963 ≈ 33mm.

(c) The 15th percentile of the defect length distribution represents the value that is larger than 15% of all defect lengths. Using the same method as in part (b), we can find the z-score that corresponds to a cumulative probability of 0.15. In this case, the z-score is approximately -1.0364. Using the z-score formula, we can find the corresponding defect length:

x = mean + (z * standard deviation) = 28 + (-1.0364 * 7.4) = 20.3576 ≈ 20.358mm.

(d) To find the values that separate the middle 80% of the defect length distribution from the smallest 10% and the largest 10%, we need to find the z-scores that correspond to cumulative probabilities of 0.10 and 0.90.

The z-score for a cumulative probability of 0.10 is approximately -1.2816. Using the z-score formula, we can find the corresponding defect length:

x1 = mean + (z * standard deviation) = 28 + (-1.2816 * 7.4) = 17.4808 ≈ 17.481mm.

The z-score for a cumulative probability of 0.90 is approximately 1.2816. Using the z-score formula, we can find the corresponding defect length:

x2 = mean + (z * standard deviation) = 28 + (1.2816 * 7.4) = 38.5192 ≈ 38.519mm.

Therefore, the values that separate the middle 80% of the defect length distribution from the smallest 10% and the largest 10% are approximately 17.481mm and 38.519mm respectively.