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

a. What is the probability that defect length is at most 20 mm? Less than 20 mm?
b. What is the 75th percentile of the defect length distribution?
c. What is the 15th percentile of the defect length distribution?
d. What values separate the middle 80% of the defect length distribution from the smallest 10% and the largest 10%?

a) 0.1003

b) 35.26

c) 21.92

d) Middle 80% smallest 10% and largest 10% = 20.02 and 39.98

Step-by-step explanation:

Given that: mean ц = 30 , standard deviation δ = 7.8

a) P(x ≤ 20)

   = P [ (x - ц ) / δ   ≤  ( 20 -30) / 7.8 ]

   = P(z ≤ -1.28)

   = 0.1003

probability = 0.1003

b) 75th percentile

   P(Z < z) = 0.75

   z = 0.674

Using z-score formula

   x = z * δ + ц

   x = 0.674 * 7.8 + 30

   x = 35.26

c) 15th percentile

   P(Z < z) = 0.15

   z = -1.036

Using z-score formula

   x = z * δ + ц

   x = (-1.036) * 7.8 + 30

   x = 21.92

d) Middle 80%

   P(-z ≤ Z ≤ z) = 0.80

   P(Z ≤ z) - P(Z ≤ -z) = 0.80

   2P(Z ≤ z) - 1 = 0.80

   2P(Z ≤ z) = 1 + 0.80 = 1.80

   P(Z ≤ z) = 1.80 / 2 = 0.90

   P(Z ≤ ± 1.28) = 0.90

   z =  -1.28, + 1.28

   x = z * δ + ц

   x = (-1.28) * 7.8 + 30

   x = 20.02

   x = z * δ + ц

   x = 1.28 * 7.8 + 30

   x = 39.98

Middle 80% two values = 20.02 and 39.98

smallest 10%

P(Z < z) = 0.10

z = -1.28

Using z-score formula

x = z * δ + ц

x = ( -1.28) * 7.8 + 30

x = 20.02

largest 10%

P(Z > z ) = 0.10

1 - P(z < z) = 0.10

P(z < z) = 1 - 0.10 = 0.90

z = 1.28

Using z-score formula

x = z * δ + ц

x = (1.28) * 7.8 + 30

x = 39.98

according to this law, when unemployment goes up by 1%, gdp drops by 2%. in the question, unemployment rate goes from 2% to 5%, meaning unemployment goes up by 3%, so the gdp drops by 6%. thus the effect is a drop in gdp by 6%.

step-by-step explanation:

24

step-by-step explanation:

if it takes 8 pounds to make 12 divide 12 by 8 and you get 1.5, 1.5 x 16 = 24