Answer:
The probability that defect length is at most 20 mm is 0.0495.
Explanation:
We are given that the defect length of a corrosion defect in a pressurized steel pipe is normally distributed with a mean value of 33 mm and a standard deviation of 7.9 mm.
Let X = the defect length of a corrosion defect in a pressurized steel pipe
The z-score probability distribution for the normal distribution is given by;
Z =
~ N(0,1)
where,
= mean defect length = 33 mm
= standard deviation = 7.9 mm
So, X ~ Normal(
)
Now, the probability that defect length is at most 20 mm is given by = P(X
20 mm)
P(X
20 mm) = P(
) = P(Z
-1.65) = 1 - P(Z < 1.65)
= 1 - 0.9505 = 0.0495
The above probability is calculated by looking at the value of x = 1.65 in the z table which has an area of 0.9505.