The defect length of a corrosion defect in a pressurized steel pipe is normally distributed with mean value 33 mm and standard deviation 7.9 mm. (a) What is the probability that defect length is at most 20 mm

0.49926

Step-by-step explanation:

We solve for this using z score formula

z-score is

z = (x-μ)/σ,

where x is the raw score

μ is the population mean

σ is the population standard deviation

The probability that defect length is at most 20 mm is calculated as:

x = 20mm, μ = 33 mm, σ = 7.9mm

z = 20 - 33/7.9

= -13/7.9

= -1.64557

Obtaining the Probability value from Z-Table:

Probability (At most 20mm) = P(x ≤ 20mm) P(z = -1.64557)

P(x ≤ 20) = 0.049926

Therefore, the probability that defect length is at most 20 mm is 0.049926

The probability that defect length is at most 20 mm is 0.0495.

Step-by-step 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.

24

step-by-step explanation:

i assuming that 2 sides are 6 the other 2 sides are 4 cm so 6 times 4 is 24 : )