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.

maybe - definition of supplementary angles. im not sure. sorry

step-by-step explanation:

arithmetic sequence

step-by-step explanation:

given that a sequence has terms as

19, 15, 11,

this is represented on a graph

the pattern indicates that the sequence is an arithmetic sequence with i term

a=19 and common difference d =-4

any general term

thus the sequence is reduced by 4 for each successive term starting from 1 and nth term is

24-5n.

first negative term would start after 5th term

5th term =24-25< 0

from 5th term onwards the sequence would be negative.