Question

Knowledge Objective: Uniform Continuous Distribution

Tensile tests of a 3D printed material show that the average yield strength of that material is about 800 MPa with a standard deviation of 40 MPa. Due to the variations of the process parameters, the yield strength can vary, and historical data show that yield strength of the material follows a uniform continuous distribution. Answer questions 1a to 1e using this information.

1a. What is the random variable (X)?

a. Value of the most critical parameter

b. Average yield strength of the material

c. 40 Mpa

d. Yield strength of the material

1b. If yield strength of the material is uniformly distributed within 730.72 to 869.28, what is the expected value of yield strength of the material?

1c. If yield strength of the material is uniformly distributed within 730.72 to 869.28, what is the variance of yield strength of the material?

1d. If yield strength of the material is uniformly distributed within 730.72 to 869.28, what is the probability that the yield strength is between 750 to 800 MPa (round to 2 digits)?

1e. If yield strength of the material is uniformly distributed within 730.72 to 869.28, find the value of yield strength x so that there is 40% chance that the yield strength does not exceed x (round to 2 digits).

Answer 1

1 a.

Random Variable X :

d. Yield strength of the material

If X is uniformly distributed within a,b(on the interval a,b);

Probability density function

Cumulative Distribution function

Mean : Expected value of X

Variance of X =

1b. If yield strength of the material is uniformly distributed within 730.72 to 869.28

i.e b = 869.28 and a = 730.72

expected value of yield strength of the material :

expected value of yield strength of the material :800

1c. If yield strength of the material is uniformly distributed within 730.72 to 869.28,

variance of yield strength of the material

variance of yield strength of the material = 1599.906133

1d. If yield strength of the material is uniformly distributed within 730.72 to 869.28,

probability that the yield strength is between 750 to 800 MPa = P(750X800) = P(X800) - P(X​​​​​​​750)

P(X​​​​​​​800) - P(X​​​​​​​750) = 0.5 - 0.1391= 0.3609

probability that the yield strength is between 750 to 800 MPa = 0.3609

1e. If yield strength of the material is uniformly distributed within 730.72 to 869.28, find the value of yield strength x so that there is 40% chance that the yield strength does not exceed x

i,e

P(Xx) = 40/100 = 0.4

i.e

F(x) = 0.4

x = 0.4 x 138.56 +730.32 = 55.424+730.32=785.744

Value of Yield strength x = 785.74