Question

{ 7875 Elongation (in %) of steel plates treated with aluminum are random with probability density function 15 sxs 30 True (Note: "True" means "Otherwise" or "Elsewere") 1) What proportion of steel plates have elongations greater than 25%? 2) Find the mean elongation. 3) Find the variance of the elongations, 4) Find the standard deviation of the elongations. 5) Find the cumulative distribution function of the elongations. 6) A particular plate elongates 20%. What proportion of plates elongate more than this? Show your explanations. Displaying only the final answer is not enough to get credit. Note: round calculated numerical values to the fourth decimal place where applicable.

Answer 1

We would be looking at the first 4 parts here:

a) The proportion of steel plates that have elongation greater than 25% is computed here as:

$P(X > 25) = \int_{25}^{30}\frac{x^2}{7875} \ dx = \frac{30^3 - 25^3}{3*7875} = 0.4815$

therefore 0.4815 is the required probability here.

b) The mean elongation is computed here as:

$E(X )= \int_{15}^{30}\frac{x^3}{7875} \ dx = \frac{30^4 - 15^4}{4*7875} = 24.1071$

Therefore 24.1071% is the mean elongation here.

c) The second moment of X here is computed as:

$E(X^2 )= \int_{15}^{30}\frac{x^4}{7875} \ dx = \frac{30^5 - 15^5}{5*7875} = 597.8571$

Now the variance of elongation here is computed as:
Var(X) = E(X²) - [E(X)]² = 597.8571 - 24.1071² = 16.7049

Therefore 16.7049 %² is the required variance here.

d) The standard deviation here is computed as:

$SD(X) = \sqrt{16.7049} = 4.0872$

Therefore 4.0872% is the required standard deviation here.