Question

Please help with parts 3, 4, 5 & 6

{ Elongation (in %) of steel plates treated with aluminum are random with probability density function 15 SX < 30 7875 0 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

(3)The expected value of the random variable is

E(X) = T X 12 -dc 7875 30 15 1 15 23625

30 E(X) = x² x 22 -dz 7875 24 30 15 24.107 15 31500

Therefore

Var(X) = E(X) - (E(X))2 = 24.107 – 12 = 23.107

(4) Standard deviation of X is given by

SD(X) = Var(X) = 23.107 = 4.8069.

(5) The cumulative distribution function of X can be defined as,

FC) = f f(x) di

For x<15,

FC) = = Odx = 0

for

15 2 < 30

I F(x) = -dc 7875 115 23625 3375 23625 15 23625

For x>30

15 -30 F(2) Odc + f(x)dx + Odc = 1 15 30

Therefore,

0 F(t) = 3 23625 3375 23625 2 <15 15 <<30 .. > 30 1

(6) Here we want to find the proportion of plates elongates more than 20%.

That is we want to find P(X>20).

$P(X>20)=1-P(X\leq 20)=1-(F(20))=1-\left ( \frac{20^3}{23625}-\frac{3375}{23625} \right )=1-0.1957=0.80423.$