Question

3. (10 points) Thickness measurements of a coating process are made to the nearest hun- dredth of a millimeter. Let X be a random variable representing the thickness measure- ments. X can take values 0.15, 0.16, 0.17, 0.18, and 0.19, and the outcomes are equally likely (i.e., P(X = x;) = 1/5 for i = 1...5). Determine E(X) and E(X2). Use these results to find E[(3X – 5)2] and Var(2.5X – 10).

Answer 1

Answers:

E(X) = 0.17

E(X²) = 0.0291

E[(3X – 5)2) = 20.1619

Var( 2.5X -10) = 0.0012

Explanation:

Let X be a random variable representing the thickness measurements.

The formula of E(X) and E(X²) for probability distribution is as follows:

E(X) = Στ και Ρ(at)

E(X?) - Σ2 και P()

It is given that the outcomes are equally likely.

Let's make the table:

X 0.15 0.16 0.17 0.18 0.19 P(x) 0.2 0.2 0.2 0.2 0.2 X*P(x) 0.03 0.032 0.034 0.036 0.038 0.17 X^2 *P(x) 0.0045 0.00512 0.00578 0.00648 0.00722 0.0291 Sum

From the above table we get

E(X) = Στ* P(x) = 0.17

Ε(x2) = Σ 2 * P(x) = 0.0291

Next we want to find

E[(3x – 5)

E|(3x – 5)2 = E (9x2 – 30X + 25) = 9E(X2) – 30E(X) + 25

Ef(3x – 5)2 =9*0.0291 – 30 * 0.17 + 25 = 20.1619

Next we want to find Var( 2.5X -10)

Var( 2.5X -10) = Var(2.5X) ...{ because the variance is not affect the change of origin

Var( 2.5X -10) = 2.5² * Var(X) = 6.25 [ E(X²) - {E(X)}²] = 6.25 (0.0291 - 0.0289) = 6*0.0002 = 0.0012

Var( 2.5X -10) = 0.0012