Mathematics

Question 1 of 10 15.0 Points

Which of the following statements is true ?

 A.If a constant c is added to each possible value of a discrete random variable X, then the variance of X will be shifted by that same constant amount.
 B.For any discrete random variable X and constants a and b, E(aX+b) = (a + b)E(X)
 C.

For any discrete random variable X and constants a and b, Var(aX+b) = (a + b)2Var(X)

 D.If a constant c is added to each possible value of a discrete random variable X, then the expected value of X will be shifted by that same constant amount.
Question 2 of 10 15.0 Points

If the probability distribution of continuous random variable X has the coefficient of skewness γ1 = 0 and the excess kurtosis γ2 = -1, than corresponding distribution curve

 A.is skewed to the right
 B.is more flat than the standard normal curve
 C.is skewed to the left
 D.looks exactly as the standard normal curve
 E.is more sharp that the standard normal curve
Question 3 of 10

15.0 Points

Let X be a discrete random variable with the second noncentral moment μ2= 19.75 and the second central moment μ’2= 16.3275, then expected value of X is

 A.3.4225
 B.±1.85
 C.±6.006
 D.4.04
Question 4 of 10 15.0 Points

Let X be an exponential random variable with parameter λ and moment generating function mX(t) = λ/(λ – t). Let random variable Y = 2X. Then the moment generating function of Y is

 A.None of the given mgf’s.
 B.

mY(t) = λ/(λ – t)2

 C.

mY(t) = λ/(λ – 2t)

 D.

mY(t) = λ/(2λ – t)

 E.

mY(t) = 2λ/(λ – t)

Question 5 of 10

15.0 Points

Random variable X has the probability mass function p(x) = p, if x=1 or p(x) = (1 – p), if x=0. The moment generating function of X is

 A.

mX(t) = (1 – p)2 + pet

 B.

mX(t) = pet – q

 C.

mX(t) = 1 + pet

 D.

mX(t) = 1 – p + pet

Question 6 of 10 15.0 Points

Determine the distribution of the random variable that has the moment generating function mX(t) = 0.5(0.5 – t)-1

 A.Uniform(-0.5, +0.5)
 B.Poisson(λ = 0.5)
 C.Geometric(p = 0.5)
 D.Exponential(λ = 1/2)
 E.Binomial(n = 1, p = 0.5)
 F.None of the given distributions.
Question 7 of 10 15.0 Points

Determine the distribution of the random variable that has the moment generating function mX(t) = (0.2et+0.8)8

 A.Normal(μ = 8, σ = 0.2)
 B.Binomial(n = 8, p = 0.2)
 C.Uniform(0.2, 0.8)
 D.Binomial(n = 8, p = 0.8)
 E.Binomial(n = 8, p = 0.2)
 F.Bernoulli(p=0.8)
Question 8 of 10

15.0 Points

If Y = a + bX, where a and b are constants, express the moment generation function of Y in terms of moment generation function of X:

 A.

mY(t) =eat mX(bt)

 B.

mY(t) =a + bmX(t)

 C.

mY(t) =ebt mX(at)

 D.

mY(t) =eat + mX(bt)

Question 9 of 10 15.0 Points

An actuary determines that the claim size for a certain class of accidents is a random variable, X, with moment generating function mx(t) = 1/(1−2500t)4. Determine the standard deviation of the claim size for this class of accidents.

 A.5,000
 B.1,340
 C.10,000
 D.8,660
 E.11,180
Question 10 of 10 15.0 Points

Use the following property of the moment generating function: If X1,X2,…Xn are independent random variables with existing moment generating functions, the random variable X = X1 + X2 +…+ Xn has the moment generating function mx(t)= mX1(t) · mX2(t)… mXn(t).

A company insures homes in three cities. Since sufficient distance separates the cities, it is reasonable to assume that the losses occurring in these cities, X, Y and Z, are independent. The moment generating functions for the loss distributions of the cities are: mX(t) = (1 – 2t)-3       mY(t) = (1 – 2t)-2.5          mZ(t) = (1 – 2t)-4.5

Let L represent the combined losses from the three cities: L = X + Y + Z. Find the third noncentral moment of random variable L.

 A.10,560
 B.5,760
 C.2,082
 D.1,320
 E.8,000

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