Question

1. Suppose X has a mean 5 and standard deviation 2. Let Y= 4x-3. Determine the mean and standard deviation of Y.

2. From an urn containing 6 red and 4 white marbles, 3 are drawn with replacement. Let X be the number of red marbles selected. Define Y as the square of the number of red marbles selected. Determine E(Y) and V(Y).

3.An examiner contains 10 multiple choice questions, each with five choices. If a student guesses the answer to each question what is the probability the students exam grade is a) at least 60%, b) at least 80& c) 50% or less d) what is the students expected grade?

4.A fair pair of dice are tossed. a) determine the probability the fifth sum of seven occurs on the 15th toss. b) How many tosses of the dice are needed to obtain the fifth sum of seven and with what standard deviation.

Answer 1

Solution

Q1

Back-up Theory

E(a + bX) = a + bE(X), ………………………………………………………….(1)

Var(a + bX) = b²V(X), ……………………………………………………….(2)

SD(a + bX) = bSD(X), ……………………………………………………….(3)

Now, to work out the solution,

Given mean(X) = 5 and SD(X) = 2 and Y = 4x – 3,

Vide (1), mean(Y) = 4{mean(X)} – 3

= 17 Answer 1

Vide (3),

SD(Y) = 4{SD(X)}

= 8 Answer 2

Q2

Back-up Theory

If a discrete random variable, X, has probability p(x), x = x1, x2, …., xn, then

Mean (average) of X = E(X) = µ = Σ{x.p(x)} summed over all possible values of x……..…. (4)

Mean of a function f(x) of variable X

= E{f(X)} = Σf(x).p(x)} summed over all possible values of x………………………………(5)

Variance of X = Var(X) = σ² = E(X²) – {E(X)}²……………………………………………..(6)

Now, to work out the solution,

Since sampling is with replacement, probability of a red marble = 6/10 = 0.6 for all draws. So, probability of a non-red marble = 0.4.

To get x red marbles in a sample of 3, there are ³C_x possibilities each with a probability of 0.6^x.0.4^{3 – x}

So, p(x red marbles) = (³C_x)0.6^x.0.4^{3 – x}, where x = 0, 1, 2 or 3.

The probability distribution of X and related calculations are shown below:

x	0	1	2	3	Total
p(x)	0.064	0.288	0.432	0.216	1
x.p(x)	0	0.288	0.864	0.648	1.8
x^2.p(x)	0	0.288	1.728	1.944	3.96
x^4.p(x)	0	0.288	6.912	17.496	24.696

Given Y = X^2
E(X) =	1.8	[vide (4)]
E(X^2) =	3.96	[vide (5)]
E(X^4) =	24.696	[vide (5)]
E(Y) = E(X^2) = 3.96 Answer 1
V(Y) = E(Y^2) - {E(y)}^2
or V(Y) = E(X^4) - {E(X^2)}^2		9.0144 Answer 2

DONE