Question

An urn has marbles of red, blue, and yellow. There are seven marbles of each of the three colors. The experiment consists of drawing a single, random marble from the urn. Define a random variable X by

X(s) = -1 if the marble has red

0, if the marble has blue

+1, if the marble has yellow

54 random samples of X are taken. Approximate the probability that the mean of the sample is between 0 and 1/20.

Answer 1

Solution

Back-up Theory

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

Expected value = Mean (average) of X = E(X) = µ = Σ{x.p(x)} summed over all possible values of x…...................…. (1)

E(X²) = Σ{x².p(x)} summed over all possible values of x…………………................................................………………..(2)

Variance of X = Var(X) = σ2 = E[{X – E(X)}²] = E(X²) – {E(X)}²…………….....................................................…………..(3)

Standard Deviation of X = SD(X) = σ = sq.rt of Var(X) …………………..…..................................................……………..(4)

If X has mean µ, and standard deviation σ, then by Central Limit Theorem, the sample average Xbar based on a sample of size n, has Normal distribution with mean µ, and standard deviation σ/√n ..................................................... (5)

Now to work out the solution,

‘There are seven marbles of each of the three colors.’ =>

P(Red) = P(Blue) = P(Yellow) = 1/3.

Also given that

X(s) = -1 if the marble is red

0, if the marble is blue

+1, if the marble is yellow,

Vide (1),

Mean of X,µ = 0

Vide (2),

E(X²) = 2/3

And hence vide (4),

standard deviation of X, σ = √(2/3) = 0.8165

So, given n = 54, vide (5),

Xbar ~ N(0, 0.1111) [0.1111 = 0.8165/√54]

The required probability

= P(0 < Xbar < 0.05)

= P{0 < Z < (0.05/0.1111), where Z ~ N(0, 1)

= P(0 < Z < 0.45)

= 0.1736 [Using Excel Function: Statistical NORMSDIST] Answer

DONE