Kaplan International Pathways Mini Test ✓ Solved

Kaplan International Pathways 1 Kaplanpathwayscommini Test

Analyze a discrete random variable X with a given probability distribution involving an unknown constant a, calculate expected values and variance for X, determine the variance of a transformed variable Y = 6 - 2X, and evaluate the probability P(X ≥ Y).

Sample Paper For Above instruction

Introduction

This paper addresses a set of statistical problems involving discrete random variables, their probability distributions, expectations, variances, transformations, and probability calculations. The tasks involve algebraic manipulation, application of probability rules, and understanding of variance and distributions in probability theory.

Problem Statement and Analytical Approach

The problem begins by considering a discrete random variable X with the probability distribution given as follows: P(X = x) = a/5 for x = 1, and P(X = x) = 1/10 for x = 2, with a being an unknown constant. The first task is to find the value of a using the probability axioms. Subsequently, the expected value E(X) and the variance Var(X) are computed. The problem then introduces a transformation of the variable, defining Y = 6 - 2X, and asks for the variance of Y, Var(Y). Finally, the probability P(X ≥ Y) is to be calculated, which involves understanding the distribution of X and the values Y can take based on X.

Step 1: Determining the Constant a

The basic probability rule states that the sum of probabilities for all possible outcomes must equal 1. Given the distribution: P(X=1) = a/5, P(X=2) = 1/10, and assuming these are comprehensive for the discrete outcomes considered, we sum to find a:

 (a/5) + (1/10) = 1

Converting to a common denominator and solving for a:

 (2a/10) + (1/10) = 1

which simplifies to:

 (2a + 1)/10 = 1

Multiplying both sides by 10:

 2a + 1 = 10

and solving for a:

 a = (10 - 1)/2 = 9/2 = 4.5

Thus, the constant a equals 4.5.

Step 2: Calculating Expected Value E(X)

The expected value of X is given by:

 E(X) = Σ x  P(X = x) = (1)(a/5) + (2)*(1/10)

Substituting a = 4.5:

 E(X) = 1(4.5/5) + 2(1/10) = (4.5/5) + (2/10)

Calculate each term:

 4.5/5 = 0.9 
 2/10 = 0.2

Adding these gives:

 E(X) = 0.9 + 0.2 = 1.1

Therefore, the expected value E(X) is 1.1.

Step 3: Variance of X, Var(X)

Variance is calculated using:

 Var(X) = E(X^2) - [E(X)]^2

First, find E(X^2):

 E(X^2) = (1^2)(a/5) + (2^2)(1/10) = 1(4.5/5) + 4(1/10) = 0.9 + 0.4 = 1.3

Then, compute Var(X):

 Var(X) = 1.3 - (1.1)^2 = 1.3 - 1.21 = 0.09

Thus, the variance Var(X) = 0.09.

Step 4: Variance of Y = 6 - 2X, Var(Y)

Using properties of variance, for a constant c and a random variable X:

 Var(c - dX) = d^2 * Var(X)

In this case, c = 6 and d = 2, thus:

 Var(Y) = 4  Var(X) = 4  0.09 = 0.36

So, the variance of Y is 0.36.

Step 5: Calculating P(X ≥ Y)

Because Y = 6 - 2X, the inequality X ≥ Y becomes:

 X ≥ 6 - 2X

Adding 2X to both sides:

 3X ≥ 6

Dividing both sides by 3:

 X ≥ 2

This indicates that P(X ≥ Y) is equivalent to P(X ≥ 2). Now, based on the earlier distribution, the outcomes were primarily for X = 1 and X = 2; if there are only these two outcomes, then:

  • If P(X=2) is known, then P(X ≥ 2) is simply P(X=2), as X=2 is the only outcome satisfying X ≥ 2.

Previously, we calculated P(X=2) as 1/10, but note that a/5 was 4.5/5=0.9, leading to total probability sum of 1. For our specific distribution, the probability P(X=2) is 1/10 = 0.1. Therefore,

 P(X ≥ 2) = P(X=2) = 0.1

Thus, the probability P(X ≥ Y) equals 0.1.

Conclusion

This analysis demonstrates the application of fundamental probability and statistics principles to discrete random variables, their transformations, and calculations of expectations, variances, and probabilities. Understanding these concepts is vital for statistical inference, modeling, and decision-making.

References

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  • Moore, D. S., McCabe, G. P., & Craig, B. A. (2012). Introduction to the Practice of Statistics (8th ed.). W. H. Freeman.
  • Freedman, D., Pisani, R., & Purves, R. (2007). Statistics (4th ed.). W. W. Norton & Company.
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