Math Problem Statement

Consider the following probability distribution of X X 1 2 3 4 5 P(X=x) 3k 5k 2k k k i) Find the value of k ii) P(x) is even iii) Calculate distribution function of X

Solution

Let's solve each part of the problem step by step.

Part (i) Find the value of kk

Given the probability distribution of XX:

X & 1 & 2 & 3 & 4 & 5 \\ \hline P(X = x) & 3k & 5k & 2k & k & k \\ \end{array}$$ The sum of all probabilities must equal 1 because they represent the complete probability distribution: $$3k + 5k + 2k + k + k = 1$$ Simplifying: $$12k = 1$$ So, $$k = \frac{1}{12}$$ ### Part (ii) $$P(X = x)$$ when $$x$$ is even The even values of $$X$$ are 2 and 4. The probability that $$X$$ is even is: $$P(X \text{ is even}) = P(X = 2) + P(X = 4)$$ Substituting the values: $$P(X \text{ is even}) = 5k + k = 6k$$ Using the value of $$k$$ found earlier: $$P(X \text{ is even}) = 6 \times \frac{1}{12} = \frac{6}{12} = \frac{1}{2}$$ ### Part (iii) Calculate the distribution function $$F(x)$$ of $$X$$ The distribution function $$F(x)$$ is defined as: $$F(x) = P(X \leq x)$$ Let's compute $$F(x)$$ for each possible value of $$x$$: $$F(1) = P(X \leq 1) = P(X = 1) = 3k = 3 \times \frac{1}{12} = \frac{3}{12} = \frac{1}{4}$$ $$F(2) = P(X \leq 2) = P(X = 1) + P(X = 2) = 3k + 5k = 8k = 8 \times \frac{1}{12} = \frac{8}{12} = \frac{2}{3}$$ $$F(3) = P(X \leq 3) = P(X = 1) + P(X = 2) + P(X = 3) = 3k + 5k + 2k = 10k = 10 \times \frac{1}{12} = \frac{10}{12} = \frac{5}{6}$$ $$F(4) = P(X \leq 4) = P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) = 3k + 5k + 2k + k = 11k = 11 \times \frac{1}{12} = \frac{11}{12}$$ $$F(5) = P(X \leq 5) = P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 12k = 12 \times \frac{1}{12} = 1$$ So, the distribution function $$F(x)$$ is: $$F(x) = \begin{cases} 0 & \text{if } x < 1 \\ \frac{1}{4} & \text{if } 1 \leq x < 2 \\ \frac{2}{3} & \text{if } 2 \leq x < 3 \\ \frac{5}{6} & \text{if } 3 \leq x < 4 \\ \frac{11}{12} & \text{if } 4 \leq x < 5 \\ 1 & \text{if } x \geq 5 \\ \end{cases}$$ ### Summary: - $$k = \frac{1}{12}$$ - $$P(X \text{ is even}) = \frac{1}{2}$$ - The distribution function $$F(x)$$ is piecewise defined as described above. Would you like more details on any part of this solution? Here are 5 related questions to deepen your understanding: 1. How do you derive the cumulative distribution function (CDF) from a probability mass function (PMF)? 2. What is the significance of the normalization condition (sum of all probabilities equals 1)? 3. How would the distribution function $$F(x)$$ change if the probability distribution were different? 4. Can you explain the relationship between the distribution function and the probability of an event? 5. How would you calculate $$P(X$$ is odd)? **Tip:** Remember, the cumulative distribution function (CDF) $$F(x)$$ gives you the probability that a random variable $$X$$ is less than or equal to $$x$$.

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Math Problem Analysis

Mathematical Concepts

Probability Theory
Probability Distribution
Cumulative Distribution Function (CDF)

Formulas

Sum of probabilities = 1
Distribution function calculation

Theorems

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Suitable Grade Level

Advanced High School

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