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) Obtain cumulative distribution function (c.d.f) of X
Solution
Let's solve each part of the problem step by step.
Part i) Find the value of
Given the probability distribution of :
3k & \text{for } X = 1 \\ 5k & \text{for } X = 2 \\ 2k & \text{for } X = 3 \\ k & \text{for } X = 4 \\ k & \text{for } X = 5 \end{cases}$$ Since the sum of all probabilities must equal 1 (the total probability must be 1): $$3k + 5k + 2k + k + k = 1$$ $$12k = 1$$ $$k = \frac{1}{12}$$ So, the value of $$k$$ is $$\frac{1}{12}$$. ### Part ii) Find $$P(X)$$ when $$X$$ is even The even values of $$X$$ are 2 and 4. We need to find the probability that $$X$$ is even: $$P(X \text{ is even}) = P(X = 2) + P(X = 4)$$ Substitute the values from the probability distribution: $$P(X \text{ is even}) = 5k + k = 6k$$ Using the value of $$k = \frac{1}{12}$$: $$P(X \text{ is even}) = 6 \times \frac{1}{12} = \frac{6}{12} = \frac{1}{2}$$ So, $$P(X \text{ is even}) = \frac{1}{2}$$. ### Part iii) Obtain the cumulative distribution function (c.d.f) of $$X$$ The cumulative distribution function (c.d.f) $$F(x)$$ is defined as: $$F(x) = P(X \leq x)$$ To obtain $$F(x)$$, we calculate the cumulative probabilities up to each value of $$X$$: - For $$X = 1$$: $$F(1) = P(X \leq 1) = P(X = 1) = 3k = 3 \times \frac{1}{12} = \frac{3}{12} = \frac{1}{4}$$ - For $$X = 2$$: $$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}$$ - For $$X = 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{5}{6}$$ - For $$X = 4$$: $$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}$$ $$F(4) = \frac{11}{12}$$ - For $$X = 5$$: $$F(5) = P(X \leq 5) = P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) = 3k + 5k + 2k + k + k = 12k = 1$$ So, the cumulative distribution function $$F(x)$$ is: $$F(x) = \begin{cases} 0 & x < 1 \\ \frac{1}{4} & 1 \leq x < 2 \\ \frac{2}{3} & 2 \leq x < 3 \\ \frac{5}{6} & 3 \leq x < 4 \\ \frac{11}{12} & 4 \leq x < 5 \\ 1 & x \geq 5 \end{cases}$$ Would you like any additional details or have any questions? Here are some related questions to consider: 1. What is the expected value (mean) of $$X$$ for this distribution? 2. How would the variance of $$X$$ be calculated for this distribution? 3. How does changing the value of $$k$$ affect the distribution? 4. How do you interpret the cumulative distribution function in practical terms? 5. What is the probability that $$X$$ is greater than 3? **Tip:** Understanding the cumulative distribution function (c.d.f) helps in determining the probability that a random variable takes on a value less than or equal to a specific value.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Probability Theory
Cumulative Distribution Function
Formulas
Probability distribution function
Theorems
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Suitable Grade Level
College Level
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