Math Problem Statement
- A random variable X has the following probability function: X -3 -2 -1 0 1 2 3 P(x) k 0.1 K 0.2 2K 0.4 2K Then find (i) k (ii) mean (iii) variance (iv) P(0 < x < 3)
Solution
We are given a random variable with a probability function. The values of and their corresponding probabilities are as follows:
\hline X & P(X) \\ \hline -3 & k \\ -2 & 0.1 \\ -1 & k \\ 0 & 0.2 \\ 1 & 2k \\ 2 & 0.4 \\ 3 & 2k \\ \hline \end{array}$$ We need to find: 1. $$k$$ 2. The mean ($$\mu$$) 3. The variance ($$\sigma^2$$) 4. $$P(0 < X < 3)$$ ### **Step 1: Finding $$k$$** We know that the sum of all probabilities for a random variable must equal 1: $$k + 0.1 + k + 0.2 + 2k + 0.4 + 2k = 1$$ Simplifying: $$6k + 0.7 = 1$$ $$6k = 1 - 0.7 = 0.3$$ $$k = \frac{0.3}{6} = 0.05$$ Thus, $$k = 0.05$$. ### **Step 2: Finding the Mean** The mean of a random variable is calculated as the expected value $$E(X)$$, which is given by: $$E(X) = \sum (X \cdot P(X))$$ Substituting the values of $$X$$ and $$P(X)$$: $$E(X) = (-3)(0.05) + (-2)(0.1) + (-1)(0.05) + (0)(0.2) + (1)(0.1) + (2)(0.4) + (3)(0.05)$$ $$E(X) = -0.15 - 0.2 - 0.05 + 0 + 0.1 + 0.8 + 0.15$$ $$E(X) = 0.65$$ Thus, the mean $$\mu = 0.65$$. ### **Step 3: Finding the Variance** The variance is given by: $$\text{Var}(X) = E(X^2) - (E(X))^2$$ First, let's calculate $$E(X^2)$$: $$E(X^2) = (-3)^2(0.05) + (-2)^2(0.1) + (-1)^2(0.05) + 0^2(0.2) + 1^2(0.1) + 2^2(0.4) + 3^2(0.05)$$ $$E(X^2) = 9(0.05) + 4(0.1) + 1(0.05) + 0(0.2) + 1(0.1) + 4(0.4) + 9(0.05)$$ $$E(X^2) = 0.45 + 0.4 + 0.05 + 0 + 0.1 + 1.6 + 0.45$$ $$E(X^2) = 3.05$$ Now, we calculate the variance: $$\text{Var}(X) = 3.05 - (0.65)^2 = 3.05 - 0.4225 = 2.6275$$ Thus, the variance $$\sigma^2 = 2.6275$$. ### **Step 4: Finding $$P(0 < X < 3)$$** We need to find the probability that $$X$$ is between 0 and 3. This is the sum of the probabilities for $$X = 1$$ and $$X = 2$$: $$P(0 < X < 3) = P(X = 1) + P(X = 2)$$ $$P(0 < X < 3) = 0.1 + 0.4 = 0.5$$ Thus, $$P(0 < X < 3) = 0.5$$. ### **Summary of Results:** 1. $$k = 0.05$$ 2. Mean ($$\mu$$) = 0.65 3. Variance ($$\sigma^2$$) = 2.6275 4. $$P(0 < X < 3) = 0.5$$ Would you like any further details or explanations on any of these steps? ### Related Questions: 1. How would you find the skewness of this distribution? 2. What would happen to the probabilities if $$k$$ were different? 3. Can we calculate the standard deviation from the variance? 4. How would you find the probability $$P(X \leq -1)$$? 5. What would the distribution look like if all values of $$P(X)$$ were scaled by 2? **Tip:** Always check that the sum of all probabilities equals 1, as this ensures that the probability function is valid.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Probability
Random Variables
Mean
Variance
Expected Value
Formulas
Sum of probabilities: ΣP(X) = 1
Mean (Expected Value): E(X) = Σ(X * P(X))
Variance: Var(X) = E(X^2) - (E(X))^2
Probability: P(a < X < b) = ΣP(X) for a < X < b
Theorems
Properties of probability distributions
Suitable Grade Level
Grades 11-12
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