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Example
Created on 2024-06-20Asked by Logan Allen (Solvelet student)
Find the expected value E(X) E(X) of a discrete random variable X X with the following probability distribution: xiP(X=xi)10.220.330.440.1 \begin{array}{c|c} x_i & P(X = x_i) \\ \hline 1 & 0.2 \\ 2 & 0.3 \\ 3 & 0.4 \\ 4 & 0.1 \\ \end{array}

Solution

To find the expected value E(X) E(X) of a discrete random variable X X : 1. Use the formula for expected value: E(X)=ixiP(X=xi). E(X) = \sum_{i} x_i P(X = x_i). 2. Substitute the given values into the formula: E(X)=10.2+20.3+30.4+40.1. E(X) = 1 \cdot 0.2 + 2 \cdot 0.3 + 3 \cdot 0.4 + 4 \cdot 0.1. 3. Simplify: E(X)=0.2+0.6+1.2+0.4=2.4. E(X) = 0.2 + 0.6 + 1.2 + 0.4 = 2.4. Therefore, the expected value E(X) E(X) is 2.4. Solved on Solvelet with Basic AI Model
Some of the related questions asked by Ella Smith on Solvelet
1. Calculate the expected value of a random variable with a probability distribution given by P(X=1)=0.2 P(X = 1) = 0.2 , P(X=2)=0.5 P(X = 2) = 0.5 , and P(X=3)=0.3 P(X = 3) = 0.3 .2. Find the variance of a random variable with a probability distribution given by P(X=1)=0.3P(X = -1) = 0.3 and P(X=1)=0.7P(X = 1) = 0.7.
DefinitionRandom variables are the numerical outcomes of random events. They can be Just a specific or individual values (called Discrete) or any values within a range (Continuous). Heads in 10 coin flips (discrete random variable) vs. height of students in a class (continuous random variable) if our random variable is counts(#) then it is discrete, heights( 4f,4.11 fts) as we can see our random variable is quantitative and it is not countable.
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