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Question 27

\begin{tabular}{|l|l|}
\hline
[tex]$X$[/tex] & [tex]$P(X)$[/tex] \\
\hline
0 & 0.228 \\
\hline
1 & 0.268 \\
\hline
2 & 0.299 \\
\hline
3 & 0.205 \\
\hline
\end{tabular}

Find the variance of the discrete random variable [tex]$X$[/tex].

Round any intermediate values to three decimals, then round your answer to two decimals.


Sagot :

To find the variance of the discrete random variable \(X\), we need to go through several steps: calculating the expected value \(E(X)\), the expected value of \(X^2\) which is \(E(X^2)\), and then applying the formula for variance. Here's the detailed, step-by-step solution:

### Step 1: Calculate the Expected Value \(E(X)\)
The expected value, \(E(X)\), of a random variable \(X\) with possible values \(x_i\) and probabilities \(P(x_i)\) is given by:
[tex]\[ E(X) = \sum (x_i \cdot P(x_i)) \][/tex]

For given values:
[tex]\[ \begin{aligned} E(X) &= 0 \cdot 0.228 + 1 \cdot 0.268 + 2 \cdot 0.299 + 3 \cdot 0.205 \\ &= 0 + 0.268 + 0.598 + 0.615 \\ &= 1.481 \end{aligned} \][/tex]

Rounding to three decimals, we get:
[tex]\[ E(X) = 1.481 \][/tex]

### Step 2: Calculate \(E(X^2)\)
Next, we calculate \(E(X^2)\):
[tex]\[ E(X^2) = \sum (x_i^2 \cdot P(x_i)) \][/tex]

For the given values:
[tex]\[ \begin{aligned} E(X^2) &= 0^2 \cdot 0.228 + 1^2 \cdot 0.268 + 2^2 \cdot 0.299 + 3^2 \cdot 0.205 \\ &= 0 \cdot 0.228 + 1 \cdot 0.268 + 4 \cdot 0.299 + 9 \cdot 0.205 \\ &= 0 + 0.268 + 1.196 + 1.845 \\ &= 3.309 \end{aligned} \][/tex]

Rounding to three decimals, we get:
[tex]\[ E(X^2) = 3.309 \][/tex]

### Step 3: Calculate the Variance
The variance of a random variable \(X\), denoted as \(Var(X)\), is given by the formula:
[tex]\[ Var(X) = E(X^2) - [E(X)]^2 \][/tex]

Substituting the values we found:
[tex]\[ \begin{aligned} Var(X) &= 3.309 - (1.481)^2 \\ &= 3.309 - 2.192561 \\ &= 1.116439 \end{aligned} \][/tex]

### Step 4: Round the Variance
Finally, rounding the variance to two decimals, we get:
[tex]\[ Var(X) = 1.12 \][/tex]

Thus, the variance of the discrete random variable [tex]\(X\)[/tex] is [tex]\(1.12\)[/tex].