Question:
Let X be a random variable with distribution.
x \(-2\) \(-1\) 3 4 6 P(X = x) \(\frac15\) a \(\frac13\) \(\frac15\) b
If the mean of X is 2.3 and variance of X is \(\sigma^2\), then \(100\sigma^2\) is equal to :
Let X be a random variable with distribution.
If the mean of X is 2.3 and variance of X is \(\sigma^2\), then \(100\sigma^2\) is equal to :
| x | \(-2\) | \(-1\) | 3 | 4 | 6 |
| P(X = x) | \(\frac15\) | a | \(\frac13\) | \(\frac15\) | b |
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Use \(\sum p=1\) and the mean 2.3 to find a and b, then \(\sigma^2=E(X^2)-[E(X)]^2\).
Updated On: Nov 4, 2026
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Correct Answer: 781
Solution and Explanation
Step 1: \(\frac15+a+\frac13+\frac15+b=1\Rightarrow a+b=\frac{4}{15}\).
Step 2: Mean: \(-\frac25-a+1+\frac45+6b=2.3\Rightarrow -a+6b=0.9\).
Step 3: Adding: \(7b=\frac{4}{15}+\frac{9}{10}=\frac{35}{30}\Rightarrow b=\frac16\), and \(a=\frac4{15}-\frac16=\frac1{10}\).
Step 4: \(E(X^2)=4\cdot\frac15+1\cdot\frac1{10}+9\cdot\frac13+16\cdot\frac15+36\cdot\frac16=0.8+0.1+3+3.2+6=13.1\).
Step 5: \(\sigma^2=13.1-(2.3)^2=13.1-5.29=7.81\), so \(100\sigma^2=781\).
Answer: 781
Step 2: Mean: \(-\frac25-a+1+\frac45+6b=2.3\Rightarrow -a+6b=0.9\).
Step 3: Adding: \(7b=\frac{4}{15}+\frac{9}{10}=\frac{35}{30}\Rightarrow b=\frac16\), and \(a=\frac4{15}-\frac16=\frac1{10}\).
Step 4: \(E(X^2)=4\cdot\frac15+1\cdot\frac1{10}+9\cdot\frac13+16\cdot\frac15+36\cdot\frac16=0.8+0.1+3+3.2+6=13.1\).
Step 5: \(\sigma^2=13.1-(2.3)^2=13.1-5.29=7.81\), so \(100\sigma^2=781\).
Answer: 781
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