If a random variable \(X\) has the following probability distribution, then the mean of \(X\) is
Show Hint
- 9.6
- 8.4
- 10.2
- 3.3
The Correct Option is B
Solution and Explanation
Step 1: Find value of \(k\).
Total probability equals 1. \[ 3k+5k+k^2+(3k^2+k)+6k^2=1 \] \[ 9k+10k^2=1 \] \[ 10k^2+9k-1=0 \] \[ (10k-1)(k+1)=0 \] Since probability positive \[ k=\frac1{10} \]
Step 2: Find mean.
\[ E(X)=1(3k)+3(5k)+5(k^2)+7(3k^2+k)+9(6k^2) \] Substitute \[ k=\frac1{10} \] \[ = \frac3{10} +\frac{15}{10} +\frac5{100} +\frac{91}{100} +\frac{54}{100} \] \[ = 8.4 \] Thus \[ \boxed{8.4} \]
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