MHT CET202015 Oct 2020Evening ShiftMathematicsProbabilityActual
For the probability distribution of X given below array |c|c|c|c|c|c| X=x & -2 & -1 & 0 & 1 & 2 P ( X =x) & 0 2 & 0 3 & 0 1 & 0 15 & 0 25 array The variance of X is
Options
- A2.4257
- B2.5427
- C2.5742
- D2.2475
Correct answer
D. 2.2475
Step-by-step solution
_ i=1 ^ n x_ i P_ i =E(X) aligned E(X) &=(-2)(0.2)+(-1)(0.3)+0(0.1)+1(0.15)+2(0.25) &=-0.4-0.3+0.15+0.50=-0.7+0.65=-0.05 E (X² ) &= _ i=1 ^ n x_ i ² P_ i &=4(0.2)+1(0.3)+0(0.1)+1(0.15)+4(0.25) &=0.8+0.3+0.15+1=2.25 &=E (X² )-[E(X)]² &=2.25-0.0025=2.2475 aligned