#### Question

The probability mass function for X = number of major defects in a randomly selected

appliance of a certain type is

X = x | 0 | 1 | 2 | 3 | 4 |

P(X = x) | 0.08 | 0.15 | 0.45 | 0.27 | 0.05 |

Find the expected value and variance of X.

clickto share

#### Solution

`E(X)=sumx_i.P(x_i)`

`=0(0.08)+1(0.15)+2(0.45)+3(0.27)+4(0.05)`

`=0+0.15+0.9+0.81+0.2=2.06`

`E(X^2)=sumx_i^2.P(x_i)`

`=0(0.08)+1^2(0.15)+2^2(0.45)+3^2(0.27)+4^2(0.05)`

`=0 + 0.15 + 1.8 + 2.43 + 0.8=5.18`

`Var(X)=E(X^2)-[E(X)]^2`

`=5.18-(2.06)^2`

`=5.18-4.2436`

`=0.9364`

Is there an error in this question or solution?

Solution for question: The probability mass function for X = number of major defects in a randomly selected appliance of a certain type is concept: Variance of Binomial Distribution (P.M.F.). For the courses HSC Science (Computer Science), HSC Science (Electronics), HSC Arts, HSC Science (General)