Answer:
Mean, E(x) = 0.17
Variance, V(x) = 0.12
Step-by-step explanation:
90% contains no defective bulbs, P(X=0) = 0.9
5% contain one defective bulbs, P(X=1) = 0.05
3% contain two defective bulbs, P(X=2) = 0.03
2% contain three defective bulbs, P(X=3) = 0.02
The mean for the number of defective bulbs:
[tex]E(X) = \sum XP(X)\\E(X) = (0*0.9) + (1*0.05) + (2*0.03) + (3* 0.02)\\E(X) = 0 + 0.05 + 0.06 + 0.06\\E(X) = 0.17[/tex]
Variance for the defective bulbs:
[tex]V(x) = (\sum x^{2}P(x) ) - \mu^2[/tex]
[tex]\sum x^{2}P(x) = (0^2*0.9) + (1^2*0.05) + (2^2*0.03) + (3^2* 0.02)\\\sum x^{2}P(x) = 0.05 + 0.12 + 0.12\\\sum x^{2}P(x) = 0.29[/tex]
V(x) = 0.29 - 0.17
V(x) = 0.12
