Question

X 0 1 2 3 4 5 P(X) 0.10 0.20 0.45 0.15 0.05 0.05 Using the above information, the standard deviation of the number of accidents is A. 1.18 B. 1.4 C. 1.81 D. 1.96

Answer 1

hope this helps.enjoy

Answer 2

1.18

Explanation:

Given the data:

X 0 1 2 3 4 5

P(X) 0.10 0.20 0.45 0.15 0.05 0.05

E(x) = (0*0.1) + (1*0.2) + (2*0.45) + (3*0.15) + (4*0.05) + (5*0.05) = 2

Var(x) = [((0^2)*0.1)+((1^2)*0.2)+((2^2)*0.45)+((3^2)*0.15)+((4^2)*0.05)+((5^2)*0.05)] - 2²

5.4 - 4

= 1.4

Standard deviation = √var(x)

Standard deviation = √1.4

Standard deviation = 1.183

Standard deviation = 1.18