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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

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Answer:

1.18

Step-by-step 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

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