The number of electrical outages in a city varies from day to day. Assume that the number of electrical outages ( x ) in the city has the following probability distribution. x f (x) 0 0.80 1 0.15 2 0.04 3 0.01 The mean and the standard deviation for the number of electrical outages (respectively) are . a. 0.26 and .577 b. 3 and .01 c. 2.6 and 5.77

0.26 and 0.577

Step-by-step explanation:

Given :

x :___0 ___ 1 2 3

f(x):_0.80 _0.15__0.04 __0.01

The mean :

E(X) = Σ(x * p(x))

E(X) = (0*0.8) + (1*0.15) + (2*0.04) + (3*0.01)

E(X) = 0 + 0.15 + 0.08 + 0.03

E(X) = 0.26

Var(X) = Σx²*p(x)) - E(x)²

Σx²*p(x)) = (0^2*0.8) + (1^2*0.15) + (2^2*0.04) + (3^2*0.01) = 0.4

Σx²*p(x)) - E(x)² = 0.4 - 0.26² = 0.3324

Std(X) = √0.334 = 0.577

you can first factor out the n to get n(1+5). we know that 1+5 is 6, so we just have 6n.

ddd

step-by-step explanation: