Every day, the number of network blackouts has a distribution (probability mass function)
1 answer:
Expected Mean, E(X), is obtained by multiplying each pair of 
and its 
and add up the answers
E(X) = (0×0.7) + (1×0.2) + (2×0.1) = 0.4
The formula to calculate the variance, Var(X), is given by E(X)² - (E(X))²
E(X²) = (0²×0.7) + (1²×0.2) + (2²×0.1) = 0+0.2+0.4 = 0.6
(E(X))² = (0.4)² = 0.16
Var(X) = 0.6 - 0.16 = 0.44
Translating these answers into the context we have
E(Y) = 0.4×500 = $200
Var(Y) = $110
E(X) = (0×0.7) + (1×0.2) + (2×0.1) = 0.4
The formula to calculate the variance, Var(X), is given by E(X)² - (E(X))²
E(X²) = (0²×0.7) + (1²×0.2) + (2²×0.1) = 0+0.2+0.4 = 0.6
(E(X))² = (0.4)² = 0.16
Var(X) = 0.6 - 0.16 = 0.44
Translating these answers into the context we have
E(Y) = 0.4×500 = $200
Var(Y) = $110
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SOLUTION
The mean is 4min
standard deviation is 1min
the z score is
where
then we have
The probability the call lasted less than 3 min will be
Therefore, the probability that (z < -1 ) is
[tex]\begin{gathered} Pr(z<-1)=Pr(0Hence, the percentage of the calls that lasted less than 3 min is 16%