SOLUTION: Let y = number of languages in which a person is fluent. According to Statistics Canada, for residents of Canada y has probability distribution P(0) = 0.02, P(1) =0.81, and P(2) =

Question 1165739: Let y = number of languages in which a person is fluent. According to Statistics Canada, for residents of
Canada y has probability distribution P(0) = 0.02, P(1) =0.81, and P(2) = 0.17, with negligible probability for
higher values of y.
(a) Is y a discrete, or a continuous, variable? Why?
(b) Construct a table showing the probability distribution of y.
(c) Find the probability that a Canadian is not multilingual.
(d) Find the mean of this probability distribution.

Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
y is discrete. Here, the number of languages in which one is fluent has to be an integer.
probability a Canadian is not multilingual is 0.83%, 1- probability of P(2)
the mean is x*p(x)
0*0.02+1*0.81+2(0.17)=1.15 languages (mean)
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