Question

The probability distribution shown represents the number of trips of five nights or more that American adults take per year. (That is, 6% do not take any trips lasting five nights or more, 70% take one trip lasting five nights or more per year, etc.) Find the mean.

Answer 1

$$\def\arraystretch{1.5} \begin{array}{c:c:c:c:c:} Number & 0 & 1 & 2 & 3 & 4 \\ of\ trips \\ \hline P(X=x) & 0.06 & 0.70 & 0.20 & 0.03& 0.01 \\ \end{array}$$

$$mean=E(X)=\sum_ix_iP(X=x_i)$$

$$=0(0.06)+1(0.70)+2(0.20)+3(0.03)+4(0.01)$$

$$=1.23$$

The mean is $1.23.$