Question

Suppose Piper eats out twice a week 15% of the time, she eats out once a week 35% of the time, and she does not eat out any time during the week 50% of the time.What is the expected value for the number of times Piper eats out during the week? Round your answer to the nearest hundredth if needed.

Answer 1

We are given

Probability of eating out twice in a week = 15% = 0.15

Probability of eating out once in a week = 35% = 0.35

Probability of not eating out in a week = 50% = 0.50

Let X be a random variable of the number of times Piper eats out in a week

So we have the table

Note: The Formula For finding the Expected value E(X) is given by

`E(X)=\sum ^{}_{}xp(x)`

Substituting we get

`\begin{gathered} E(X)=0(0.50)+1(0.35)+2(0.15) \\ E(X)=0+0.35+0.30 \\ E(X)=0.65 \end{gathered}`

Therefore, the expected value is

`E(X)=0.65`