Question

The manager of a video store has compiled the following table, which gives the probabilities that a customer will buy 0, 1, 2, 3, or 4 DVDs. How many DVDs can a given customer be expected to buy? DVDs - 0 1 2 3 4 Probability - .42 .36 .14 .05 .03

Answer 1

Answer:

0.91 DVDs

Step-by-step explanation:

The expected number of DVDs is the sum of products of the number of DVDs and their probability:

  E(n) = 0·0.42 + 1·0.36 + 2·0.14 + 3·0.05 + 4·0.03

  = 0 + 0.36 + 0.28 + 0.15 + 0.12

  = 0.91

A given customer can be expected to buy 0.91 DVDs.