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18 January, 11:38

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

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  1. 18 January, 14:48
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    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.
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