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A newsstand has ordered five copies of a certain issue of a photography magazine. let x 5 the number of individuals who come in to purchase

A newsstand has ordered five copies of a certain issue of a photography magazine. let x 5 the number of individuals who come in to purchase this magazine. if x has a poisson distribution with parameter m 5 4, what is the expected number of copies that are sold

Asked by: Guest | Views: 55
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Guest [Entry]

"Hi!

To get the expected number of copies that are sold, we have to add up the products of the expected value * probability.

So first by expected value, I mean the expected number of copies to be sold. Our possible number of copies sold is anywhere from 0, 1, 2, 3, 4, 5. Five is the maximum since the newsstand only ordered 5 copies of the magazine. Even if more than 5 people come in to purchase a magazine, there is only 5 in stock.

Now, to get the probability of each x happening, that is, the probability that x ( k in the table and formula, to avoid confusion) people will come in to purchase a magazine, we use Poisson's probability mass function:

where , as given.

The answers to this, when k = 0, 1, 2, 3, 4 are in the table under the ""Probability"" column. To get the probability of k > 4, we simply subtract the total of the probabilities from 1. This is since it is a property of probabilities that their sum is 1.

Why did we separate k>4? This is because the expected value is still 5 either way, again because of the order. So the table so far goes like this:

k = number of people who come in to purchase a magazine

probability = probability that k people will come in

units sold = units sold if k people come in

The last column is just the product when you multiply the 2nd and 3rd columns. The expected value is what you get when you add all the entries of this last column. You can think of what's happening here as just taking the average number of units sold, but weighted based on their probabilities.

Thus, the answer is 3.59.

Hope this helps!"