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January 30, 2015

January 30, 2015

Posted by **Ryan** on Monday, February 7, 2011 at 7:50am.

According to the Census Bureau publication Current Population Reports, the probability distribution for household size (number of people per household, say X) in the United States is as follows. For the purpose of the next two questions, the experiment consists of randomly selecting a household in the United States and observing the number of people living there.

x 1 2 3 4 5 6 7

P(X = x) 0.16 0.32 0.05 0.31 0.10 0.05 0.01

How many people can we expect to see living in the randomly chosen household?

Answer

4.14

3.06

4.00

1.79

1.24

What is the probability that the randomly selected household will contain more than 4 people?

Answer

0.47

0.31

0.16

0.84

0.53

For the first one i got 1.79 and for the second i don't know if i did it right but i got .53

- Stats- need someone to check -
**drwls**, Monday, February 7, 2011 at 10:23amThe expectation value, or mean value, is the sum of the x*P(x) products, which is 3.06

Of course, one would not expect to find 6% of a person in a house. On the other hand, 3 is a relatively unlikely number for this distribution. 2 and 4 are seen much more often.

I do not like the way the question is worded. If they want the mean vaue, they should say so.

For the second, you do not need to use a normal distribution and compute the standard deviation etc.. Just add P(5), P(6) and P(7). You get 0.16

- Stats- need someone to check -
**Ryan**, Tuesday, February 8, 2011 at 3:17amThanks drwls

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