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The distribution of the number of children per household for households receiving aid to dependent children (ADC) in a large eastern city is as follows: 5% of ADC households have one child, 35% of ADC households have two children, 30% of ADC households have 3 children, 20% of ADC households have 4 children, and 10% of ADC households have 5 children.

Calculate the mean, variance and Standard Deviation for this data.

Attempt:

$\text{Mean} = \sum X P(X)$

$\text{Mean} = 0.05\times1 + 0.35\times2 + 0.30\times3 + 0.20\times4 + 0.10\times5 = 2.95$

$\text{Variance} = \sum X^2 P(X) - \text{Mean}^2$

$\text{Variance} = \left(1^2\times0.05 + 2^2\times0.35 + 3^2\times0.30 + 4^2\times0.20 + 5^2\times0.10\right) - (2.95^2) = 1.1475 $

$\text{Standard Deviation} = \sqrt{\text{Variance}}$

$\text{Standard Deviation} = \sqrt{1.1475} = 1.071$

What concerns me is that I have not calculated the probability correct here perhaps. If I have, kindly check my answers and let me know if they are right or wrong :)

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I haven't checked the sums, but it looks right. What has this got to do with the hypergeometric distribution? – Lucas Jun 7 '13 at 2:41
    
Yeah I just realized it really didn't. The question was just in the same chapter as Hypergeometric Distribution so I thought I might have missed something. Thanks for the help! – user81167 Jun 7 '13 at 2:52

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