Suppose that the probability that a child living in an urban

Suppose that the probability that a child living in an urban area in the US is obese is 20%. A social worker sees 10 children living in an urban area. What is the probability that no more than 3 are obese?

Solution

Answer: .879

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Using the binomial formula:

X = # who are obese

Since X is a count of Yes/No outcomes in n independent trials with a constant probability,

X has a binomial distribution.

binomial formuala: P(x) = nCx * p^x * (1-p)^(n-x), where nCx = n!/(x!*(n-x)!)
n = 10
p = .20
1-p = .80

P(no more than 3)
= P(X < or = 3)
= P(3) + P(2) + P(1) + P(0)

P(3) = 10C3 * .20^3 + .80^7 = .2013266
P(2) = 10C2 * .20^2 + .80^8 = .3019899
P(1) = 10C1 * .20^1 + .80^9 = .2684355
P(0) = 10C0 * .20^0 + .80^10 = .1073742

= .2013266 + ... + .1073742
= .879

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I hope this helped. If you have any questions, please ask them in the comment section. :)