Question
Asked Oct 8, 2019
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According to a government study, 15% of all children live in a household that has an income below the poverty level. If a random sample of 15 children is selected:
a) what is the probability that 5 or more live in poverty?
b) what is the probability that 5 live in poverty?
c) what is the expected number (mean) that live in poverty? What is the variance? What is the standard deviation?

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Expert Answer

Step 1

The binomial distribution gives the probability of number of successes out of n trials in a series of Bernoulli trials.

The probability mass function (pmf) of a binomial random variable X is given as:

Let X is the number of children who live in poverty, which follows the binomial distribution.

 

It is given that the probability of a child who live in poverty is 0.15 and the sample size is 15.

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p*q*, x= 0,1,..n,0<p<1;q 1- p P(x)= otherwise |0 n=15 p 0.15

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Step 2

a).

The probability that 5 or more live in poverty is calculated as follows:

Now, the probability that X less than 5 can be calculated as follows.

The required probability is,

Therefore, the probability that 5 or more live in poverty is 0.064.

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P(X2 5) 1-P(X<5) Р(х <5)- JP(х - 0) + P(X -1) + P(X -2) |-P(x -3) + P(х -4) JO190-1509 .15(0)0.1 (0.5)" 15 15 (0.15 3 = {0.087 +0.231+ 0.28 5 + 0.218+0.115 0.936 P(X25) 1-P(X< 5) 1-0.936 0.064

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Step 3

b).

The probability that 5 live in poverty can be calculated as follows:

T...

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15 (0.15(0.85) 10 P(X=5) 5 =0.044

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