Question

According to a recent Census Bureau report, 12.7% of Americans live below the poverty level. Suppose you plan to sample at random 100 Americans and count the number of people who live below the poverty level. a. What is the probability that you count exactly 10 in poverty? b. What is the probability that you start taking the random sample and you find

Answer 1

Answer:

a) $P(X=10)=(100C10)(0.127)^{10} (1-0.127)^{100-10}=0.0928$

b) $P(X \leq 10) = 0.2614$

c) $(1-0.127)^7 (0.127) =0.0491$

Step-by-step explanation:

Previous concepts

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".

Solution to the problem

Let X the random variable of interest, on this case we now that:

$X \sim Binom(n=100, p=0.127)$

The probability mass function for the Binomial distribution is given as:

$P(X)=(nCx)(p)^x (1-p)^{n-x}$

Where (nCx) means combinatory and it's given by this formula:

$nCx=\frac{n!}{(n-x)! x!}$

a. What is the probability that you count exactly 10 in poverty?

For this case we want this probability P(X=10)

$P(X=10)=(100C10)(0.127)^{10} (1-0.127)^{100-10}=0.0928$

b. What is the probability that you count 10 or less in poverty?  .2614

For this case we want this probability $P(X=\leq10)$

$P(X\leq10)=P(X=1)+P(X=2)+P(X=3)+P(X=4)+P(X=5)+P(X=6)+P(X=7)+P(X=8)+P(X=9)+P(X=10)$

And we can find the individual probabilities like this:

$P(X=0)=(100C0)(0.127)^{0} (1-0.127)^{100-0}=1.263x10^{-6}$

$P(X=1)=(100C1)(0.127)^{1} (1-0.127)^{100-1}=1.837x10^{-5}$

$P(X=2)=(100C2)(0.127)^{2} (1-0.127)^{100-2}=0.000132$

$P(X=3)=(100C3)(0.127)^{3} (1-0.127)^{100-3}=0.000629$

$P(X=4)=(100C4)(0.127)^{4} (1-0.127)^{100-4}=0.00222$

$P(X=5)=(100C5)(0.127)^{5} (1-0.127)^{100-5}=0.00620$

$P(X=6)=(100C6)(0.127)^{6} (1-0.127)^{100-6}=0.0143$

$P(X=7)=(100C7)(0.127)^{7} (1-0.127)^{100-7}=0.0279$

$P(X=8)=(100C8)(0.127)^{8} (1-0.127)^{100-8}=0.0471$

$P(X=9)=(100C9)(0.127)^{9} (1-0.127)^{100-9}=0.0701$

$P(X=10)=(100C10)(0.127)^{10} (1-0.127)^{100-10}=0.0928$

And then repplacing we got:

$P(X \leq 10) = 0.2614$

c. What is the probability that you start taking the random sample and you find the first person in poverty on the 8th person selected?  .0491

For this case we need after 7 people , 1 in poverty so we can find this probability like this:

$(1-0.127)^7 (0.127) =0.0491$