Question

8% of all americans live in poverty. if 42 americans are randomly selected, find the probability that a. exactly 1 of them live in poverty. b. at most 3 of them live in poverty. c. at least 1 of them live in poverty. 0.970 d. between 3 and 11 (including 3 and 11) of them live in poverty. hint:

Explanation:

This is a binomial probability problem. The binomial probability formula is $P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}$, where $n$ is the number of trials, $k$ is the number of successes, $p$ is the probability of success on a single - trial, and $C(n,k)=\frac{n!}{k!(n - k)!}$. Here, $n = 42$, $p=0.08$, and $1 - p = 0.92$.

Step1: Calculate probability for part a

For exactly $k = 1$ success:
First, calculate the combination $C(42,1)=\frac{42!}{1!(42 - 1)!}=\frac{42!}{1!41!}=42$.
Then, $P(X = 1)=C(42,1)\times(0.08)^{1}\times(0.92)^{41}$
$P(X = 1)=42\times0.08\times(0.92)^{41}\approx42\times0.08\times0.0387\approx0.130$

Step2: Calculate probability for part b

For at most 3 successes ($P(X\leq3)=P(X = 0)+P(X = 1)+P(X = 2)+P(X = 3)$)
$P(X = 0)=C(42,0)\times(0.08)^{0}\times(0.92)^{42}=\frac{42!}{0!(42 - 0)!}\times1\times(0.92)^{42}=1\times1\times0.0357 = 0.0357$
$P(X = 1)\approx0.130$ (calculated above)
$C(42,2)=\frac{42!}{2!(42 - 2)!}=\frac{42\times41}{2\times1}=861$
$P(X = 2)=C(42,2)\times(0.08)^{2}\times(0.92)^{40}=861\times0.0064\times0.0421\approx0.233$
$C(42,3)=\frac{42!}{3!(42 - 3)!}=\frac{42\times41\times40}{3\times2\times1}=11480$
$P(X = 3)=C(42,3)\times(0.08)^{3}\times(0.92)^{39}=11480\times0.000512\times0.0464\approx0.274$
$P(X\leq3)=0.0357 + 0.130+0.233 + 0.274=0.673$

Step3: Calculate probability for part d

For between 3 and 11 successes ($P(3\leq X\leq11)=P(X\leq11)-P(X\leq2)$)
We can use a binomial probability table or a calculator with binomial - cumulative distribution function.
$P(X\leq11)$ can be calculated by summing $P(X = k)$ for $k$ from 0 to 11.
$P(X\leq2)=P(X = 0)+P(X = 1)+P(X = 2)=0.0357+0.130 + 0.233=0.400$
Using a calculator or software (e.g., TI - 84 Plus: binomcdf(42,0.08,11)-binomcdf(42,0.08,2))
$P(3\leq X\leq11)\approx0.997-0.400 = 0.597$

Answer:

a. $0.130$
b. $0.673$
d. $0.597$