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:

Step1: Identify binomial parameters

n=42, p=0.08, q=0.92

Step2: a. P(X=1) using binomial formula

$P(1)=\binom{42}{1}(0.08)^1(0.92)^{41}\approx42*0.08*0.0327\approx0.110$

Step3: b. P(X≤3)=P(0)+P(1)+P(2)+P(3)

$P(0)\approx0.0301$, $P(1)\approx0.110$, $P(2)\approx0.196$, $P(3)\approx0.227$; Sum≈0.563

Step4: c. Given P(X≥1)=0.970

Confirmed as 1-P(0)=1-0.030≈0.970

Step5: d. P(3≤X≤11)=Sum P(3) to P(11)

Sum≈0.227+0.193+0.128+0.068+0.031+0.012+0.004+0.001≈0.664

Answer:

a. 0.110
b. 0.563
c. 0.970
d. 0.664