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:

Part a: Exactly 1 success

Use binomial formula: \( P(X=1) = C(42,1)(0.08)^1(0.92)^{41} \)
\( C(42,1)=42 \), \( 42 \times 0.08 \times 0.92^{41} \approx 0.110 \)

Part b: At most 3 successes

\( P(X \leq 3) = P(0)+P(1)+P(2)+P(3) \)
\( P(0)=0.92^{42}\approx0.030 \), \( P(1)\approx0.110 \), \( P(2)\approx0.196 \), \( P(3)\approx0.227 \)
Sum: \( 0.030+0.110+0.196+0.227\approx0.563 \)

Part c: At least 1 success

\( P(X \geq 1) = 1 - P(X=0) \approx 1 - 0.030 = 0.970 \) (given)

Part d: 3 to 11 successes (inclusive)

\( P(3 \leq X \leq 11) = P(X \leq 11) - P(X \leq 2) \)
\( P(X \leq 2)\approx0.030+0.110+0.196=0.336 \), \( P(X \leq 11)\approx1 \)
Thus: \( 1 - 0.336\approx0.664 \)

Answer:

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