14% of all Americans live in poverty. If 49 Americans are randomly selected, find the probability that

Question

14% of all Americans live in poverty. If 49 Americans are randomly selected, find the probability that

a. Exactly 9 of them live in poverty. ?

b. At most 9 of them live in poverty. ?

c. At least 9 of them live in poverty. ?

d. Between 4 and 11 (including 4 and 11) of them live in poverty. ?

Answer 1

Hi Jenessa,

This is the same procedure as the questions you posted about bird life and felons.

General Formula: P(X=x) = C(n,x)p^x*q^n-x

a. P(X=9) = C(49,9)*0.14⁹*0.86⁴⁰ = 0.102

b. Compute all probabilities from X=0 to X=9, which is already done:

P(X=0) = C(49,0)0.14⁰*0.86⁴⁹

P(X=1) = C(49,1)0.14¹*0.86⁴⁸

^....

P(X=9) = C(49,9)*0.14⁹*0.86⁴⁰ = 0.102

Add together to get probability of at least 9 living in poverty.

c. Subtract 0.102 from your answer in part b. This is P(X=9). Then, subtract the result from 1.

P(X>=9) = 1 - P(X<9)

d. Same procedure, essentially, as in b but for x=4 to x=11:

P(X=4) = C(49,4)0.14⁴*0.86⁴⁵

P(X=5) = C(49,5)0.14⁵*0.86⁴⁴

P(X=6) = C(49,5)0.14⁵*0.86⁴⁴

^...

Continue until you reach X=11, then add the probabilities together. I hope this helps.