Question

18% of all Americans live in poverty. If 43 Americans are randomly selected, find the probability that a. Exactly 8 of them live in poverty. b. At most 9 of them live in poverty. c. At least 8 of them live in poverty. d. Between 4 and 10 (including 4 and 10) of them live in poverty.

Answer 1

Total number of people tht are randomly selected (n) = 43

Now, 18% of the total population of America live in poverty

Hence, p = probability of people living in poverty = 0.18

q = 1 - p = 1 - 0.18 = 0.82

Let X = numbers of people living in poverty

(a) Exactly 8 people live under poverty

(b) At most 9 people live under poverty

(c) At least 8 people live in poverty

(d) Between 4 and 10 people live in poverty (including 4 and 10)

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