SOLUTION: 8% of all Americans live in poverty. If 30 Americans are randomly selected, find the probability that a. Exactly 2 of them live in poverty. b. At most 2 of them live in po

Question 1173032: 8% of all Americans live in poverty. If 30 Americans are randomly selected, find the probability that
a. Exactly 2 of them live in poverty.
b. At most 2 of them live in poverty.
c. At least 1 of them live in poverty.
d. Between 1 and 5 (including 1 and 5) of them live in poverty.

Answer by ewatrrr(24785) (Show Source): You can put this solution on YOUR website!

 

Hi, 

p = .08  n = 30 

P (x = 2) = binompdf(30,.08,2) = .2696

P( x < 2) = binomcdf(30,.08,1) = .2958

P ( x ≤ 1 = binomcdf(30,.08,1)= .2958

P (1 ≤ x ≤ 5) = P(x ≤ 5) - P(x< 1) = .9707 - .082 = .8887 

Recommend until You are comfortable using YOUR statistical Calculator,

to check YOUR findings with the stattrek calculator on line.

Wish You the Best in your Studies. 

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