If 41 Americans are randomly selected, find the probability that

Question

65% of all Americans live in cities with population greater than 100,000 people. If 41 Americans are randomly selected, find the probability that

a. Exactly 24 of them live in cities with population greater than 100,000 people. 

b. At most 27 of them live in cities with population greater than 100,000 people. 

c. At least 26 of them live in cities with population greater than 100,000 people. 

d. Between 24 and 28 (including 24 and 28) of them live in cities with population greater than 100,000 people. 

Answer 1

Hello Anni,

This is a binomial experiment with n =41 independent trials. Let success = lives in a city with a population greater than 100,000 people. The probability of success, p = 0.65. So that the probability of exactly x = 24 successes is given by the following binomial distribution function:

P(x) = _nC_x (p)^x (1 - p)^n-x

a) P(exactly 24 successes) = P(24)

= ₄₁C₂₄ (0.65)²⁴(1-0.65)^41-24 = [41! ÷ (24! * (41-24)!)] (0.65)²⁴(0.35)¹⁷≈ 0.09

Which means that if we repeat the experiment (randomly selecting 41 Americans) 100 times, about 9 of the experiments will result in exactly 24 Americans living in a city with a population greater than 100,000 people.

Note: if you are using a TI graphing calculator you may use the binompdf function with the following arguments: binompdf(n,p,x) = binompdf(41,0.65,24) ≈0.09

The other parts b - d, please use binomcdf, that is, the binomial cumulative distribution function as follows:

b) binomcdf(41,0.65,27) ≈ 0.60

c) binomcdf(41,0.65,41)-binomcdf(41,0.65,26) ≈0.53

d) binomcdf(41,0.65,28) - binomcdf(41,0.65,24) ≈ 0.49

Hope this helps!