Question

in this problem, round to four decimals when possible. 79% of all americans live in cities with population greater than 100,000 people. if 39 americans are randomly selected, find the probability that a. exactly 31 of them live in cities with population greater than 100,000 people. b. at most 31 of them live in cities with population greater than 100,000 people. c. at least 30 of them live in cities with population greater than 100,000 people. d. between 29 and 37 (including 29 and 37) of them live in cities with population greater than 100,000 people. hint: hint video on finding binomial probabilities +

Explanation:

Step1: Identify binomial parameters

Let \(n = 39\) (number of trials/sample size), \(p=0.79\) (probability of success). The binomial probability formula is \(P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}\), where \(C(n,k)=\frac{n!}{k!(n - k)!}\).

Step2: Calculate part a

For \(k = 31\), \(C(39,31)=\frac{39!}{31!(39 - 31)!}=\frac{39!}{31!8!}=\frac{39\times38\times37\times36\times35\times34\times33\times32}{8\times7\times6\times5\times4\times3\times2\times1}=63574539\)
\(P(X = 31)=C(39,31)\times(0.79)^{31}\times(1 - 0.79)^{39 - 31}\)
\(=63574539\times(0.79)^{31}\times(0.21)^{8}\)
\(\approx63574539\times0.001747\times0.000038\)
\(\approx0.4237\)

Step3: Calculate part b

\(P(X\leq31)=\sum_{k = 0}^{31}C(39,k)\times(0.79)^{k}\times(0.21)^{39 - k}\). Using a binomial - probability calculator or software (e.g., TI - 84 Plus: binomcdf(39,0.79,31)), we get \(P(X\leq31)\approx0.6797\)

Step4: Calculate part c

\(P(X\geq30)=1 - P(X\lt30)=1-\sum_{k = 0}^{29}C(39,k)\times(0.79)^{k}\times(0.21)^{39 - k}\)
Using a binomial - probability calculator or software (e.g., TI - 84 Plus: 1 - binomcdf(39,0.79,29)), we get \(P(X\geq30)\approx0.7152\)

Step5: Calculate part d

\(P(29\leq X\leq37)=\sum_{k = 29}^{37}C(39,k)\times(0.79)^{k}\times(0.21)^{39 - k}\)
Using a binomial - probability calculator or software (e.g., TI - 84 Plus: binomcdf(39,0.79,37)-binomcdf(39,0.79,28)), we get \(P(29\leq X\leq37)\approx0.9477\)

Answer:

a. \(0.4237\)
b. \(0.6797\)
c. \(0.7152\)
d. \(0.9477\)