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:

This is a binomial probability problem where the binomial probability formula is $P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}$, $P(X\leq k)=\sum_{i = 0}^{k}C(n,i)\times p^{i}\times(1 - p)^{n - i}$, $P(X\geq k)=1 - P(X

Step1: Calculate binomial coefficient for part a

For $k = 31$, $C(39,31)=\frac{39!}{31!(39 - 31)!}=\frac{39!}{31!8!}=\frac{39\times38\times37\times36\times35\times34\times33\times32}{8\times7\times6\times5\times4\times3\times2\times1}=63574539$.
$P(X = 31)=C(39,31)\times(0.79)^{31}\times(0.21)^{8}$
$P(X = 31)=63574539\times(0.79)^{31}\times(0.21)^{8}\approx0.1499$.

Step2: Calculate cumulative probability for part b

We need to find $P(X\leq31)=\sum_{i = 0}^{31}C(39,i)\times(0.79)^{i}\times(0.21)^{39 - i}$. Using a binomial probability calculator or software (e.g., in R: pbinom(31,39,0.79)), $P(X\leq31)\approx0.5886$.

Step3: Calculate probability for part c

$P(X\geq30)=1 - P(X<30)=1-\sum_{i = 0}^{29}C(39,i)\times(0.79)^{i}\times(0.21)^{39 - i}$. Using a binomial probability calculator or software (e.g., in R: 1 - pbinom(29,39,0.79)), $P(X\geq30)\approx0.7873$.

Step4: Calculate probability for part d

$P(29\leq X\leq37)=\sum_{i = 29}^{37}C(39,i)\times(0.79)^{i}\times(0.21)^{39 - i}$. Using a binomial probability calculator or software (e.g., in R: pbinom(37,39,0.79)-pbinom(28,39,0.79)), $P(29\leq X\leq37)\approx0.9585$.

Answer:

a. $0.1499$
b. $0.5886$
c. $0.7873$
d. $0.9585$