Question

in this problem, round to four decimals when possible.
77% 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:
videos on finding binomial probabilities (1)

Explanation:

Step1: Identify binomial probability formula

The binomial probability formula is $P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}$, where $n$ is the number of trials, $k$ is the number of successes, $p$ is the probability of success on a single - trial, and $C(n,k)=\frac{n!}{k!(n - k)!}$. Assume $n$ (number of selected Americans) is known, and $p = 0.77$ (probability that an American lives in a city with population greater than 100000).

Step2: Calculate part a

For exactly $k = 14$ successes, first calculate the combination $C(n,14)=\frac{n!}{14!(n - 14)!}$, then $P(X = 14)=C(n,14)\times(0.77)^{14}\times(0.23)^{n - 14}$.

Step3: Calculate part b

For at most $k = 11$ successes, $P(X\leq11)=\sum_{k = 0}^{11}C(n,k)\times(0.77)^{k}\times(0.23)^{n - k}$.

Step4: Calculate part c

For at least $k = 20$ successes, $P(X\geq20)=\sum_{k = 20}^{n}C(n,k)\times(0.77)^{k}\times(0.23)^{n - k}=1 - P(X\lt20)=1-\sum_{k = 0}^{19}C(n,k)\times(0.77)^{k}\times(0.23)^{n - k}$.

Step5: Calculate part d

For between $k_1 = 29$ and $k_2 = 37$ successes, $P(29\leq X\leq37)=\sum_{k = 29}^{37}C(n,k)\times(0.77)^{k}\times(0.23)^{n - k}$.

Since the value of $n$ (the number of selected Americans) is not given in the problem - statement, we cannot give numerical answers. But the general procedures are as above.

Answer:

a. $P(X = 14)=C(n,14)\times(0.77)^{14}\times(0.23)^{n - 14}$
b. $P(X\leq11)=\sum_{k = 0}^{11}C(n,k)\times(0.77)^{k}\times(0.23)^{n - k}$
c. $P(X\geq20)=1-\sum_{k = 0}^{19}C(n,k)\times(0.77)^{k}\times(0.23)^{n - k}$
d. $P(29\leq X\leq37)=\sum_{k = 29}^{37}C(n,k)\times(0.77)^{k}\times(0.23)^{n - k}$