SOLUTION: 73% of all Americans live in cities with population greater than 100,000 people. If 35 Americans are randomly selected, find the probability that a. Exactly 26 of them live in c

Question 1168409: 73% of all Americans live in cities with population greater than 100,000 people. If 35 Americans are randomly selected, find the probability that
a. Exactly 26 of them live in cities with population greater than 100,000 people.
b. At most 26 of them live in cities with population greater than 100,000 people.
c. At least 25 of them live in cities with population greater than 100,000 people.
d. Between 21 and 25 (including 21 and 25) of them live in cities with population greater than 100,000 people.
Found 2 solutions by CPhill, ikleyn:
Answer by CPhill(1959)   (Show Source): You can put this solution on YOUR website!
Let's solve this problem using the binomial probability distribution.
**Given:**
* Probability of an American living in a city with population > 100,000 (p) = 73% = 0.73
* Number of Americans selected (n) = 35
**a) Exactly 26 of them live in cities with population greater than 100,000 people.**
We want to find P(X = 26), where X is the number of Americans living in cities with population > 100,000.
The binomial probability formula is:
P(X = k) = (nCk) * p^k * (1-p)^(n-k)
Where:
* nCk = n! / (k! * (n-k)!)
* n = 35
* k = 26
* p = 0.73
P(X = 26) = (35C26) * (0.73)^26 * (0.27)^9
* 35C26 = 35! / (26! * 9!) = 54,835,425
* (0.73)^26 ≈ 0.000304
* (0.27)^9 ≈ 0.000001966
P(X = 26) = 54,835,425 * 0.000304 * 0.000001966
P(X = 26) ≈ 0.0327
**b) At most 26 of them live in cities with population greater than 100,000 people.**
We want to find P(X ≤ 26), which is the sum of P(X = k) for k = 0 to 26.
P(X ≤ 26) = Σ [ (35Ck) * (0.73)^k * (0.27)^(35-k) ] for k = 0 to 26
This is tedious to calculate manually. We can use a calculator or software.
P(X ≤ 26) ≈ 0.449
**c) At least 25 of them live in cities with population greater than 100,000 people.**
We want to find P(X ≥ 25), which is the sum of P(X = k) for k = 25 to 35.
P(X ≥ 25) = Σ [ (35Ck) * (0.73)^k * (0.27)^(35-k) ] for k = 25 to 35
Using a calculator or software:
P(X ≥ 25) ≈ 0.141
**d) Between 21 and 25 (including 21 and 25) of them live in cities with population greater than 100,000 people.**
We want to find P(21 ≤ X ≤ 25), which is the sum of P(X = k) for k = 21 to 25.
P(21 ≤ X ≤ 25) = Σ [ (35Ck) * (0.73)^k * (0.27)^(35-k) ] for k = 21 to 25
Using a calculator or software:
P(21 ≤ X ≤ 25) ≈ 0.297
**Answers:**
a) Approximately 0.0327
b) Approximately 0.449
c) Approximately 0.141
d) Approximately 0.297

Answer by ikleyn(52786)   (Show Source): You can put this solution on YOUR website!
.
73% of all Americans live in cities with population greater than 100,000 people.
If 35 Americans are randomly selected, find the probability that
a. Exactly 26 of them live in cities with population greater than 100,000 people.
b. At most 26 of them live in cities with population greater than 100,000 people.
c. At least 25 of them live in cities with population greater than 100,000 people.
d. Between 21 and 25 (including 21 and 25) of them live in cities with population greater than 100,000 people.
~~~~~~~~~~~~~~~~~~~~~~~~~~

        As it regularly happens with @CPhill' solutions on Binomial distribution,
        all calculations (a), (b), (c) and (d) in the post by @CPhill are incorrect.

        I came to bring correct answers.

(a)   P(X = 26) = 0.1505

(b)   P(X <= 26) = 0.6311

(c)   P(X >= 25) = 0.6641

(d)   P(21 <= X <= 25) = P(X <= 25) - P(X <= 20) = 0.4806 - 0.0311 = 0.4495.

To check your/my calculations, use online calculator

https://stattrek.com/online-calculator/binomial

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