Question 1168409
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