SOLUTION: 62% of all Americans are home owners. If 35 Americans are randomly selected, find the probability that
a. Exactly 22 of them are are home owners.
b. At most 24 of them are a
Algebra.Com
Question 1199759: 62% of all Americans are home owners. If 35 Americans are randomly selected, find the probability that
a. Exactly 22 of them are are home owners.
b. At most 24 of them are are home owners.
c. At least 22 of them are home owners.
d. Between 17 and 21 (including 17 and 21) of them are home owners.
Answer by textot(100) (Show Source): You can put this solution on YOUR website!
**a. Probability of exactly 22 homeowners:**
* This follows a binomial distribution with:
* n = 35 (sample size)
* p = 0.62 (probability of success - homeowner)
* q = 1 - p = 0.38 (probability of failure - not a homeowner)
* k = 22 (number of successes)
* Using the binomial probability formula:
P(X = k) = (nCk) * p^k * q^(n-k)
where nCk = n! / (k! * (n-k)!)
* P(X = 22) = (35C22) * (0.62)^22 * (0.38)^(35-22)
P(X = 22) ≈ 0.0898
**b. Probability of at most 24 homeowners:**
* P(X ≤ 24) = P(X = 0) + P(X = 1) + ... + P(X = 24)
* This requires calculating the binomial probability for each value of X from 0 to 24 and summing them.
* Using a calculator or statistical software (like Excel or a TI-84 calculator with the binomcdf function):
P(X ≤ 24) ≈ 0.9588
**c. Probability of at least 22 homeowners:**
* P(X ≥ 22) = P(X = 22) + P(X = 23) + ... + P(X = 35)
* This requires calculating the binomial probability for each value of X from 22 to 35 and summing them.
* Alternatively:
P(X ≥ 22) = 1 - P(X ≤ 21)
* Using a calculator or statistical software:
P(X ≥ 22) ≈ 0.2493
**d. Probability of between 17 and 21 homeowners (inclusive):**
* P(17 ≤ X ≤ 21) = P(X = 17) + P(X = 18) + ... + P(X = 21)
* Calculate each probability using the binomial probability formula and sum them up.
* Using a calculator or statistical software:
P(17 ≤ X ≤ 21) ≈ 0.6550
**In summary:**
* a) Probability of exactly 22 homeowners: 0.0898
* b) Probability of at most 24 homeowners: 0.9588
* c) Probability of at least 22 homeowners: 0.2493
* d) Probability of between 17 and 21 homeowners: 0.6550
RELATED QUESTIONS
62% of all Americans are home owners. If 35 Americans are randomly selected, find the... (answered by Solver92311)
66% of all Americans are home owners. If 50 Americans are randomly selected, find the... (answered by ikleyn)
69% of all Americans are home owners. If 37 Americans are randomly selected, find the... (answered by CPhill,ikleyn)
If 18% of all Americans are underweight, find the probability two randomly selected... (answered by ikleyn)
8% of all Americans live in poverty. If 30 Americans are randomly selected, find the... (answered by ewatrrr)
8% of all Americans live in poverty. If 37 Americans are randomly selected, find the... (answered by Solver92311)
16% of all Americans live in poverty. If 47 Americans are randomly selected, find the... (answered by Boreal)
8% of all Americans live in poverty. If 37 Americans are randomly selected, find the... (answered by CPhill)
The percentage of Americans who are vegetarians is 3.2%. Four people are randomly... (answered by stanbon)