Question

66% of all americans are home owners. if 41 americans are randomly selected, find the following probabilities. round answers to 4 decimal places.
a. probability that exactly 26 of them are home owners.
b. probability that at most 29 of them are home owners.
c. probability that at least 26 of them are home owners.
d. probability that between 25 and 29 (including 25 and 29) of them are home owners.
hint:
hint
video on finding binomial probabilities +

Explanation:

Step1: Identify binomial distribution parameters

Let $n = 41$ (number of trials/sample size), $p=0.66$ (probability of success - being a homeowner), $q = 1 - p=0.34$. The binomial probability formula is $P(X = k)=C(n,k)\times p^{k}\times q^{n - k}$, where $C(n,k)=\frac{n!}{k!(n - k)!}$.

Step2: Calculate part a

For $k = 26$:
$C(41,26)=\frac{41!}{26!(41 - 26)!}=\frac{41!}{26!15!}$
$P(X = 26)=C(41,26)\times(0.66)^{26}\times(0.34)^{15}$
$C(41,26)=\frac{41!}{26!15!}=\frac{41\times40\times\cdots\times16}{15!}\approx1.14\times10^{11}$
$P(X = 26)\approx1.14\times10^{11}\times(0.66)^{26}\times(0.34)^{15}\approx0.1370$

Step3: Calculate part b

$P(X\leq29)=\sum_{k = 0}^{29}C(41,k)\times(0.66)^{k}\times(0.34)^{41 - k}$. Using a binomial - probability calculator or software (e.g., Excel's BINOM.DIST function: =BINOM.DIST(29,41,0.66,TRUE)), we get $P(X\leq29)\approx0.7984$.

Step4: Calculate part c

$P(X\geq26)=1 - P(X\leq25)=\ 1-\sum_{k = 0}^{25}C(41,k)\times(0.66)^{k}\times(0.34)^{41 - k}$. Using a binomial - probability calculator or software (e.g., =1 - BINOM.DIST(25,41,0.66,TRUE)), we get $P(X\geq26)\approx0.7540$.

Step5: Calculate part d

$P(25\leq X\leq29)=\sum_{k = 25}^{29}C(41,k)\times(0.66)^{k}\times(0.34)^{41 - k}=BINOM.DIST(29,41,0.66,TRUE)-BINOM.DIST(24,41,0.66,TRUE)\approx0.5564$

Answer:

a. $0.1370$
b. $0.7984$
c. $0.7540$
d. $0.5564$