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 parameters

Let $n = 41$ (number of trials, i.e., number of Americans selected), $p=0.66$ (probability of success, i.e., probability of an American being a homeowner), and $q = 1 - p=1 - 0.66 = 0.34$. 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 probability for part a

For $k = 26$, $C(41,26)=\frac{41!}{26!(41 - 26)!}=\frac{41!}{26!15!}$.
\[

$$\begin{align*} P(X = 26)&=C(41,26)\times(0.66)^{26}\times(0.34)^{15}\\ &=\frac{41!}{26!15!}\times(0.66)^{26}\times(0.34)^{15}\approx0.1379 \end{align*}$$

\]

Step3: Calculate probability for 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 or a TI - 84 Plus calculator's binomialcdf function), $P(X\leq29)\approx0.7984$.

Step4: Calculate probability for 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, $P(X\geq26)\approx0.7540$.

Step5: Calculate probability for part d

$P(25\leq X\leq29)=\sum_{k = 25}^{29}C(41,k)\times(0.66)^{k}\times(0.34)^{41 - k}=P(X\leq29)-P(X\leq24)$. Using a binomial - probability calculator, $P(25\leq X\leq29)\approx0.5564$.

Answer:

a. $0.1379$
b. $0.7984$
c. $0.7540$
d. $0.5564$