Question

70% of all americans are home owners. if 44 americans are randomly selected, find the following probabilities. round answers to 4 decimal places.
a. probability that exactly 30 of them are home owners.
b. probability that at most 31 of them are home owners.
c. probability that at least 33 of them are home owners.
d. probability that between 27 and 34 (including 27 and 34) of them are home owners.
hint:
hint
video on finding binomial probabilities +

Explanation:

Step1: Identify binomial distribution parameters

Let $n = 44$ (number of trials/sample size), $p=0.7$ (probability of success - being a home - owner), $q = 1 - p=0.3$. 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 $k = 30$:
$C(44,30)=\frac{44!}{30!(44 - 30)!}=\frac{44!}{30!14!}=\frac{44\times43\times\cdots\times31}{14\times13\times\cdots\times1}\approx1.077\times10^{12}$
$P(X = 30)=C(44,30)\times(0.7)^{30}\times(0.3)^{14}$
$P(X = 30)\approx1.077\times10^{12}\times2.25\times10^{- 7}\times2.82\times10^{-8}\approx0.0697$

Step3: Calculate part b

$P(X\leq31)=\sum_{k = 0}^{31}C(44,k)\times(0.7)^{k}\times(0.3)^{44 - k}$. Using a binomial - probability calculator or software (e.g., in R: pbinom(31,44,0.7)), $P(X\leq31)\approx0.4779$

Step4: Calculate part c

$P(X\geq33)=1 - P(X\leq32)$. First, find $P(X\leq32)$ using a binomial - probability calculator (e.g., in R: pbinom(32,44,0.7)). Then $P(X\geq33)=1 - P(X\leq32)\approx1-0.7939 = 0.2061$

Step5: Calculate part d

$P(27\leq X\leq34)=\sum_{k = 27}^{34}C(44,k)\times(0.7)^{k}\times(0.3)^{44 - k}$. Using a binomial - probability calculator (e.g., in R: pbinom(34,44,0.7)-pbinom(26,44,0.7)), $P(27\leq X\leq34)\approx0.7850$

Answer:

a. $0.0697$
b. $0.4779$
c. $0.2061$
d. $0.7850$