Question

section 6.2 homework
score: 4/14 answered: 4/14
progress saved done
question 5
0/1 pt 3 99 details
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 parameters

Let $n = 44$ (number of trials/sample size), $p=0.7$ (probability of success - being a homeowner), and $q = 1 - p=0.3$. 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 probability for part a

For $k = 30$, $C(44,30)=\frac{44!}{30!(44 - 30)!}=\frac{44!}{30!14!}=\frac{44\times43\times\cdots\times31}{14\times13\times\cdots\times1}\approx1.4897\times 10^{12}$
$P(X = 30)=C(44,30)\times(0.7)^{30}\times(0.3)^{14}$
$P(X = 30)\approx1.4897\times 10^{12}\times2.2518\times10^{- 11}\times2.8243\times10^{-8}\approx0.0958$

Step3: Calculate probability for 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., Excel's BINOM.DIST function), $P(X\leq31)\approx0.6288$

Step4: Calculate probability for part c

$P(X\geq33)=1 - P(X\leq32)$. Using a binomial - probability calculator, $P(X\leq32)\approx0.7939$, so $P(X\geq33)=1 - 0.7939 = 0.2061$

Step5: Calculate probability for 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, $P(27\leq X\leq34)\approx0.7285$

Answer:

a. $0.0958$
b. $0.6288$
c. $0.2061$
d. $0.7285$