Question

fifty percent of consumers prefer to purchase electronics online. you randomly select 8 consumers. find the probability that the number of consumers who prefer to purchase electronics online is (a) exactly five, (b) more than five, and (c) at most five. (a) find the probability that the number that prefer to purchase electronics online is exactly five. p(5)=0.219 (round to three decimal places as needed.) (b) find the probability that the number that prefer to purchase electronics online is more than five. p(x>5)= (round to three decimal places as needed.)

Explanation:

Step1: Identify binomial probability formula

The binomial probability formula is $P(X = k)=C(n,k)\times p^{k}\times(1 - p)^{n - k}$, where $n$ is the number of trials, $k$ is the number of successes, $p$ is the probability of success on a single - trial, and $C(n,k)=\frac{n!}{k!(n - k)!}$. Here, $n = 8$, $p=0.5$, and $1 - p = 0.5$.

Step2: Calculate $P(X>5)$

$P(X>5)=P(X = 6)+P(X = 7)+P(X = 8)$.
For $k = 6$:
$C(8,6)=\frac{8!}{6!(8 - 6)!}=\frac{8!}{6!2!}=\frac{8\times7}{2\times1}=28$
$P(X = 6)=C(8,6)\times(0.5)^{6}\times(0.5)^{2}=28\times(0.5)^{8}=28\times\frac{1}{256}\approx0.109$
For $k = 7$:
$C(8,7)=\frac{8!}{7!(8 - 7)!}=8$
$P(X = 7)=C(8,7)\times(0.5)^{7}\times(0.5)^{1}=8\times(0.5)^{8}=8\times\frac{1}{256}\approx0.031$
For $k = 8$:
$C(8,8)=\frac{8!}{8!(8 - 8)!}=1$
$P(X = 8)=C(8,8)\times(0.5)^{8}\times(0.5)^{0}=(0.5)^{8}=\frac{1}{256}\approx0.004$
$P(X>5)=P(X = 6)+P(X = 7)+P(X = 8)=0.109 + 0.031+0.004=0.144$

Step3: Calculate $P(X\leq5)$

We know that $P(X\leq5)=1 - P(X>5)$. Since $P(X>5)=0.144$, then $P(X\leq5)=1 - 0.144 = 0.856$

Answer:

(a) $0.219$
(b) $0.144$
(c) $0.856$