Question

Sixty 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.

Answer 1

Answer:

$a. \ P(X=5)=0.2787\\\\b.\ \ P(X>5)=0.3154\\\\c. \ P(X\leq 5)=0.6846$

Step-by-step explanation:

a. This a binomial probability distribution problem expressed as:

$P(X=x)={n\choose x}p^x(1-p)^{n-x}$

#Given the probability of success, p=0.60 and the number of trials, n=8, the probability of exactly 5 purchases is calculated as:

$P(X=x)={n\choose x}p^x(1-p)^{n-x}\\\\\\P(X=5)={8\choose 5}0.6^5(1-0.6)^3\\\\=0.2787$

Hence, the probability of exactly 5 purchases is 0.2787

b. Given n=8 , p=0.60, the probability of more than 5 purchases is calculated as:

$P(X=x)={n\choose x}p^x(1-p)^{n-x}\\\\\\P(X>5)=P(X=6)+P(X=7)+P(X=8)\\\\={8\choose 6}0.6^6(1-0.6)^2+{8\choose 7}0.6^7(1-0.6)^1+{8\choose 8}0.6^8(1-0.6)^0\\\\=0.2090+0.0896+0.0168\\\\=0.3154$

Hence, the probability of more than 5 purchases is 0.3154

c. The probability of at most 5 purchases is equivalent to 1-P(x>5) and is calculated as:

#From b above, P(X>5)=0.3154;

$P(X\leq 5)=1-P(X>5)\\\\=1-0.3154\\\\=0.6846$

Hence, the probability of at most 5 purchases is 0.6846