Ask your own question, for FREE!
Mathematics 7 Online
OpenStudy (anonymous):

A study determined that 15% of a population use a certain brand of laundry detergent. What is the probability that more than 2 shoppers selected at random from 12 shoppers use that brand of detergent?

OpenStudy (kirbykirby):

You can recognize this being a binomial problem. If you let \(X\) be the number of shoppers selected that use the brand of detergent, then \(X\) follows a binomial distribution. The question asks for \(P(X>2)\), which is the same as \(P(X \ge 3)\) since the random variable is discrete. This is a bit long to calculate, so it's easier to calculate this with the aid of the complement, and find \(1-P(X \le 2)\). \[P(X \le 2)= P(X=0)+P(X=1)+P(X=2)\] and you can find \(P(X = x)\) as \[\large {12 \choose x} 0.15^x(1-0.15)^{12-x}\], replacing \(x\) with 0, 1 and 2

OpenStudy (anonymous):

ive tried plugging it into my calculator as a binomial but i keep getting the wrong answer

OpenStudy (kirbykirby):

\[ P(X \le 2)= {12 \choose 0} 0.15^0(0.85)^{12}+ {12 \choose 1} 0.15^1(0.85)^{11}+{12 \choose 2} 0.15^2(0.85)^{10} \\=0.7358\]

OpenStudy (kirbykirby):

?

OpenStudy (anonymous):

i keep getting .2924 with my calculator but i know that isnt right, i think yours is!

OpenStudy (kirbykirby):

And then since we needed to use the complement, 1 - 0.7358 = 0.2642

OpenStudy (anonymous):

so .2642 is the final?

OpenStudy (kirbykirby):

Yes, should be. Do they give you an answer?

OpenStudy (anonymous):

no I wish but I'm sure you're right!

OpenStudy (kirbykirby):

I'm fairly certain about my answer, but to be honest I am quite tired at this time so maybe I made a mistake and don't realize it o.o!

OpenStudy (anonymous):

better than what i have and i know mine isn;t right! haha thank you Kirby!

OpenStudy (kirbykirby):

your welcome! :)

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!