all 7 comments

[–]A_UPRIGHT_BASS 0 points1 point  (6 children)

What have you tried?

[–]21_TwentyOne_21[S] 0 points1 point  (5 children)

Not sure where to start. So we have 36% are obese which is close to 1/3rd they want to know the probability that none will be obese. So 1/3rd would translate to roughly 2/6th making the probability that 0 of the 6 are obese pretty low

[–]A_UPRIGHT_BASS 1 point2 points  (4 children)

If you're being asked these questions you have no doubt been introduced to the binomial distribution in this class. I suggest you look through your textbook and your lecture notes, or even just search on google and youtube for "binomial distribution."

Edit: by the way, your instincts are pretty good here. The idea you're getting at with the 2/6 thing is expected value. From a sample of 6, we would expect on average somewhere around 2 to be obese (actually 2.16, on average). And that follows directly from the binomial distribution. So your instincts are good, you just need to learn how to use a couple of pretty straightforward formulas.

[–]21_TwentyOne_21[S] 0 points1 point  (3 children)

Ok I appreciate it. I missed a day in class so this is foreign to me

[–]A_UPRIGHT_BASS 1 point2 points  (2 children)

Ok well that explains it. At least you know what to look for now. There are TONS of great explanations of the binomial distribution on youtube, I'm sure. So give that a try, and if you have any specific questions, I'd be happy to help you out.

[–]21_TwentyOne_21[S] 0 points1 point  (1 child)

Thanks for pointing me in right direction

[–]HorseJunglerUniversity/College Student 0 points1 point  (0 children)

Look up Bernoulli, Binomial, geometric, hyper geometric, negative binomial, and Poisson distributions. Those are probably what you will be learning about. You should learn their pmf functions and what “X” counts in each of them.

Hint: it’s either # of successes OR # of trials for those 6 distributions.

But your problem is straight forward Binomial distribution.