Probability distribution help

[doc_doc]

hello,
please i'm, taking statistic this term and i really have a problem with this question:
which is
for the distribution B(3,0.5),
a) how many outcomes are there to each trial?
b) how many trials are there?
c) how many possible values can the variable take?
d) what are the mean and standard daviations of this distribution?
please help me. if you can't just tell me who can help me ?
please
i'm beggining you
thanks alot
jody

 

[theFuture]

I assume you are talking about a binomial distribuition for n=3 and p=.5? Think about what a binomial distribution is. It counts the number of sucesses in n independent bernoulli trials that have sucess with probability p, right?

a) Well, what are the outcomes of a bernoulli trial?
b) Well, what is n?
c) Think about the sample space, how many possible successes can you get in n trials?
d)Well, the mean is the expectation and the SD is the root of the variance.

That should get you started.

 

[tacman]

Hi Jody,

I am happy to help you with your question about probability distributions. The distribution B(3,0.5) is a binomial distribution, where the number of trials is 3 and the probability of success for each trial is 0.5.

a) There are two possible outcomes for each trial, either success or failure.

b) As mentioned above, there are 3 trials in this distribution.

c) The variable can take on 4 possible values: 0, 1, 2, or 3.

d) The mean of this distribution is given by the formula n*p, where n is the number of trials and p is the probability of success. In this case, the mean is 3*0.5 = 1.5. The standard deviation can be calculated using the formula sqrt(n*p*(1-p)), which in this case is sqrt(3*0.5*(1-0.5)) = sqrt(0.75) = 0.866.

If you are still having trouble with understanding probability distributions, I recommend seeking help from your professor, teaching assistant, or a tutor. They will be able to provide you with more personalized and in-depth assistance.

Best of luck in your studies!