Which of the following could be the set of all possible outcomes for a random variable that follows a binomial distribution? A) (0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11). B) (-1, 0, 1). C) (0, 0.5, 1, 1.5, 2, 2.5, 3). D) (1, 2).
D. But it is just a guess.
a binomial distribution has only two possible outcomes, like the flip of a coin, the outcome is either heads or tails. answers A-C have greater than 2 outcomes so they’re not binomial distributions.
D agree with slave probably one of the few quant questions i know how to answer
the answer is actually A… i think it should be D aswell, but i’ve done this exact question numerous times on qbank, and the ‘correct’ answer is always A… maybe Schweser is wrong (again) ?
I remember this from qbank as well. I think the explanation said something about the answer has to contain zero, and can only be positive whole #s. I may be wrong though.
yeh i just tried looking thru the schweser notes then, but still have no idea how to explain this one…
The answer makes sense if you think of the Binomial distribution as a coin toss. In 11 tosses, the random variable could take on any value between 0 to 11 with respect to the number of times heads shows up. Fractions, zeros and negative numbers don’t work.
chebychev Wrote: ------------------------------------------------------- […] Fractions, zeros and > negative numbers don’t work. good point. I guess you always need a zero though. That’s likely the reason D is not the right answer?
ahhh thanks for that chebychev… makes sense now…
You’re right Lola – I mean to say “fractions, absence of zero, and negatives don’t work”
This “absence of zero” thing fills me with existential angst. Might be some Zen thing.
Just a quesiotn here. So if i am taking SAT in which there is a negative marking, say -0.25 for a wrong ans and 0.25 for rt. This wont be a binomial distribution considering there are only two outcomes?
Reread the thread. chebyshev: The answer makes sense if you think of the Binomial distribution as a coin toss. In 11 tosses, the random variable could take on any value between 0 to 11 with respect to the number of times heads shows up. Fractions, [an absence of/lack of/intersection of {0} and set of outcomes = {} etc] zeros and negative numbers don’t work.
SO what will we call a distribution that has -ve nos
As long as there are two and only two outcomes in each trial, I don’t think it makes a difference if the outcomes are labeled with fractions. For example, if we tossed a coin with .25 written on one side and -.25 on the other, the number of times .25 shows up in a set of tosses is still binomial. Similarly, the number of wrong versus right answers in your SAT would be a binomial distribution. The distribution would be quantified in the number wrong, however, rather than the points lost. The expected amount of lost points could then be calculated by applying the .25 to the expected number of wrong answers.
Could be more or less anything - this was defined by what it couldn’t be. The only possible answer to the question was A, as it contained 0, didn’t contain a negative number, and only contained integers. Just remember that a binomially distributed variable is a count of successful trials, and you’ll be fine - that’s the key to this question.
smeet Wrote: ------------------------------------------------------- > Just a quesiotn here. So if i am taking SAT in > which there is a negative marking, say -0.25 for a > wrong ans and 0.25 for rt. This wont be a binomial > distribution considering there are only two > outcomes? X = The number right has a binomial distribution (not really because the SAT has different p’s for each question, but we’ll pretend). The score would be some linear function of taht.
Yes I agree with that. I was just talking wrt every question . I either get 0.25 or lose 0.25 points.