75% of children have a systolic blood pressure lower than 136 mm of mercury. What is the probability that a sample of 12 children will include:
A) exactly 4 who have a blood pressure greater than 136?
B) no more than 4 who have a blood pressure greater than 136?
C) at least 4 who have a blood pressure greater than 136?
I'm working on this question for a long time now, ant it really bothers me because I have no idea how to get to the right answer. I need some guidance, please.
What I tried to do for part A was:
Since it's exactly 4 children: $Pr(Y=4) = {12 \choose 4}(.75)^4(0.25)^8$
However, it gives me incredibly large number that doesn’t make sense. For part B and C, I don’t even know how to start… How to calculate something when wordings “no more” and “at least” are included?
Thank you for your help.
$\endgroup$ 53 Answers
$\begingroup$Your calculation for exactly four is almost correct if by (12/4) you mean ${12 \choose 4}$, the number ways to select four things out of twelve. The exponents are switched, because the problem switched between lower and higher. I don't get an incredibly large number, Alpha gives $\frac {3247695}{16777216}\approx 0.1936$ For no more than four, you would do the same calculation replacing $4$ by $0,1,2,3$ and add them up.
$\endgroup$ 4 $\begingroup$A) Your answer is (almost) correct, and it shows that you know what you are doing. The only problem here that since they ask you about 4 who has pressure greater than 136, and 0.75 is the probability of having pressure below 136, you should swap the probabilities (I assume no one has pressure exactly 136, right?).
B) "no more" means exactly what it means: what's the probability that there are 0,1,2,3 or 4 who have a blood pressure. Calculate similar expressions for these numbers and sum them up (as these events are independent).
C) Similar with "at least": 4,5,....
$\endgroup$ $\begingroup$Your solution for part a is correct, except replace $.75$ by $.25$
For b) we need $P(Y \leq 4)$. Since $Y$ follows the binomial distribution with parameter $n=12$ and $p=0.25$ we have that
$P(Y \leq 4) = \sum\limits_{k=0}^{4}{12 \choose k}(0.25)^{k}(0.75)^{12-k}.$
For c) we need $P(Y \geq 4)= 1- P(Y < 4)$. Proceed as in b)
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