What is the probability that exactly 8 of them are over the age of 65

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The probability of exactly 8 out of 15 coronary bypass patients being over the age of 65 can be calculated using the binomial distribution formula. Given that the probability (p) of a patient being over 65 is 0.53, the binomial coefficient is represented as \(\begin{pmatrix}n \\ i\end{pmatrix}\), where n=15 and i=8. The calculations for probabilities of fewer than 10, more than 10, and 11 or fewer patients over 65 are also derived from this distribution, utilizing complementary counting for efficiency.

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2) The Agency for Healthcare Research and Quality reported that 53% of people who had coronary bypass surgery in 2008 were over the age of 65. Fifteen coronary bypass patients are sampled.

a) What is the probability that exactly 8 of them are over the age of 65?

b) P (less than 10 are over 65) =

c) P (more than 10 are over 65) =

d) P (11 or fewer are over 65) =

e) P ( more than 11 are over 65) =
 
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I am puzzled by this thread. If you have never taken a course in "probability and statistics" where did you get these exercises? If you have, or are now taking such a course, why have you shown no attempt to answer these yourself?

They are all applications of the basic "binomial distribution": if the probability a particular event will result in "a" is p and the probability it will result is "b" is 1- p, the probability that, in n events, it will result in "a" i times and "b" n-i times with probability [math]\begin{pmatrix}n \\ I \end{pmatrix}p^i(1- p)^{n-I}[/math].

Here, "a" is "a person who had coronary bypass surgery is over 65", p= 0.53, "b" is "a person who had coronary bypass surgery is NOT over 65", and 1- p= 1- 0.53= 0.47. n= 15. [math\begin{pmatrix} n \\ i\/end{pmatrix}[/math] is the "binomial coefficient", [math]\frac{n!}{i!(n-i)!}[/math].

In problem (a) i= 8.
in problem (b) it is simplest to calculate that for i= 10, 11, 12, 13, 14, and 15, add them (to determine the probability "10 or more are over 65") and subtract from 1. The harder way is to calculate that for i= 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9 and add.
In problem (c) you can use the number you got before you subtracted from 1!
In problem (d) "11 or fewer" is the same as "less than 12" so you can do the same as (c), calculate the probability for i= 12, 13, 14, 15 and subtract from 1.
 
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