I wish we had LaTex. It would be handy for this problem.
10
#1. sum(C(10,k)p^kq^(10-k)), p=0.92 and q=0.08
k=8
#2. C(10,2)p^2q^8
a pharmaceutical firm claims that a certain medication is 92% effective in relieving arthric pain. if ten people who suffer from arthritis are randomly selected and given the medicine, and assuming that the claim is correct, find the probability that
A) At least eight of them will be relieved of their arthritic pain
B) Exactly two of them will not be relieved of their arthritic pain
I wish we had LaTex. It would be handy for this problem.
10
#1. sum(C(10,k)p^kq^(10-k)), p=0.92 and q=0.08
k=8
#2. C(10,2)p^2q^8
thank you for the help. I will try my best.
A)
sum(C(10,8) 0.92 ^8 X 0.08^(10-8)), p=0.92 and q=0.08 k=8
7.36 X 0.64(2)=9.42
b) C(10,2)0.92^2X0.92 ^8=
I THINK IM REALLY REALLY LOST.
(a)
use formula for r = 8, 9, 10 and sum them:
C(10,r) * 0.92^r * 0.08^(10-r)
(b)
C(10,2) * 0.08^2 * 0.92^8
Thank you. I will try it
Hi, when i see something like this (10,2) does that mean i should multiply. i dont know what the comma means.
For question A i got 0.410
Can someone pls tell me if im correct with this answer. I worked it out again and for
A) I GOT
80*7.36*6.4=3768.32
PLS JUST TELL ME IF IM CORRECT OR NOT
We cover these types of problems, using the binomial distribution, in the Examples forum:
http://www.talkstats.com/showthread.php?t=146
NO, you're not correct. You're looking for probability. It must be between 0 and 1. Think about it, 3768.32?,Originally Posted by sweetlady
C(n,r) means combinations. A selection of r objects from a group of n objects without regard to order. These are the coefficients in your binomial expansion. Given by n!/[(n-r)!r!].
Since they ask for at least 8, they want 8,9, and 10.
You have: C(10,8)p^8q^2+C(10,9)p^9q+C(10,10)p^10, where p=0.92 and q=0.08
thanks john. I will try it.
thank you very very much
Galactus,
Take it easy. We're trying to help out beginners here.
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Sweetlady,
Whenever you see C(n,r) it is short-hand for the formula for the number of combinations of n items, taking r at a time:
n!/[(n-r)!r!]
So, for C(10,8), we get: 10! / [ (10 - 8)! * 8! ] which computes to 45
for C(10,9), we get: 10! / [ (10 - 9)! * 9! ] which computes to 10
for C(10,10), we get: 10! / [ (10 - 10)! * 10! ] which computes to 1
So, to compute the probability of "at least 8" we add up the probability of 8, 9, and 10
P(r) = C(n,r) * p^r * q^(n-r)
P(8) = C(10,8) * 0.92^8 * 0.08^2 = .1478
P(9) = C(10,9) * 0.92^9 * 0.08^1 = .3778
P(10) = C(10,10) * 0.92^10 * 0.08^0 = .4344
So, P(at least 8) = .1478 + .3778 + .4344 = .96
Originally Posted by JohnM
I'm sorry, I didn't mean to come off as impatient or nasty.
Note, there were no exclamation points.
No worries. Thank you very much for your help
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