C(9,2) = 9!/[2!(7!)]
= 9*8/2 = 9*4 = 36
so 36(.35)^2(1-.35)^7
= 36 * .1225 * .049
= .216
I do not know why by PDF means (By probability distribution function?)
Assume that a procedure yields a binomial distribution with a trial repeated n times. use the binomial probability formula to find the probability of x success given the probability p of success on a single trial.
1. by formula- n=9, x=2, p=0.35
2. by PDF- n=15, x=13, p=1/3
5 answers
yes thank you I believe that's what it stands for...
Well, I suppose I could look it up in a table of the distribution for n = 15, x =13, p =.3333... by symmetry
it is the same as for n = 15, x = 2, p = .33333....
but I do not have a good table
I have only up to n = 10
so have o do it the same old way
C(15,13) = 105
so
105 * (1/3)^13 * (2/3)^2
= 2.927*10^-5
c(15,13)
it is the same as for n = 15, x = 2, p = .33333....
but I do not have a good table
I have only up to n = 10
so have o do it the same old way
C(15,13) = 105
so
105 * (1/3)^13 * (2/3)^2
= 2.927*10^-5
c(15,13)
Now maybe they mean using the normal function which is the limit of binomial for large n
u = n p = 15*1/3 = 5
s^2 = 5(2/3) = 10/3 = 3.333...
s = 1.83
z = (x-u)/s = (13-5)/1.83 = 4.38
f(z) = (1/sqrt(2pi) e^-(4.38^2/2)
= 2.72 * 10^-5
gee, not too far off
u = n p = 15*1/3 = 5
s^2 = 5(2/3) = 10/3 = 3.333...
s = 1.83
z = (x-u)/s = (13-5)/1.83 = 4.38
f(z) = (1/sqrt(2pi) e^-(4.38^2/2)
= 2.72 * 10^-5
gee, not too far off
For a binomial distribution with parameters
n = 5 , p = 0.3 . Find the probabilities of
getting :
(i) Atleast 3 successes.
(ii) Atmost 3 successes.
n = 5 , p = 0.3
n = 5 , p = 0.3 . Find the probabilities of
getting :
(i) Atleast 3 successes.
(ii) Atmost 3 successes.
n = 5 , p = 0.3