1.)
Assume that a procedure yields a binomial distribution with a
trial repeated n=5n=5 times. Use technology to find the probability
distribution given the probability p=0.702p=0.702 of success on a
single trial.
(Report answers accurate to 4 decimal places.)
k | P(X = k) |
---|---|
0 | |
1 | |
2 | |
3 | |
4 | |
5 |
2.)
Assume that a procedure yields a binomial distribution with a
trial repeated n=5n=5 times. Find the probability distribution
given the probability p=0.799p=0.799 of success on a single trial.
Report answers accurate to 4 decimal places.
k | P(X = k) |
---|---|
0 | |
1 | |
2 | |
3 | |
4 | |
5 |
Solution:
Given in the question
P(success)= 0.702
P(failure)= 0.298
No. Of trails =5
P(X=0)= 5C0*(0.702)^0*(0.298)^5 =0.0024
P(X=1)= 5C1*(0.702)^1*(0.298)^4= 0.0277
P(X=2)= 5C2*(0.702)^2*(0.298)^3= 0.1304
P(X=3)= 5C3*(0.702)^3*(0.298)^2= 0.3072
P(X=4)=5C4*(0.702)^4*(0.298)^1= 0.3619
P(X=5)= 5C5*(0.702)^5*(0.298)^0= 0.1705
Solution(b)
Given in the question
P(success)= 0.799
P(failure)= 0.201
No. Of trials = 5
P(X=0)= 5C0*(0.799)^0*(0.201)^5 = 0.0003
P(X=1)= 5C1*(0.799)^1*(0.201)^4 = 0.0065
P(X=2)= 5C2*(0.799)^2*(0.201)^3 = 0.0518
P(X=3)= 5C3*(0.799)^3*(0.201)^2 = 0.2061
P(X=4)= 5C4*(0.799)^4*(0.201)^1 = 0.4096
P(X=5)= 5C5*(0.799)^5*(0.201)^0 = 0.3256
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