Question

13. Assume that a procedure yields a binomial distribution with a trial repeated n=5n=5 times. Use some form of technology to find the cumulative probability distribution given the probability p=0.155p=0.155 of success on a single trial.

(Report answers accurate to 4 decimal places.)

k	P(X < k)
0
1
2
3
4
5

Answer 1

a)

Here, n = 5, p = 0.155, (1 - p) = 0.845 and x = 0
As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x)

We need to calculate P(X <= 0).
P(X <= 0) = (5C0 * 0.155^0 * 0.845^5)
P(X <= 0) = 0.4308
P(X <= 0) = 0.4308

b)
Here, n = 5, p = 0.155, (1 - p) = 0.845 and x = 0
As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x)

We need to calculate P(X <= 0).
P(X <= 0) = (5C0 * 0.155^0 * 0.845^5)
P(X <= 0) = 0.4308
P(X <= 0) = 0.4308

c)

Here, n = 5, p = 0.155, (1 - p) = 0.845 and x = 1
As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x)

We need to calculate P(X <= 1).
P(X <= 1) = (5C0 * 0.155^0 * 0.845^5) + (5C1 * 0.155^1 * 0.845^4)
P(X <= 1) = 0.4308 + 0.3951
P(X <= 1) = 0.8259

d)

Here, n = 5, p = 0.155, (1 - p) = 0.845 and x = 2
As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x)

We need to calculate P(X <= 2).
P(X <= 2) = (5C0 * 0.155^0 * 0.845^5) + (5C1 * 0.155^1 * 0.845^4) + (5C2 * 0.155^2 * 0.845^3)
P(X <= 2) = 0.4308 + 0.3951 + 0.145
P(X <= 2) = 0.9709

e)

Here, n = 5, p = 0.155, (1 - p) = 0.845 and x = 3
As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x)

We need to calculate P(X <= 3).
P(X <= 3) = (5C0 * 0.155^0 * 0.845^5) + (5C1 * 0.155^1 * 0.845^4) + (5C2 * 0.155^2 * 0.845^3) + (5C3 * 0.155^3 * 0.845^2)
P(X <= 3) = 0.4308 + 0.3951 + 0.145 + 0.0266
P(X <= 3) = 0.9975

f)

Here, n = 5, p = 0.155, (1 - p) = 0.845 and x = 4
As per binomial distribution formula P(X = x) = nCx * p^x * (1 - p)^(n - x)

We need to calculate P(X <= 4).
P(X <= 4) = (5C0 * 0.155^0 * 0.845^5) + (5C1 * 0.155^1 * 0.845^4) + (5C2 * 0.155^2 * 0.845^3) + (5C3 * 0.155^3 * 0.845^2) + (5C4 * 0.155^4 * 0.845^1)
P(X <= 4) = 0.4308 + 0.3951 + 0.145 + 0.0266 + 0.0024
P(X <= 4) = 0.9999

0

k	P(X < k)
0	0.4308
1	0.4308
2	0.8259
3	0.9709
4	0.9975
5	0.9999