Question

Assume that a procedure yields a binomial distribution with a trial repeated n = 5 times. Use some form of technology to find the probability distribution given the probability p = 0.463 of success on a single trial (Report answers accurate to 4 decimal places.) P(X = k) 0 132 321321 1321 Submit Question

Answer 1

1. For Binomial distribution

P(X = k) = (1.),*(1 – pja-ke

n= 5 , p =0.463

Thus ,

$P(X=k)= \binom{5}{k}0.463^{k}(1-0.463)^{5-k}$

putting k=0 , 1 , 2 etc we can find the respective probabilities

k	P(X=k)
0	0.0447
1	0.1925
2	0.3320
3	0.2862
4	0.1234
5	0.0213

Note : Used excel find the probabilities , formula for P(X=0) = "=BINOM.DIST(0,5,0.463,FALSE)"

for P(X=1) = =BINOM.DIST(1,5,0.463,FALSE) , etc

2. Let X be the number of person favor the new building project

X follow Binomial with n =10 , p =0.55

P(X = 1) = (19)0.55*(1 – 0.55):0–2

P(X = 1) = (10) 0.55'(1 – 0.55)10-1

= 0.0042

Probability that exactly one of then favor the new building project = 0.0042

3. Let X be the number of hits

X follow Binomial with n=7, p = not known to us

Given , mean =0.36

→np= 0.36

p=0.0514

To find P(X=2 )

$P(X=2)= \binom{7}{2}0.05^{2}(1-0.05)^{7-2}$

= 0.0426

4.

Let X be the number of hits

X follow Binomial with n=9, p = not known to us

Given , mean =0.226

$\Rightarrow p=0.0251$

To find P(X=3 )

$P(X=3)= \binom{9}{3}0.0251^{3}(1-0.0251)^{9-3}$

= 0.0011