Question

Assume that a procedure yields a binomial distribution with a trial repeated n=5n=5 times. Use some form of technology like Excel or StatDisk to find the probability distribution given the probability p=0.533p=0.533 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

The values provided in the above question are as below

n = 5

p = 0.533

We find the above probability using following Excel function are as below

=BINOM.DIST(number_s, trials, probability_s, cumulative)

We find the probability of k = 0 using Excel function are as

Here, number_s = 0, trials = 5, probability_s = 0.533, cumulative = FALSE

Using Excel function of above values as below

=BINOM.DIST(0, 5, 0.533, FALSE) then press Enter we get

= 0.022211833 0.0222 (Round up to 4 decimal places)

Similarly if probability of k = 1 using Excel function are as

number_s = 1, trials = 5, probability_s = 0.533, cumulative = FALSE

=BINOM.DIST(1, 5, 0.533, FALSE) then press Enter we get

= 0.126754894 0.1268 (Round up to 4 decimal places)

Similarly using the above Excel function we calculate the probabilities of k = 2, 3, 4, 5

We calculate the probability of all values and round up to 4 decimal places in the third column of given table

k	P(X=k)	P(X=k)
0	0.022211833	0.0222
1	0.126754894	0.1268
2	0.289337723	0.2893
3	0.330229136	0.3302
4	0.188449817	0.1884
5	0.043016596	0.0430

We report answers accurate to 4 decimal places are as below

k	P(X=k)
0	0.0222
1	0.1268
2	0.2893
3	0.3302
4	0.1884
5	0.0430

Summary :-

k	P(X=k)
0	0.0222
1	0.1268
2	0.2893
3	0.3302
4	0.1884
5	0.0430