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 = 0.517 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

Answer:

$P(X = 0) = 0.0263$

$P(X = 1) = 0.1407$

$P(X = 2) = 0.3012$

$P(X = 3) = 0.3224$

$P(X = 4) = 0.1725$

$P(X = 5) = 0.0369$

Step-by-step explanation:

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

In which $C_{n,x}$ is the number of different combinations of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And p is the probability of X happening.

In this problem we have that:

$n = 5, p = 0.517$

Distribution

$P(X = x) = C_{n,x}.p^{x}.(1-p)^{n-x}$

$P(X = 0) = C_{5,0}.(0.517)^{0}.(0.483)^{5} = 0.0263$

$P(X = 1) = C_{5,1}.(0.517)^{1}.(0.483)^{4} = 0.1407$

$P(X = 2) = C_{5,2}.(0.517)^{2}.(0.483)^{3} = 0.3012$

$P(X = 3) = C_{5,3}.(0.517)^{3}.(0.483)^{2} = 0.3224$

$P(X = 4) = C_{5,4}.(0.517)^{4}.(0.483)^{1} = 0.1725$

$P(X = 5) = C_{5,5}.(0.517)^{5}.(0.483)^{0} = 0.0369$