Question

Let X be a Binomial Random Variable with 8 trials and a probability of success as 0.37. Suppose you want to find k such that P(X < k) = 0.6625. What is the value of k?

Answer 1

Answer:

k = 4

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:

$n = 8, p = 0.37$. So

P(X < k) = 0.6625

k only assumes discrete values, so

$P(X < k) = P(X \leq k-1) = 0.6625$

We have to find the cummulative distribution until we hit 0.6625. So

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

$P(X = 0) = C_{8,0}.(0.37)^{0}.(0.63)^{8} = 0.0248$

$P(X \leq 0) = P(X = 0) = 0.0248$

$P(X = 1) = C_{8,1}.(0.37)^{1}.(0.63)^{7} = 0.1166$

$P(X \leq 1) = P(X = 0) + P(X = 1) = 0.0248 + 0.1166 = 0.1414$

$P(X = 2) = C_{8,2}.(0.37)^{2}.(0.63)^{6} = 0.2397$

$P(X \leq 2) = P(X = 0) + P(X = 1) + P(X = 2) = 0.0248 + 0.1166 + 0.2397 = 0.3811$

$P(X = 3) = C_{8,3}.(0.37)^{3}.(0.63)^{5} = 0.2814$

$P(X \leq 3) = P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) = 0.0248 + 0.1166 + 0.2397 + 0.2814 = 0.6625$

$P(X \leq 3) = P(X < 4) = 0.665$

So k = 4