Question

Assume the random variable X has a binomial distribution with the given probability of obtaining a success. Find the following probability, given the number of trials and the probability of obtaining a success. Round your answer to four decimal places. P(X

Answer 1

Answer:

$P(X=14)=(18C14)(0.8)^{14} (1-0.8)^{18-14}=0.2153$

$P(X=15)=(18C15)(0.8)^{15} (1-0.8)^{18-15}=0.2297$

$P(X=16)=(18C16)(0.8)^{16} (1-0.8)^{18-16}=0.1722$

$P(X=17)=(18C17)(0.8)^{17} (1-0.8)^{18-17}=0.0811$

$P(X=18)=(18C18)(0.8)^{18} (1-0.8)^{18-18}=0.0180$

And adding the values we got:

$P(X \geq 14)= 0.2153 +0.2297+0.1722+0.0811+0.0180=0.7164$

Step-by-step explanation:

Assuming this question: P(X≥14), n=18, p=0.8

Previous concepts  

The binomial distribution is a "DISCRETE probability distribution that summarizes the probability that a value will take one of two independent values under a given set of parameters. The assumptions for the binomial distribution are that there is only one outcome for each trial, each trial has the same probability of success, and each trial is mutually exclusive, or independent of each other".  

Solution to the problem

Let X the random variable of interest, on this case we now that:  

$X \sim Binom(n=18, p=0.8)$  

The probability mass function for the Binomial distribution is given as:  

$P(X)=(nCx)(p)^x (1-p)^{n-x}$  

Where (nCx) means combinatory and it's given by this formula:  

$nCx=\frac{n!}{(n-x)! x!}$  

And we want this probability:

$P(X \geq 14) = P(X=14) +P(X=15) +P(X=16)+P(X=17)+P(X=18)$

And we can find the individual proabilities using the probability mass function:

$P(X=14)=(18C14)(0.8)^{14} (1-0.8)^{18-14}=0.2153$

$P(X=15)=(18C15)(0.8)^{15} (1-0.8)^{18-15}=0.2297$

$P(X=16)=(18C16)(0.8)^{16} (1-0.8)^{18-16}=0.1722$

$P(X=17)=(18C17)(0.8)^{17} (1-0.8)^{18-17}=0.0811$

$P(X=18)=(18C18)(0.8)^{18} (1-0.8)^{18-18}=0.0180$

And adding the values we got:

$P(X \geq 14)= 0.2153 +0.2297+0.1722+0.0811+0.0180=0.7164$