Question

The Australian sheep dog is a breed renowned for its intelligence and work ethic. It is estimated that 40% of adult Australian sheep dogs weigh 65 pounds or more. A sample of 12 adult dogs is studied. What is the probability that exactly 3 of them weigh 65 lb or more

Answer 1

Answer:

$X \sim Binom(n=12, p=0.4)$

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 to find this probability:

$P(X=3)=(12C3)(0.4)^3 (1-0.4)^{12-3}=0.14189$

Step-by-step explanation:

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:

$X \sim Binom(n=12, p=0.4)$

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 to find this probability:

$P(X=3)=(12C3)(0.4)^3 (1-0.4)^{12-3}=0.14189$