Question

The Australian sheep dog is a breed renowned for its intelligence and work ethic. It is estimated that 30% of adult Australian sheep dogs weigh 65 pounds or more. A sample
of 10 adult dogs is studied. What is the probability that more than 7 of them weigh 65 lb

Answer 1

Answer:

0.15% probability that more than 7 of them weigh 65 lb

Step-by-step explanation:

For each dog, there are only two possible outcomes. Either they weigh 65 pounds, or more, or they do not. The probability of a dog weighing 65 pounds or more is independent of other dogs. So we use the binomial probability distribution to solve this question.

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.

30% of adult Australian sheep dogs weigh 65 pounds or more.

This means that $p = 0.3$

Sample of 10 adults dogs.

This means that $n = 10$

What is the probability that more than 7 of them weigh 65 lb

This is

$P(X > 7) = P(X = 8) + P(X = 9) + P(X = 10)$

In which

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

$P(X = 8) = C_{10,8}.(0.3)^{8}.(0.7)^{2} = 0.0014$

$P(X = 9) = C_{10,9}.(0.3)^{9}.(0.7)^{1} = 0.0001$

$P(X = 10) = C_{10,10}.(0.3)^{10}.(0.7)^{0} \cong 0$

$P(X > 7) = P(X = 8) + P(X = 9) + P(X = 10) = 0.0014 + 0.0001 + 0 = 0.0015$

0.15% probability that more than 7 of them weigh 65 lb