The Australian sheep dog is a breed renowned for its intelligence and work ethic. It is estimated that 45% of adult Australian sheep dogs weigh 65 pounds or more. A sample of 12 adult dogs is studied. What is the probability that more than 9 of them weigh 65 lb or more?

Probability of more than 9 adult Australian sheep dogs out of 12 weighing 65 lb or more

P(X > 9) = 0.00788

Step-by-step explanation:

The only assumption required for the question is that all 12 adult dogs sampled must all be Australian sheep dogs.

This is a binomial distribution problem

Binomial distribution function is represented by

P(X = x) = ⁿCₓ pˣ qⁿ⁻ˣ

n = total number of sample spaces = number of adult dogs to be sampled = 12

x = Number of successes required = number of dogs that weigh 65 lb or more

= more than 9; >9

p = probability of success = probability of a dog weighing 65 lb or more = 0.45

q = probability of failure = probability of a dog NOT weighing 65 lb or more = 1 - 0.45 = 0.55

P(X > 9) = P(X=10) + P(X=11) + P(X=12)

Solving each of these probabilities, using the binomial distribution formula

P(X = x) = ¹²Cₓ (0.45)ˣ (0.55)¹²⁻ˣ with x = 10, 11 and 12

P(X > 9) = P(X=10) + P(X=11) + P(X=12)

= 0.00679820806 + 0.00101130368 + 0.00006895252

= 0.00787846427

= 0.00788 to 3 s.f

Hope this Helps!!!

(- 1, 0 )

step-by-step explanation:

an x- intercept is where the function crosses the x- axis.

to find the x- intercept/s look for the value of x when f(x) = 0

from the given table there are 3 values of x where f(x) = 0

when x = - 1, x = 2, x = 3

the one that is listed is (- 1, 0 )

pls ask