Question

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?

Answer 1

Answer:

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!!!

Answer 2

Answer:

The probability that more than 9 of them weigh 65 lb or more is 7.8785 $\times 10^{-3}$.

Step-by-step explanation:

We are given that 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.

Also,  sample of 12 adult dogs is studied.

The above situation can be represented through Binomial distribution;

$P(X=r) = \binom{n}{r}p^{r} (1-p)^{n-r} ; x = 0,1,2,3,.....$

where, n = number of trials (samples) taken = 12 dogs

             r = number of success = more than 9

            p = probability of success which in our question is % of adult

                  Australian sheep dogs who weigh 65 pounds or more, i.e; 45%

LET X = Number of dogs who weigh 65 pounds or more

So, it means X ~ $Binom(n=12, p=0.45)$

Now, Probability that more than 9 of them weigh 65 lb or more is given by = P(X > 9)

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

= $\binom{12}{10}\times 0.45^{10} \times (1-0.45)^{12-10} + \binom{12}{11}\times 0.45^{11} \times (1-0.45)^{12-11} + \binom{12}{12}\times 0.45^{12} \times (1-0.45)^{12-12}$

= $66 \times 0.45^{10} \times 0.55^{2}+ 12 \times 0.45^{11} \times 0.55^{1}+ 1 \times 0.45^{12} \times 0.55^{0}$

= 7.8785 $\times 10^{-3}$

Therefore, Probability that more than 9 of them weigh 65 lb or more is 7.8785 $\times 10^{-3}$.