For this section, use the following distributions: 1) A direwolf has a 70% probability of killing a stag upon encounter; 2) a dire wolf awakens the village on average 1.2 times per six weeks; 3) a dire wolf will eat an amount that is uniformly distributed between 10-24 pounds per week; 4) the weight of an average dire wolf is normally distributed with an average of 150 lbs and a standard deviation of 21.43 pounds.

Show which is more likely:
A dire wolf eating exactly 15lbs of food next week or a direwolf weighing exactly 170 lbs.

I know the answer is that both are 0, but I'm not sure how to get there.
Thanks!!

Neither probability = 0.

Weight

Z = (score-mean)/SD

Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability related to the Z score.

Food

24 - 10 = 14

(15-10)/14 = ?

To determine which event is more likely, we need to calculate the probability for each event.

Let's start with the probability of a dire wolf eating exactly 15 lbs of food next week. According to the given information, the amount of food a dire wolf eats per week follows a uniform distribution between 10-24 pounds. Since the distribution is uniform, we can calculate the probability of eating exactly 15 lbs as the ratio of the width of the interval (which is 24 - 10 = 14) to the total range of the distribution. Therefore, the probability is 1/14 or approximately 0.0714.

Now let's calculate the probability of a dire wolf weighing exactly 170 lbs. We are given that the weight of an average dire wolf follows a normal distribution with a mean µ of 150 lbs and a standard deviation σ of 21.43 lbs. We can use the formula for the probability density function (PDF) of the normal distribution to calculate the probability.

P(X = 170) = 1 / (σ * sqrt(2π)) * e^(-(170 - µ)^2 / (2σ^2))

Using the given values, we can plug in the numbers and calculate the probability.

P(X = 170) = 1 / (21.43 * sqrt(2π)) * e^(-(170 - 150)^2 / (2 * 21.43^2))

However, when we plug in the values, we will notice that the PDF of the normal distribution will return a very small value, close to 0. Therefore, the probability of a dire wolf weighing exactly 170 lbs is practically 0.

Hence, both events (a dire wolf eating exactly 15 lbs of food next week and a dire wolf weighing exactly 170 lbs) have a probability close to 0, making them highly unlikely.