First, we need to find the probability that a single jaguar weighs more than 76 kg. Given that the mean weight is 64 kg and the variance is 361 kg^2 (so the standard deviation is sqrt(361) = 19 kg), we can standardize the weight of 76 kg using the z-score formula: z = (76 - 64) / 19 = 0.63 Using a standard normal distribution table or a calculator, we find that P(Z 0.63) = 1 - P(Z 2) = 1 - P(X ≤ 2) To find P(X ≤ 2), we can use the binomial distribution formula: P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2) P(X = k) = C(n, k) * p^k * (1 - p)^(n - k) where n is the number of trials (12 jaguars), k is the number of successes (number of jaguars that weigh more than 76 kg), and p is the probability of success (0.2643). P(X = 0) = C(12, 0) * 0.2643^0 * (1 - 0.2643)^(12 - 0) = 0.035 P(X = 1) = C(12, 1) * 0.2643^1 * (1 - 0.2643)^(12 - 1) = 0.132 P(X = 2) = C(12, 2) * 0.2643^2 * (1 - 0.2643)^(12 - 2) = 0.264 P(X ≤ 2) = 0.035 + 0.132 + 0.264 = 0.431 Therefore, P(X 2) = 1 - P(X ≤ 2) = 1 - 0.431 = 0.569 So, the probability that more than 2 out of 12 jaguars will need to go on the special healthy diet is approximately 0.569.

Question

First, we need to find the probability that a single jaguar weighs more than 76 kg.

Given that the mean weight is 64 kg and the variance is 361 kg^2 (so the standard deviation is sqrt(361) = 19 kg), we can standardize the weight of 76 kg using the z-score formula:

z = (76 - 64) / 19 = 0.63

Using a standard normal distribution table or a calculator, we find that P(Z > 0.63) = 1 - P(Z < 0.63) = 1 - 0.7357 = 0.2643.

This is the probability that a single jaguar weighs more than 76 kg.

Next, we want to find the probability that more than 2 out of 12 jaguars weigh more than 76 kg. This is a binomial problem, where the number of trials is 12, the number of successes is more than 2, and the probability of success on each trial is 0.2643.

The probability that more than 2 jaguars weigh more than 76 kg is 1 minus the probability that 2 or fewer jaguars weigh more than 76 kg:

P(X > 2) = 1 - P(X ≤ 2)

To find P(X ≤ 2), we can use the binomial distribution formula:

P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2)

P(X = k) = C(n, k) * p^k * (1 - p)^(n - k)

where n is the number of trials (12 jaguars), k is the number of successes (number of jaguars that weigh more than 76 kg), and p is the probability of success (0.2643).

P(X = 0) = C(12, 0) * 0.2643^0 * (1 - 0.2643)^(12 - 0) = 0.035

P(X = 1) = C(12, 1) * 0.2643^1 * (1 - 0.2643)^(12 - 1) = 0.132

P(X = 2) = C(12, 2) * 0.2643^2 * (1 - 0.2643)^(12 - 2) = 0.264

P(X ≤ 2) = 0.035 + 0.132 + 0.264 = 0.431

Therefore, P(X > 2) = 1 - P(X ≤ 2) = 1 - 0.431 = 0.569

So, the probability that more than 2 out of 12 jaguars will need to go on the special healthy diet is approximately 0.569.

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