Question

Candace has an ash tree that has been infected with emerald ash borer. Suppose that the remaining life (yrs) of a treated ash tree like Candace's can be modeled as Y∼N(13,6) and the remaining life (yrs) of an untreated ash tree like Candace's can be modeled as X∼N(8,3). The two variables X and Y are independent. 5(a). Compute the chances that the untreated ash tree will last less than 48 months. That is, compute P(12 X < 48). 5(b). Compute the chances that the treated ash tree would last at least 10 years longer than the untreated ash tree. That is, compute P(Y >= X+10). Show your work.

Answer 1

For part (a), the untreated ash tree will have a probability of 0 of lasting less than 48 months. For part (b), the treated ash tree has a 0.7711 probability of lasting at least 10 years longer than the untreated ash tree.

Step-by-step explanation:

5(a). Given the information that X follows a normal distribution with mean 8 and standard deviation 3, we can standardize the variable to calculate the probability that X is less than 48. From the standard normal distribution table, we can find that the z-score of 48 is (48 - 8) / 3 = 13.3333. Looking up this z-score, we find a probability of approximately 1. However, since we want the probability that X < 48, we subtract this value from 1 to get 1 - 1 = 0.

5(b). To calculate the probability that Y >= X + 10, we first need to find the distribution of the difference Z = Y - X. Since both Y and X are independent normal random variables, the difference Z will also follow a normal distribution with mean 13 - 8 = 5 and standard deviation sqrt(6^2 + 3^2) = sqrt(45) = 6.7082. To find the probability that Z >= 10, we can standardize Z by subtracting the mean of 5 and dividing by the standard deviation of 6.7082. This gives us a z-score of (10 - 5) / 6.7082 = 0.7463. Looking up this z-score, we find a probability of approximately 0.7711. Therefore, the chances that the treated ash tree will last at least 10 years longer than the untreated ash tree is 0.7711.