Questions: The number of ants per acre in the forest is normally distributed with mean 44,000 and standard deviation 12,019. Let X= number of ants in a randomly selected acre of the forest. Round all answers to 4 decimal places where possible. a. What is the distribution of X ? X ~ N( , , b. Find the probability that a randomly selected acre in the forest has fewer than 49,514 ants. 0.6768 c. Find the probability that a randomly selected acre has between 48,009 and 62,608 ants. 0.3086 d. Find the first quartile. ants (round your answer to a whole number)

Solution Steps

The number of ants per acre in the forest is normally distributed with a mean (\( \mu \)) of 44,000 and a standard deviation (\( \sigma \)) of 12,019. Therefore, we can express the distribution of \( X \) as: \[ X \sim N(44000, 144456361) \]

To find the probability that a randomly selected acre has fewer than 49,514 ants, we calculate: \[ P(X < 49514) = \Phi(Z_{end}) - \Phi(Z_{start}) = \Phi(0.4588) - \Phi(-\infty) = 0.6768 \] Thus, the probability that a randomly selected acre has fewer than 49,514 ants is: \[ P(X < 49514) = 0.6768 \]

Next, we calculate the probability that a randomly selected acre has between 48,009 and 62,608 ants: \[ P(48009 < X < 62608) = \Phi(Z_{end}) - \Phi(Z_{start}) = \Phi(1.5482) - \Phi(0.3336) = 0.3086 \] Thus, the probability that a randomly selected acre has between 48,009 and 62,608 ants is: \[ P(48009 < X < 62608) = 0.3086 \]

Final Answer