Question

the number of ants per acre in the forest is normally distributed with mean 45,289 and standard deviation 12,340. a. find the probability that a randomly acre in the forest has fewer than 40,000 ants. select b. find the probability that a randomly selected acre has between 35,000 and 50,000 ants. select c. find the z - score of a forest with 57,629 ants per acre. select

Explanation:

Step1: Calculate z-score for 40,000

$z = \frac{X - \mu}{\sigma} = \frac{40000 - 45289}{12340} \approx -0.43$

Step2: Find $P(Z < -0.43)$

Using z-table, $P(Z < -0.43) \approx 0.3336$

Step3: Calculate z-scores for 35,000 and 50,000

$z_1 = \frac{35000 - 45289}{12340} \approx -0.83$, $z_2 = \frac{50000 - 45289}{12340} \approx 0.38$

Step4: Find $P(-0.83 < Z < 0.38)$

$P(Z < 0.38) - P(Z < -0.83) \approx 0.6480 - 0.2033 = 0.4447$

Step5: Calculate z-score for 57,629

$z = \frac{57629 - 45289}{12340} = \frac{12340}{12340} = 1$

Answer:

A. 0.3336
B. 0.4447
C. 1