Question

The number of ants per acre in the forest is normally distributed with mean 43,000 and standard deviation 12,014. 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 46,226 ants.

c. Find the probability that a randomly selected acre has between 44,570 and 49,795 ants.

d. Find the first quartile. ants (round your answer to a whole number)

Answer 1

a)
X ~ N(43000, 12014)

b)
Here, μ = 43000, σ = 12014 and x = 46226. We need to compute P(X <= 46226). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z = (46226 - 43000)/12014 = 0.27

Therefore,
P(X <= 46226) = P(z <= (46226 - 43000)/12014)
= P(z <= 0.27)
= 0.6064

c)
P(44570 <= X <= 49795) = P((49795 - 43000)/12014) <= z <= (49795 - 43000)/12014)
= P(0.13 <= z <= 0.57) = P(z <= 0.57) - P(z <= 0.13)
= 0.7157 - 0.5517
= 0.1640

d)
z-value = -0.67

x = 43000 -0.67*12014 = 34950.62

Ans: 34950