Question

The number of ants per acre in the forest is normally distributed with mean 45,000 and standard deviation 12,278. 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 57,834 ants.

c. Find the probability that a randomly selected acre has between 44,152 and 55,612 ants.

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

Answer 1

Solution :

Given that ,

mean = = 45000

standard deviation = = 12278

a) The distribution of x is normal X ~ N(45000, 12278)

b) P(x < 57834) = P[(x - ) / < (57834 - 45000) / 12278]

= P(z < 1.05)

Using z table,

= 0.8531

c) P(44152 < x < 55612) = P[(44152 - 45000)/ 12278) < (x - ) /  < (55612 - 45000) / 12278) ]

= P(-0.07 < z < 0.86)

= P(z < 0.86) - P(z < -0.07)

Using z table,

= 0.8051 - 0.4721

= 0.3330

d) Using standard normal table,

The z dist'n First quartile is,

P(Z < z) = 25%

= P(Z < z) = 0.25  

= P(Z < -0.6745 ) = 0.25

z = -0.6745

Using z-score formula,

x = z * +

x = -0.6745 * 12278 + 45000

x = 36718.489

First quartile =Q1 = 36,718