Question

Due Thu 07/30/2020 11:59 pm The number of ants per acre in the forest is normally distributed with mean 43,000 and standard deviation 12.046. 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 29,536 ants. c. Find the probability that a randomly selected acre has between 29,180 and 42,014 ants. d. Find the first quartile. ants (round your answer to a whole number)

Answer 1

a)

Distribution of X ~ N(43000, 12046)

b)

µ = 43000, σ = 12046

P(X < 29536) =

= P( (X-µ)/σ < (29536-43000)/12046 )

= P(z < -1.1177)

Using excel function:

= NORM.S.DIST(-1.1177, 1)

= 0.1318

c)

P(29180 < X < 42014) =

= P( (29180-43000)/12046 < (X-µ)/σ < (42014-43000)/12046 )

= P(-1.1473 < z < -0.0819)

= P(z < -0.0819) - P(z < -1.1473)

Using excel function:

= NORM.S.DIST(-0.0819, 1) - NORM.S.DIST(-1.1473, 1)

= 0.3417

d)

P(x < a) = 0.25

Z score at p = 0.25 using excel = NORM.S.INV(0.25) = -0.6745

Value of X = µ + z*σ = 43000 + (-0.6745)*12046 = 34875