Question

The number of ants per acre in the forest is normally distributed with mean 42,000 and standard deviation 12.263. 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-NO b. Find the probability that a randomly selected acre in the forest has fewer than 28,594 ants. c. Find the probability that a randomly selected acre has between 32,065 and 41,894 ants. ants (round your answer to a d. Find the first quartile. whole number)

Answer 1

Given Mean u = 42000 and standard deviation o = 12263 a) The distribution of X-N(u=42000,02 =150381169) Some text books are followed X-N(u=42000,0=12263) b) P(x<28594) = P(*** < 28599743000) =P(Z<-1.09) =0.1379 (using z value table) Normdist(28594,42000,12263,true)=0.13715 (using excel) c) We have P(32065<x<41894) = P200 - D132065-42000X- 4 12263 41894-42000 12263 =P(-0.81<Z<0.01) =P(z<-0.01)-p(z<-0.81) =0.4960-0.2090 (using z value table) =0.2870 Normdist(32065,42000,12263,true)-Normdist(41894,42000,12263,true) =0.287627704 (using excel)

d) P(X<x)=0.25 P(* <*792000) =0.25 z value for the P(0.25) is = -0.67 x-42000 =-0.67 12263 x=42000-0.67*12263=33783.79 =42000+NORMSINV(0.25)*12263 =33728.73 (using excel)