Question 4857
 Looks like the 1st question is related to Poission Distribution.
 Every 20 min, lambda = 4. Let the random varible be X with 
 (I) Poission Distribution with mean Lambda = 4. 
 Pr(X =k) = (Lambda^k/k!) *e^(-Lambda) 
 A) exactly five insects will be caught?
 B) no more than two insects will be caught?
 c) at least four insects will be caught?
 D) over a randomly chosen one hour period what is the probability that exactly 
 10 insects will be caught? 
 Sol: A)  Lambda = 4, k= 5, Pr(X = 5) = (4^5/5!) *e^(-4) = 0.156
      B)  Lambda = 4, Pr(X <= 2) = 0.238
      C)  Lambda = 4, Pr(X => 4) = 1 - Pr(X <= 3) = 0.567
      D) When time = 1 hr = 3*20 min, the mean Lambda is 3*4 = 12 .
       Lambda = 12, Pr(X = 10) = 0.105

 (II) Mu mean = 37.2 and a standard deviation sigma = 0.35 degress.
  Convert it to the dstandard normal distribution by
  z = (X- 37.2)/0.35 and look up the table to find the probability

 i) more than 39 degrees ?
  X = 39, Pr(X> 39)=1- NORMDIST(39,37.2,0.35,1) = 1.35541E-07
 
 ii) less than 36.8 degrees?
  X = 36.8, Pr(X< 36.8)= NORMDIST(36.8,37.2,0.35,1) = 0.127
iii) between 36.5 degrees and 38.5 degrees? 
  Pr(36.5 < X< 38.5)= Pr( X< 38.5) - Pr( X< 36.5)
  = 0.98

 Suggestion : Use MS Excel or Tables in the textbook.

 Kenny