SOLUTION: An insect trap catches an average of four insects per 20 minute. Assuming that insects are caught randomly and independently, what is the probability that in a randomly chosen 20

Question 4857: An insect trap catches an average of four insects per 20 minute. Assuming that insects are caught randomly and independently, what is the probability that in a randomly chosen 20 minute period:
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?
The body temperature of a population of mammals are known to be normally distributed with a mean of 37.2 degrees and a standard deviation of 0.35 degress.
A) what proportion of the population have a tempreture of
i) more than 39 degrees ?
ii) less than 36.8 degrees?
iii) between 36.5 degrees and 38.5 degrees?
I will be very grateful if you could help me with these questions please.
Thank you very much
Birgul....

Answer by khwang(438) (Show Source): You can put this solution on YOUR website!
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

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