Problem #8: Data from the Central Hudson Laboratory determined that the mean num

Question

Problem #8: Data from the Central Hudson Laboratory determined that the mean number of insect fragments in 225-gram chocolate bars was 14.4 (http://www.centralhudsonlab.com/chocolates.shtml). In a 41-gram bar the mean number of insect fragments would then be 2.62. Assume that the number of insect fragments follows a Poissorn distribution (a) If you eat a 41-gram chocolate bar, find the probability that you will have eaten at least 3 insect fragments. (b) If you eat a 41-gram chocolate bar every week for 12 weeks, find the probability that you will have eaten no insect fragments in exactly 6 of those weeks.

Explanation / Answer

possion distribution
pmf of p.d is = f ( k ) = e- x / x!
where
= parameter of the distribution.   
x = is the number of independent trials
mean rate of possion variable is = = 2.62

PART A.
probability that you will have eaten atleast 3 insects fragments is,
p( x < 3) = p(x=2) + p(x=1) + p(x=0)
= e^-2.62 * 4 ^ 2 / 2! + e^-2.62 * 3 ^ 1 / 1! + e^-2.62 * 4 ^ 0 / 0!   
= 0.5134,
p( x > = 3 ) = 1 - p (x < 3) = 0.4866

PART B.
before finding the resulst for 12 week, first we calculate one week result to which eaten no insect fragments and we apply on the rest there after.
p( x = 0 ) = e ^-2.62 * 2.62^0 / 0! = 0.0728

we have result of p= 0.0728, n = 12, x = 6, Here X ~ B ( 12, 0.0728 )
pmf of b.d is = f ( k ) = ( n k ) p^k * ( 1- p) ^ n-k
where
k = number of successes in trials   
n = is the number of independent trials   
p = probability of success on each trial

P( X = 6 ) = ( 12 6 ) * ( 0.0728^6) * ( 1 - 0.0728 )^6
probability that you will have eaten no
insect fragments in exactly 6 of those weeks is = 0.000087

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