Question

A. The number of insects in peanut butter follows a Poisson distribution. Suppose that a sample of five grams peanut butter contains 5.9 insect fragments. Compute the probabilities of the following cases: At least six insect fragments in an eight-gram sample.

B.The number of insects in peanut butter follows a Poisson distribution. Suppose that a sample of five grams peanut butter contains 5.9 insect fragments. Compute the probabilities of the following cases: In between 10 and 15 insect fragments in a 12-gram sample.

C. The number of insects in peanut butter follows a Poisson distribution. Suppose that a sample of five grams peanut butter contains 5.9 insect fragments. Compute the probabilities of the following cases: The number of insect fragments in a 15-gram sample are within 10 insect fragments each side from the mean.

Answer 1

A,

Let X be the number of insect fragments in k grams of peanut butter. Then X ~ Poisson(k)

where is the average number of insect fragments per gram of peanut butter.

Given,  five grams peanut butter contains 5.9 insect fragments

= 5.9 / 5 = 1.18

For k = 8 grams, X ~ Poisson(1.18 * 8 = 9.44)

Probability that at least six insect fragments in an eight-gram sample = P(X 6)

= 1 - P(X 5)

= 1 - [P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5) ]

= 1 - [ exp(-9.44) 9.44⁰ /0! + exp(-9.44) 9.44¹ /1! + exp(-9.44) 9.44² /2! + exp(-9.44) 9.44³ /3! + exp(-9.44) 9.44⁴ /4! + exp(-9.44) 9.44⁵/5! ]

= 1 - (0.0000 + 0.0008 + 0.0035 + 0.0114 + 0.0263 + 0.0497)

= 0.9083

B.

For k = 12 grams, X ~ Poisson(1.18 * 12 = 14.16)

Probability that in between 10 and 15 insect fragments in a 12-gram sample = P(10 X 15)

= [P(X = 10) + P(X = 11) + P(X = 12) + P(X = 13) + P(X = 14) + P(X = 15) ]

= exp(-14.16) 14.16¹⁰ /10! + exp(-14.16) 14.16¹¹ /11! + exp(-14.16) 14.16¹² /12! + exp(-14.16) 14.16¹³ /13! + exp(-14.16) 14.16¹⁴ /14! + exp(-14.16) 14.16¹⁵/15! ]

= 0.06327900 + 0.08145734 + 0.09611966 + 0.10469649 + 0.10589302 + 0.09996301

= 0.5514

C.

For k = 15 grams, X ~ Poisson(1.18 * 15 = 17.7)

Probability that in a 15-gram sample are within 10 insect fragments each side from the mean = P(17.7 - 10 X 17.7 + 10)

= P(7.7   X 27.7)

Using Normal approximation to the Poisson distribution, X ~ Normal( = 17.7, = 4.207)

P(7.7   X 27.7) = P(X 27.7) - P(X 7.7)

= P(X < 28) - P(X < 8)

= P(X < 27.5) - P(X < 7.5)

= P[Z < (27.5 - 17.7)/4.207] - P[Z < (7.5 - 17.7)/4.207]

= P[Z < 2.33] - P[Z < -2.42]

= 0.99 - 0.0078

= 0.9822