Consider the following scenario. Some US bank notes, n = 100, are seized on a su
ID: 3133456 • Letter: C
Question
Consider the following scenario. Some US bank notes, n = 100, are seized on a suspect. The probability that a randomly selected bank note from the general population contains cocaine is about 57%. The probability that a bank note contains cocaine if it has been involved in drug trafficking is about 83%. What is the probability that more than 50 notes contain cocaine if they were not involved in drug trafficking? What is the probability that more than 50 notes contain cocaine if they were involved in drug trafficking? In 2015 there were about 38.1 billion US bank notes in circulation. How many on average are contaminated with cocaine? What is the standard deviation in the number of US bank notes that contained cocaine in 2015? What is the probability that a bank note in your wallet contains cocaine?Explanation / Answer
A)
Here, n = 100, p = 0.57.
We first get the z score for the critical value:
x = critical value = 50.5
u = mean = np = 57
s = standard deviation = sqrt(np(1-p)) = 4.950757518
Thus, the corresponding z score is
z = (x-u)/s = -1.31293039
Thus, the left tailed area is
P(z < -1.31293039 ) = 0.905396797 [ANSWER]
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b)
We first get the z score for the critical value:
x = critical value = 50.5
u = mean = np = 83
s = standard deviation = sqrt(np(1-p)) = 3.756327994
Thus, the corresponding z score is
z = (x-u)/s = -8.652066606
Thus, the left tailed area is
P(z < -8.652066606 ) = 1 [ANSWER]
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c)
As 57% contain cocaine,
E(x) = n p = (38.1 B)*0.57 = 21.717 Billion [ANSWER]
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d)
As n = 3.81 B, p = 0.57,then
s = standard deviation = sqrt(np(1-p)) = 23186.00871 [ANSWER]
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e)
As given
P = 0.57 [ANSWER]