A retail store has implemented procedures aimed at reducing the number of bad ch

Question

A retail store has implemented procedures aimed at reducing the number of bad checks cashed by its cashiers. The store's goal is to cash no more than eight bad checks per week. The average number of bad checks cashed is 2 per week. Let x denote the number of bad checks cashed per week. Assuming that x has a Poisson distribution:

Round your answers to 4 decimal places.

(a) Find the probability that the store's cashiers will not cash any bad checks in a particular week.

Probability                        

(b) Find the probability that the store will meet its goal during a particular week.

Probability            

(c) Find the probability that the store will not meet its goal during a particular week.

Probability            

(d) Find the probability that the store's cashiers will cash no more than ten bad checks per two-week period.

Probability            

(e) Find the probability that the store's cashiers will cash no more than five bad checks per three-week period.

Probability            

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
I.
mean =
= 2

a.
EQUAL
P( X = 0 ) = e ^-2 * 2^0 / 0! = 0.13534

b.
P( X < = 8) = P(X=8) + P(X=7) + P(X=6) + P(X=5) + P(X=4) + P(X=3) + P(X=2) + P(X=1) + P(X=0)
= e^-2 * 2 ^ 8 / 8! + e^-2 * 0 ^ 7 / 7! + e^-2 * ^ 6 / 6! + e^-2 * ^ 5 / 5! + e^-2 * ^ 4 / 4! + e^-2 * ^ 3 / 3! +
e^-2 * ^ 2 / 2! + e^-2 * ^ 1 / 1! + e^-2 * ^ 0 / 0!
= 0.99976

c.
P( X > 8) = 1 -P ( X <= 8) = 1 - 0.9998 = 0.00024

d.
per two week period the mean rate will be 2*2 = 4
P( X < = 10) = P(X=10) + P(X=9) + P(X=8) + P(X=7) + P(X=6) + P(X=5) + P(X=4) + P(X=3) + P(X=2) + P(X=1) + P(X=0)  
= e^-4 * 4 ^ 10 / 10! + e^-4 * 0 ^ 9 / 9! + e^-4 * ^ 8 / 8! + e^-4 * ^ 7 / 7! + e^-4 * ^ 6 / 6! + e^-4 * ^ 5 / 5! +
e^-4 * ^ 4 / 4! + e^-4 * ^ 3 / 3! + e^-4 * ^ 2 / 2!
e^-4 * ^ 1 / 1! + e^-4 * ^ 0 / 0! + = 0.99716,
P( X > 10) = 1 -P ( X <= 10) = 1 - 0.9972 = 0.00284

e.
per three week period = 2*3 = 6
P( X < = 5) = P(X=5) + P(X=4) + P(X=3) + P(X=2) + P(X=1) + P(X=0)
= e^-6 * 6 ^ 5 / 5! + e^-6 * 0 ^ 4 / 4! + e^-6 * ^ 3 / 3! + e^-6 * ^ 2 / 2! + e^-6 * ^ 1 / 1! + e^-6 * ^ 0 / 0!
= 0.44568,
P( X > 5) = 1 -P ( X <= 5) = 1 - 0.4457 = 0.55432

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