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 18 per week. Let x denote the number of bad checks cashed per week. Assuming that x has a Poisson distribution:

A) Find the probability that the store's cashiers will not cash any bad checks in a particular week. (Round your answer to 4 decimal places. Leave no cells blank - be certain to enter "0" wherever required.)

(b) Find the probability that the store will meet its goal during a particular week. (Round your answer to 4 decimal places. Leave no cells blank - be certain to enter "0" wherever required.)

(c) Find the probability that the store will not meet its goal during a particular week. (Round your answer to 4 decimal places. Leave no cells blank - be certain to enter "0" wherever required.)

(d) Find the probability that the store's cashiers will cash no more than 10 bad checks per two-week period. (Round your answer to 4 decimal places. Leave no cells blank - be certain to enter "0" wherever required.)

(e) Find the probability that the store's cashiers will cash no more than five bad checks per three-week period. (Round your answer to 4 decimal places. Leave no cells blank - be certain to enter "0" wherever required.)

Answer 1

We have, x = the number of bad checks cashed per week

By the problem,

X ~ Poisson (), being the parameter of the distribution

Then the p.m.f. of X, denoted by, f(x) can be written as

Since for Poisson distribution Mean = Variance =

Therefore, for this distribution = 18.

a) The probability that the store's cashiers will not cash any bad checks in a particular week is

  

= 0 (approx)

Ans: The probability that the store's cashiers will not cash any bad checks in a particular week is 0.

b) The store's goal is to cash no more than eight bad checks per week. So, the probability that the store will meet its goal during a particular week is

  

= 0.0071 (rounded to 4 decimal places) [using biometrica table for Poisson distribution]

Ans: The probability that the store will meet its goal during a particular week is 0.0071.

c) The probability that the store will not meet its goal during a particular week is

= 1 - 0.0071 [using the answer of b)]

= 0.9929

Ans: The probability that the store will not meet its goal during a particular week is 0.9929.

d) The probability that the store's cashiers will cash no more than 10 bad checks per two-week period is

= The probability of encashing less than or equal to 10 bad checks in a span of two-weeks.

= (The probability of encashing 0 bad checks in first week X the probability of encashing less than or equal to 10 bad checks in the second week) + (The probability of encashing 1 bad checks in first week X the probability of encashing less than or equal to 9 bad checks in the second week) + (The probability of encashing 2 bad checks in first week X the probability of encashing less than or equal to 8 bad checks in the second week) + ... + (The probability of encashing 10 bad checks in first week X the probability of encashing 0 bad checks in the second week)

  

   [the rest of the terms are zero as both and are zero for x = 0, 1, 2, 3 and 4]

  

= 0 (approx) [using biometrica table for Poisson distribution]

Ans: The probability that the store's cashiers will cash no more than 10 bad checks per two-week period is 0.

e) The probability that the store's cashiers will cash no more than 5 bad checks per three-week period is

  

= 0 [since at least one factor in all the terms is 0]

Ans: The probability that the store's cashiers will cash no more than 5 bad checks per three-week period is 0.