QA 233 (Basic Business Statistics)

Solutions to Practice Problem Set V

Pillar Foods, Inc. produces Choco-Poofies, a breakfast cereal popular with young children. They currently produce ten ounce, twenty ounce, and three pound (thirty-six ounce economy size) boxes of this cereal. After a slurry of the ingredients is mixed, it is extruded through a star-shaped mold, cut into quarter-inch wide pieces, and baked until crunchy. Every production employee has the capability of bringing production to an immediate stop if he/she observes a problem with the process. Once the baking process is complete, the final product is brought by conveyor belt into machines that funnel the product into boxes. A ticket with a point value of 1, 2, 3, 4, or 5 (redeemable for a plastic giraffe, a plastic elephant, a plastic hippopotamus, a plastic three-toed sloth, or a plastic hyena once a certain number of points is accumulated) is then dropped into the box, and the box is sealed.

1. Suppose that the average number of times the production process is stopped by an employee throughout a single shift (eight hours) is 3.5. What is the probability that the process will not be stopped during a one-hour period? What is the probability that the process will be stopped at least twice during a one-hour period?

Our random variable in this problem is the number of times the production process is stopped by an employee in an hour, which is a discrete random variable. Since we are dealing with the number of times a relatively rare event (the production process is stopped by an employee in an hour) occurs, we should consider using the Poisson Probability Distribution to answer these questions (QA 233 Students - this is the answer to your question - stop HERE for this problem!). The first question can be written as:

P(X=0)

where x is the number of times the production process is stopped by an employee in an hour. We also know, for a discrete random variable, that

P(X=0)=f(0)

And for a Poison random variable we have that

We have been given that the mean number of times the production process is stopped by an employee throughout a single shift (eight hours) is 3.5. Since our question deals with the number of times the production process is stopped by an employee in an hour, we need to convert our mean to an hourly value as well. Thus we have that:

In other words, we expect the production line to be shut down by an employee 0.4375 per hour. Now we can address the first question. Using the Poisson Probability Distribution Function we have that:

i.e., the probability that the production line will not be stopped over a one-hour period is 0.646.

We can answer the second question in a similar manner. We write this question as:

P(X³2)

where x is the number of times the production process is stopped by an employee in an hour. Again, because x is a discrete random variable, we have that

P(X³2)=f(2)+f(3)+f(4)+f(5)+^¼

which can be rewritten as

P(X³2)=1-P(X<2)=1-P(X£1)=1-[f(0)+f(1)]

and for a Poison random variable we have that

We again must convert our mean to an hourly value as well. Thus we have that m=0.4375. Now we can address the second question. Using the Poisson Probability Distribution Function we have that:

from the first part of this problem and

so the probability that the process will be stopped at least twice during a one-hour period is

P(X³2)=1-P(X<2)=1-P(X£1)=1-[f(0)+f(1)]=1-(0.646+0.282)=0.072

 

2. If tickets of the various point values (1, 2, 3, 4, 5) are distributed throughout the boxes evenly, what is the probability that a randomly selected box will include a ticket with a point value of at least four? Under these conditions, what is the probability that a randomly selected box will include a ticket with a point value of no more than two?

Our random variable in this problem is the point value of the ticket placed in a box of Choco-Poofies, which is a discrete random variable. Since all five outcomes (1, 2, 3, 4, 5) are equally likely, we should consider using the Discrete Uniform Probability Distribution to answer the question. The question we have been asked can be written as:

P(X³4)

where x is the point value of the ticket placed in a box of Choco-Poofies. We also know, since we have a Discrete Uniform random variable that can only assume the values 1, 2, 3, 4, and 5, that

P(X³4)=f(4)+f(5)

and for a Discrete Uniform random variable we have that

for any feasible value of x, where n is the number of unique feasible values that x could assume (or 5 for this problem). Thus we have that:

and

so our final answer is

P(X³4)=f(4)+f(5)=0.20+0.20=0.40

For the second part of this problem, again our random variable in this problem is the point value of the ticket placed in a box of Choco-Poofies, which is a discrete random variable. Since all five outcomes (1, 2, 3, 4, 5) are equally likely, we should consider using the Discrete Uniform Probability Distribution to answer the question. The question we have been asked can be written as:

P(X£2)

where x is the point value of the ticket placed in a box of Choco-Poofies. We also know, since we have a Discrete Uniform random variable that can only assume the values 1, 2, 3, 4, and 5, that

P(X£2)=f(1)+f(2)

and for a Discrete Uniform random variable we have that

for any feasible value of x, where n is the number of unique feasible values that x could assume (or 5 for this problem). Thus we have that:

and

so our final answer is

P(X£2)=f(1)+f(2)=0.20+0.20=0.40

 

3. Again assume the conditions outlined in problem 2 (tickets of each point value are equally likely) are true. If you randomly select twelve boxes of Choco-Poofies, what is the probability that at least three of the boxes will contain a five-point ticket? What is the probability that none of the boxes will contain a one-point ticket?

Our random variable in the first question of this problem is the number of the twelve selected boxes that contain a five-point ticket, which is a discrete random variable. Since we are dealing with the number of successes (boxes containing a five-point ticket) in a set number of trials (the twelve randomly selected boxes), we should consider using the Binomial Probability Distribution to answer this question. The first question can be written as:

P(X³3)

where x is the number of the twelve boxes that contain a five-point ticket. We also know, for this discrete random variable, that

P(X³3)=f(3)+f(4)+f(5)+f(6)+f(7)+f(8)+f(9)+f(10)+f(11)+f(12)

which can be written more simply as:

 P(X³3)=1-P(X<3)=1-P(X£2)=1-[f(0)+f(1)+f(2)]

and for a Binomial random variable we have that

We have been given that the number of trials or sampled boxes of cereal (n) is 12, and that the probability of a success (a five-point ticket) or p is 0.20. We can use this information to now address the first question. Using the Binomial Probability Distribution Function we have that:

,

,

and

.

so the probability that at least three of the twelve selected boxes will contain a five-point ticket is

P(X³3)=1-P(X£2)=1-[f(0)+f(1)+f(2)]=1-(0.0686+0.2062+0.2835)=0.4417

We can answer the second question in a similar manner. Our random variable in this question is the number of the twelve selected boxes that contain a one-point ticket, which is a discrete random variable. Since we are dealing with the number of successes (boxes containing a one-point ticket) in a set number of trials (the twelve randomly selected boxes), we should consider using the Binomial Probability Distribution to answer this question. This question can be written as:

P(X=0)

where x is the number of the twelve boxes that contain a one-point ticket. We also know, for this discrete random variable, that

P(X=0)=f(0)

and for a Binomial random variable we have that

We have been given that the number of trials or sampled boxes of cereal (n) is 12, and that the probability of a success (a one-point ticket) or p is 0.20. We can use this information to now address the first question. Using the Binomial Probability Distribution Function we have that:

so the probability that none of the twelve selected boxes will contain a one-point ticket is

P(X=0)=f(0)=0.0686.

4. Each day Pillar Foods randomly selects ten of their 1250 production employees to perform quality control inspections throughout the production process. In order to cover all quality control-related tasks, at least eight of the ten randomly selected employees must be at work on that day. If the probability that any worker will call in sick on a given day is 0.05, what is the probability Pillar Foods will not have enough employees from the ten who have been randomly selected to perform their quality control tasks (i.e., that at least two of the selected workers will call in sick on a given day)?

Our random variable in this problem is the number of the ten workers who call in sick on a particular day, which is a discrete random variable. Since we are dealing with the number of successes (workers who call in sick) in a set number of trials (the ten workers), we should consider using the Binomial Probability Distribution to answer this problem. The first problem can be written as:

P(X³2)

where x is the number of the ten workers who call in sick on a particular day. We also know, for this discrete random variable, that

P(X³2)=f(2)+f(3)+f(4)+f(5)+f(6)+f(7)+f(8)+f(9)+f(10)

which can be written more simply as:

P(X³2)=1-P(X<2)=1-P(X£1)=1-f(0)+f(1)

and for a Binomial random variable we have that

We have been given that the number of trials or workers (n) is 10, and that the probability of a success (a worker calls in sick) or p is 0.05. We can use this information to now address the question. Using the Binomial Probability Distribution Function we have that:

and

so the probability that at least two workers will call in sick on a given day is

P(X³2)=1-P(X£1)=1-[f(0)+f(1)]=1-(0.5987+0.3151)=0.0861

Note that we may wish to question the assumption of independence of events in this problem (are the physical conditions of the employees related or unrelated) because of epidemics (flu, colds, etc.) which can be spread by one employee to another.

 

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