QA 233 (Basic Business Statistics)

Solutions to Practice Problem Set VII

QCTV is a cable television provider for the greater Little Rock metropolitan area. The company receives an average of 2 non-weather related service calls per hour during regular business hours. Twenty five percent of all service calls require replacement of the cable box (converter), and the calls are distributed evenly across the four suburbs (East Little Rock, Centerville, Edgewood, and Mahomes). The distance traveled for a non-weather related service call is evenly distributed between 0.0 and 7.5 miles.

1. Suppose that QCTV can handle no more than twenty non-weather related service calls during regular business hours (9:00 a.m. to 5:00 p.m.). What is the probability that QCTV will receive more non-weather related service calls than they can handle today?

Our random variable in this problem is the number of non-weather related service calls during regular business hours (9:00 a.m. to 5:00 p.m.), 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 this question. The question can be written as:

P(x>20)

Which can be rewritten as

P(x>20)=P(x³21)=1-P(x£20)

we also know, for a discrete random variable, that

P(X=0)=f(0)

so we can again rewrite the problem as

1-P(x£20)=1-[f(0)+f(1)+¼ +f(20)]

And for a Poison random variable we have that

We have been given that the mean number of non-weather related service calls is 2.0 per hour during regular business hours, or 16.0 during our normal business day (9:00 a.m. to 5:00 p.m.). In other words, we expect 16 non-weather related service calls is 2.0 per hour during regular business hours. Now we can address the question. Using the Poisson Probability Distribution Function we have that:

i.e., the probability that we will receive 0 non-weather related service calls is 2.0 per hour during regular business hours is 0.000000113. We can perform a similar calculation for the probability of 1 non-weather related service call per hour during regular business hours through 20 non-weather related service calls per hour during regular business hours. We then sum these 21 probabilities to find the overall probability that we will have no more than 20 non-weather related service call per hour during regular business hours. If we subtract this result from one, we then have the probability of at least 21 non-weather related service calls per hour during regular business hours (i.e., the probability QCTV will receive more non-weather related service calls than they can handle today. The final answer is

P(x>20)=P(x³21)=1-P(x£20)=1-0.8681=0.1319

Of course, we could find the same result using the cumulative option for Excel’s Poisson paste function.

2. Suppose that the QCTV switchboard operator takes his one-hour lunch break immediately after he takes a non-weather related service call at 12:05 p.m. What is the probability that he will not miss a non-weather related service call while he is at lunch? What is the probability that he will miss at least two non-weather related service calls while he is at lunch?

Our random variable in this problem is the amount of time (in minutes or in hours) between non-weather related service calls during regular business hours (9:00 a.m. to 5:00 p.m.), which is a continuous random variable. Since we are dealing with the amount of time between a Poisson rare event (we receive a non-weather related service call) occurs, we should consider using the Exponential Probability Distribution to answer these questions. The first question can be written as:

P(x³60) if x is in minutes

or

P(x³1) if x is in hours

which can be found by using

Now we have a mean time between non-weather related service calls of

or

so our probability is

 (using minutes)

or

 (using hours)

Note that this problem could also be solved using the Poisson distribution. Our random variable could be considered to be the number of non-weather related service calls during a one-hour segment (12:05 p.m. to 1:05 p.m.) of regular business hours (9:00 a.m. to 5:00 p.m.), which is a discrete random variable. Since we are dealing with the number of occurrences of a relatively rare event (non-weather related service calls) during a one-hour segment (12:05 p.m. to 1:05 p.m.) of regular business hours, we can consider using the Poisson Probability Distribution to answer the first question. The first question can be written as:

P(x=0)

where x is the number of non-weather related service calls during a one-hour segment (12:05 p.m. to 1:05 p.m.) of regular business hours (9:00 a.m. to 5:00 p.m.). 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 non-weather related service calls received per hour is 2.0. Since our question deals with the number of non-weather related service calls received in an hour, we can proceed directly to the solution. Using the Poisson Probability Distribution Function we have that:

i.e., the probability that QCTV will receive no non-weather related service calls received during a one-hour period is 0.1353.

The second part of this problem definitely involves the Poisson distribution. Our random variable is the number of non-weather related service calls during a one-hour segment (12:05 p.m. to 1:05 p.m.) of regular business hours (9:00 a.m. to 5:00 p.m.), which is a discrete random variable. Since we are dealing with the number of occurrences of a relatively rare event (non-weather related service calls) during a one-hour segment (12:05 p.m. to 1:05 p.m.) of regular business hours, we can consider using the Poisson Probability Distribution to answer the second question. The second question can be written as:

P(x³2)

where x is the number of non-weather related service calls during a one-hour segment (12:05 p.m. to 1:05 p.m.) of regular business hours (9:00 a.m. to 5:00 p.m.). This can be rewritten as

P(x³2)=1-[P(x=0)+P(x=1)]

We also know, for a discrete random variable, that

P(X=0)=f(0)

so we again rewritten the question as

1-[P(x=0)+P(x=1)]=1-[f(0)+f(1)]

And for a Poison random variable we have that

We have been given that the mean number of non-weather related service calls received per hour is 2.0. Since our question deals with the number of non-weather related service calls received in an hour, we can proceed directly to the solution. Using the Poisson Probability Distribution Function we have that:

i.e., the probability that QCTV will receive no non-weather related service calls received during a one-hour period is 0.1353. Additionally, we have that

i.e., the probability that QCTV will receive no non-weather related service calls received during a one-hour period is 0.1353. Thus, the probability that the QCTV switchboard operator will miss at least two non-weather related service calls while he is at lunch is

P(x³2)=1-[P(x=0)+P(x=1)]=1-[f(0)+f(1)]1-(0.1353+0.2707)=0.5940

3. Suppose that one of the service technicians is scheduled to make 8 service calls today, and that he has five cable boxes on his service truck. If he will not be able to return to the QCTV offices until the end of his day, what is the probability that he will not have enough cable boxes today?

Our random variable in this problem is the number of his eight service calls on which the technician will use a cable box today, which is a discrete random variable. Since we are dealing with the number of successes (sales calls on which a cable box is used) in a given number of trials (sales calls), we should consider using the Binomial Probability Distribution to answer this question. The question can be written as:

P(x>5)

where x is the number of the eight service calls on which a cable box will be used. This can be rewritten as

P(x>5)=P(x³6)

We also know, for a discrete random variable, that

P(X=0)=f(0)

So again, we rewrite the question as

P(x³6)=f(6)+f(7)+f(8)

And for a Binomial random variable we have that

We know that we have n=eight trials (service calls), and the probability of success (use of a cable box on a service call) has been established to be p=0.25. Using the Binomial Probability Distribution Function we have that:

,

,

and

So we have that

P(x³6)=f(6)+f(7)+f(8)=0.0038+0.0004+0.00002=0.00422

i.e., the probability that the service technician will not have enough cable boxes to last throughout the day is 0.00422.

4. What is the probability that the next non-weather related service call will not come from the suburb of Mahomes?

Our random variable in this problem is the suburb from which a non-weather related service call is placed, which is a discrete random variable. Since we are dealing with four equally likely outcomes, we should consider using the Discrete Uniform Probability Distribution to answer this question. With four equally likely suburbs, the solution is

P(Mahomes=0.25)

5. What is the probability that each of the next three non-weather related service calls will require a replacement cable box?

Our random variable in this problem is the number of the next three service calls on which the technician will use a cable box today, which is a discrete random variable. Since we are dealing with the number of successes (sales calls on which a cable box is used) in a given number of trials (sales calls), we should consider using the Binomial Probability Distribution to answer this question. The question can be written as:

P(x=3)

where x is the number of the next three service calls on which a cable box will be used. We know, for a discrete random variable, that

P(X=3)=f(3)

And for a Binomial random variable we have that

We know that we have n=three trials (service calls), and the probability of success (use of a cable box on a service call) has been established to be p=0.25. Using the Binomial Probability Distribution Function we have that:

So we have that

P(x=6)=f(3)=0.015625

i.e., the probability that the next three service calls will all require cable boxes is 0.015625.

6. What is the probability that a truck will have to travel more than 6.3 miles to the next non-weather related service call? What is the probability that a truck will have to travel between 3.0 and 5.0 miles to the next non-weather related service call? What is the probability that a truck will have to travel between 6.5 and 10.0 miles to the next non-weather related service call?

Our random variable in this problem is the distance traveled to the next non-weather related service call, which is a continuous random variable. Since all feasible outcomes (everything from 0.0 to 7.5 miles) are equally likely, we should consider using the (Continuous) Uniform Probability Distribution to answer these questions. The first question can be written as:

P(3£x£5)

where x is the distance traveled to the next non-weather related service call,. We know, for a (continuous) uniform random variable, that

  And  a=minimum feasible value of x,

            b=maximum feasible value of x.

Since, for our first problem (c=3 and d=5), we have that a£c£d£b, the probability is

So the probability that we must travel between 3 and 5 miles to the next service is 0.2667.

For our second problem,

P(6.5£x£10.0)

c=6.5 and d=10.0, so we don’t have that a£c£d£b. We can, however, rewrite this problem as

P(6.5£x£10.0)=P(6.5£x£7.5)+P(7.5£x£10.0)

Recognizing that the distance traveled cannot exceed 7.5, we simplify the problem to

P(6.5£x£10.0)=P(6.5£x£7.5)

and so the probability is

so the probability that we must travel between 6.5 and 7.5 miles to the next service is 0.1333.

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