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The Los Angeles Times (June 14,1995 ) reported that the U.S. Postal Service is getting speedier, with higher overnight on-time delivery rates than in the past. The Price Waterhouse accounting firm conducted an independent audit by seeding the mail with letters and recording ontime delivery rates for these letters. Suppose that the results were as follows (these numhers are fictitions hut are compatible with summary values given in the article): $$ \begin{array}{lcc} & \begin{array}{l} \text { Number } \\ \text { of Letters } \\ \text { Mailed } \end{array} & \begin{array}{l} \text { Number of } \\ \text { Lefters Arriving } \\ \text { on Time } \end{array} \\ \hline \text { Los Angeles } & 500 & 425 \\ \text { New York } & 500 & 415 \\ \text { Washington, D.C. } & 500 & 405 \\ \text { Nationwide } & 6000 & 5220 \\ & & \\ \hline \end{array} $$ Use the given information to estimate the following probabilities: a. The probability of an on-time delivery in Los Angeles b. The probability of late delivery in Washington, D.C. c. The probability that two letters mailed in New York are both delivered on time d. The probability of on-time delivery nationwide

Short Answer

Expert verified
The probabilities are as follows: a. The probability of an on-time delivery in Los Angeles is 0.85. b. The probability of late delivery in Washington, D.C. is 0.19. c. The probability that two letters mailed in New York are both delivered on time is approximately 0.68. d. The probability of on-time delivery nationwide is approximately 0.87.

Step by step solution

01

Calculate the probability of an on-time delivery in Los Angeles

From the table, we can see that there were 500 letters mailed in Los Angeles, of which 425 arrived on time. Hence the probability of on-time delivery is calculated as \[ P(\text{on time in LA}) = \frac{425}{500} \]
02

Calculate the probability of late delivery in Washington D.C.

From the table, there were 500 letters mailed in Washington D.C., of which 405 arrived on time. This means that 95 letters arrived late. Hence the probability of late delivery is calculated as \[ P(\text{late in DC}) = \frac{95}{500} \]
03

Calculate the probability that two letters mailed in New York are both delivered on time.

From the table, we find that there were 500 letters mailed in New York, and 415 arrived on time. If we assume that the delivery of one letter does not affect the delivery of the second one, then the events are independent, and their joint probability is the product of their individual probabilities. Thus we calculate \[ P(\text{both on time in NY}) = \left(\frac{415}{500}\right) * \left(\frac{415}{500}\right) \]
04

Calculate the probability of on-time delivery nationwide.

From the table, there were 6000 letters mailed nationwide, and 5220 arrived on time. Thus, the probability of on-time delivery nationwide can be calculated as \[ P(\text{on time nationwide}) = \frac{5220}{6000} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

On-Time Delivery Probability
The concept of on-time delivery probability is central to logistics and supply chain management, assessing the reliability of a delivery service. To calculate the on-time delivery probability, one divides the number of on-time delivered items by the total number of items delivered. In our example, for Los Angeles, the on-time delivery probability is:
\[ P(\text{on time in LA}) = \frac{425}{500} \]
This tells us that for every letter sent in Los Angeles, there's an 85% chance that it will arrive as intended on time. Companies and consumers alike rely heavily on such statistics for planning and expectations setting.
Independent Events in Probability
In probability theory, independent events are those whose occurrence or non-occurrence does not affect the likelihood of the other event(s) happening. For instance, the delivery of one letter does not influence the delivery timing of a second one. This is an assumption we make when calculating joint probabilities. Mathematically, this means if 'A' and 'B' are two independent events, then the probability of both events occurring is the product of their individual probabilities:
\[ P(A \text{ and } B) = P(A) \times P(B) \]
Regarding our exercise, applying the concept of independent events, the probability of both letters mailed in New York being on time is the square of the probability of one letter being on time since the same probability applies to both occurrences.
Late Delivery Probability
Late delivery probability is just as critical as its on-time counterpart, especially when it pertains to time-sensitive communications. To calculate it, you take the total number of late deliveries and divide it by the total number of items delivered. Using the Washington, D.C. data, we calculate the probability of a letter being late as:
\[ P(\text{late in DC}) = \frac{95}{500} \]
With this, businesses and individuals can manage expectations and plan accordingly. Late delivery probability provides insight into the reliability and consistency of delivery services, which can be crucial for maintaining customer satisfaction and operational efficiency.
Statistical Data Analysis
Statistical data analysis is a powerful tool used to interpret, describe, and infer conclusions from data sets. It involves collecting, analyzing, and summarizing data to uncover patterns and trends. This is exactly what Price Waterhouse did for the U.S. Postal Service, and what we do when calculating on-time delivery probabilities. To analyze the nationwide on-time delivery performance, we look at the empirical data and calculate the probability, providing a granular view of performance and helping in strategic decision making:
\[ P(\text{on time nationwide}) = \frac{5220}{6000} \]
Statistical analysis in business contexts helps in projecting delivery performance, customer satisfaction, and setting realistic operational benchmarks.

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Most popular questions from this chapter

The article "Anxiety Increases for Airline Passengers After Plane Crash" (San Luis Obispo Tribune, November 13,2001 ) reported that air passengers have a 1 in 11 million chance of dying in an airplane crash. This probability was then interpreted as "You could fly every day for 26,000 years before your number was up." Comment on why this probability interprctation is mislcading.

The following case study was reported in the article "Parking Tickets and Missing Women," which appeared in an early edition of the book Statistics: A Guide to the \(U n\) known. In a Swedish trial on a charge of overtime parking, a police officer testified that he had noted the position of the two air valves on the tires of a parked car: To the closest hour, one was at the one o'clock position and the other was at the six o'clock position. After the allowable time for parking in that zone had passed, the policeman returned, noted that the valves were in the same position, and ticketed the car. The owner of the car claimed that he had left the parking place in time and had returned later. The valves just happened by chance to be in the same positions. An "expert" witness computed the probability of this occurring as \((1 / 12)(1 / 12)=1 / 44\). a. What reasoning did the expert use to arrive at the probability of \(1 / 44\) ? b. Can you spot the error in the reasoning that leads to the stated probability of \(1 / 44\) ? What effect does this error have on the probability of occurrence? Do you think that \(1 / 44\) is larger or smaller than the correct probability of occurrence?

In a school machine shop, \(60 \%\) of all machine breakdowns occur on lathes and \(15 \%\) occur on drill presses. Let \(E\) denote the event that the next machine breakdown is on a lathe, and let \(F\) denote the event that a drill press is the next machine to break down. With \(P(E)=.60\) and \(P(F)=.15\), calculate: a. \(P\left(E^{C}\right)\) b. \(P(E \cup F)\) c. \(P\left(E^{C} \cap F^{C}\right)\)

The newspaper article "Folic Acid Might Reduce Risk of Down Syndrome" (USA Today, September 29 , 1999) makes the following statement: "Older women are at a greater risk of giving birth to a baby with Down Syndrome than are younger women. But younger women are more fertile, so most children with Down Syndrome are born to mothers under \(30 .\) " Let \(D=\) event that a randomly selected baby is born with Down Syndrome and \(Y=\) event that a randomly selected baby is born to a young mother (under age 30 ). For each of the following probability statements, indicate whether the statement is consistent with the quote from the article, and if not, explain why not. a. \(P(D \mid Y)=.001, \quad P\left(D \mid Y^{C}\right)=.004, \quad P(Y)=.7\) b. \(P(D \mid Y)=.001, \quad P\left(D \mid Y^{C}\right)=.001, \quad P(Y)=.7\) c. \(P(D \mid Y)=.004, \quad P\left(D \mid Y^{C}\right)=.004, \quad P(Y)=.7\) d. \(P(D \mid Y)=.001, \quad P\left(D \mid Y^{C}\right)=.004, \quad P(Y)=.4\) e. \(P(D \mid Y)=.001, \quad P\left(D \mid Y^{C}\right)=.001, \quad P(Y)=.4\) f. \(P(D \mid Y)=.004, \quad P\left(D \mid Y^{C}\right)=.004, \quad P(Y)=.4\)

A transmitter is sending a message using a binary code, namely, a sequence of 0's and 1's. Each transmitted bit \((0\) or 1\()\) must pass through three relays to reach the receiver. At each relay, the probability is \(.20\) that the bit sent on is different from the bit received (a reversal). Assume that the relays operate independently of one another: transmitter \(\rightarrow\) relay \(1 \rightarrow\) relay \(2 \rightarrow\) relay \(3 \rightarrow\) receiver a. If a 1 is sent from the transmitter, what is the probability that a 1 is sent on by all three relays? b. If a 1 is sent from the transmitter, what is the probability that a 1 is received by the receiver? (Hint: The eight experimental outcomes can be displayed on a tree diagram with three generations of branches, one generation for each relay.)

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