/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 67 The article "Birth Beats Long Od... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The article "Birth Beats Long Odds for Leap Year Mom, Baby" (San Luis Obispo Tribune, March 2, 1996) reported that a leap year baby (someone born on February 29 ) became a leap year mom when she gave birth to a baby on February \(29,1996 .\) The article stated that a hospital spokesperson said that the probability of a leap year baby giving birth on her birthday was one in \(2.1\) million (approximately .00000047). a. In computing the given probability, the hospital spokesperson used the fact that a leap day occurs only once in 1461 days. Write a few sentences explaining how the hospital spokesperson computed the stated probability. b. To compute the stated probability, the hospital spokesperson had to assume that the birth was equally likely to occur on any of the 1461 days in a four- year period. Do you think that this is a reasonable assumption? Explain. c. Based on your answer to Part (b), do you think that the probability given by the hospital spokesperson is too small, about right, or too large? Explain.

Short Answer

Expert verified
The spokesperson calculated the probability by multiplying the chances of each independent event (being born and giving birth on a leap day). The assumption of equal birth likelihood across the 1461 days may not be entirely accurate due to factors influencing birth rates, but it's reasonably used for simplicity. Consequently, the provided probability of being both a leap year baby and mom could be roughly correct under these assumptions.

Step by step solution

01

Understanding the Probability Calculation

As stated in the exercise, a leap day occurs once in 1461 days (which are the total days in 4 years). Therefore, the chance of a person being born on a leap day, and then giving birth on a leap day would be \(1/1461 * 1/1461 = 1/2.134*10^6 = 0.0000004683\). This is approximately equal to \(0.00000047\), as stated by the hospital spokesperson. Hence, the spokesperson is merely multiplying the probabilities of the two independent events.
02

Evaluating the Assumption

The assumption made by the hospital spokesperson is that a birth is equally likely to occur on any of the 1461 days in a four-year period. Realistically, this might not be completely accurate since birth rates can vary due to factors such as season, weather, and cultural practices among others. However, without specific birth data to suggest otherwise and for simplicity, it can be seen as a reasonable assumption.
03

Evaluating the Provided Probability

The calculated probability already assumes that births are equally distributed throughout the year. Given our discussion in Step 2, if we consider variations in birth rates due to factors like seasonality, the probability could be slightly higher or lower. However, without specific data, we cannot make a definitive statement and thus the provided probability seems roughly correct given the used assumption.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Leap Year
Leap years are unique years occurring every four years in the Gregorian calendar. They include an extra day—February 29th—to help synchronize the calendar year with the solar year. This adjustment compensates for the fact that the Earth doesn't orbit the sun in exactly 365 days. Instead, it takes approximately 365.25 days to complete one cycle. To account for this discrepancy, an extra day is added every four years.

In leap years, the occurrence of February 29th, or "Leap Day," becomes a statistical curiosity. Because leap years recur every four years, any specific event or occurrence tied to this day—such as a birth—is considerably rarer than events happening on any regular day that appears annually.
Independent Events
In probability theory, two events are considered independent if the occurrence of one does not affect the occurrence of the other. This means the probability of both events happening is the product of their individual probabilities. Mathematically, if event A and event B are independent, the probability of both A and B occurring is denoted as:
  • \[ P(A \cap B) = P(A) \times P(B) \]
In the context of the leap year exercise, giving birth on February 29th is considered independent from being born on that date. Hence, the probability of a leap year baby giving birth on their own birthday is determined by multiplying the probability of each separate event.
Uniform Distribution
A uniform distribution assigns equal probability to all outcomes. In a more practical sense, if we say an event follows a uniform distribution, each day or instance within the defined interval is equally likely to occur. For example, when rolling a fair die, each of the six faces is equally likely to result, representing a uniform distribution. In the given exercise, the assumption made is that births occur uniformly throughout the 1461 days in a four-year cycle. Under this assumption, the chance of a birth on any specific day—like February 29th—is the same across the entire period. While this simplifies calculations, it's vital to acknowledge that real-world data, influenced by various factors, might deviate from this simplistic model.
Birth Rates
Birth rates refer to the number of births occurring within a specific population over a certain period of time. Various elements, such as cultural practices, climate, and even societal norms, can influence these rates, possibly affecting when and how frequently births occur. In examining the leap year exercise, assuming a constant birth rate across all days may not accurately reflect reality. Factors such as planned pregnancies, cultural preferences for certain birth timings, or even natural phenomena like seasonal variations could shift the frequency of births. These elements could impact the probability distribution and cause deviations from an idealized uniform distribution.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A medical research team wishes to evaluate two different treatments for a disease. Subjects are selected two at a time, and then one of the pair is assigned to each of the two treatments. The treatments are applied, and each is either a success (S) or a failure (F). The researchers keep track of the total number of successes for each treatment. They plan to continue the chance experiment until the number of successes for one treatment exceeds the number of successes for the other treatment by \(2 .\) For example, they might observe the results in the table below. The chance experiment would stop after the sixth pair, because Treatment 1 has 2 more successes than Treatment \(2 .\) The researchers would conclude that Treatment 1 is preferable to Treatment \(2 .\) Suppose that Treatment 1 has a success rate of \(.7\) (i.e., \(P(\) success \()=.7\) for Treatment 1 ) and that Treatment 2 has a success rate of \(.4\). Use simulation to estimate the probabilities in Parts (a) and (b). (Hint: Use a pair of random digits to simulate one pair of subjects. Let the first digit represent Treatment 1 and use \(1-7\) as an indication of a

Only \(0.1 \%\) of the individuals in a certain population have a particular disease (an incidence rate of .001). Of those who have the disease, \(95 \%\) test positive when a certain diagnostic test is applied. Of those who do not have the disease, \(90 \%\) test negative when the test is applied. Suppose that an individual from this population is randomly selected and given the test. a. Construct a tree diagram having two first-generation branches, for has disease and doesn't have disease, and two second-generation branches leading out from each of these, for positive test and negative test. Then enter appropriate probabilities on the four branches. b. Use the general multiplication rule to calculate \(P(\) has disease and positive test). c. Calculate \(P\) (positive test). d. Calculate \(P\) (has disease \(\mid\) positive test). Does the result surprise you? Give an intuitive explanation for the size of this probability.

A certain company sends \(40 \%\) of its overnight mail parcels by means of express mail service \(A_{1}\). Of these parcels, \(2 \%\) arrive after the guaranteed delivery time (use \(L\) to denote the event late delivery). If a record of an overnight mailing is randomly selected from the company's files, what is the probability that the parcel went by means of \(A_{1}\) and was late?

A deck of 52 cards is mixed well, and 5 cards are dealt. a. It can be shown that (disregarding the order in which the cards are dealt) there are \(2,598,960\) possible hands, of which only 1287 are hands consisting entirely of spades. What is the probability that a hand will consist entirely of spades? What is the probability that a hand will consist entirely of a single suit? b. It can be shown that exactly 63,206 hands contain only spades and clubs, with both suits represented. What is the probability that a hand consists entirely of spades and clubs with both suits represented? c. Using the result of Part (b), what is the probability that a hand contains cards from exactly two suits?

Consider a system consisting of four components, as pictured in the following diagram: Components 1 and 2 form a series subsystem, as do Components 3 and 4 . The two subsystems are connected in parallel. Suppose that \(P(1\) works \()=.9, P(2\) works \()=.9\), \(P(3\) works \()=.9\), and \(P(4\) works \()=.9\) and that the four components work independently of one another. a. The \(1-2\) subsystem works only if both components work. What is the probability of this happening? b. What is the probability that the \(1-2\) subsystem doesn't work? that the \(3-4\) subsystem doesn't work? c. The system won't work if the \(1-2\) subsystem doesn't work and if the \(3-4\) subsystem also doesn't work. What is the probability that the system won't work? that it will work? d. How would the probability of the system working change if a \(5-6\) subsystem were added in parallel with the other two subsystems? e. How would the probability that the system works change if there were three components in series in each of the two subsystems?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.