/*! 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 74 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 only about 1 in \(2.1\) million births is a leap year baby born to a leap year mom (a probability of 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 probability is computed by using the chance of being born on a leap year day. If we divide one by the total number of days in four years (1461), the result is the probability of any birth occurring on February 29. This calculation is squared to get the chance of both mother and child being born on the leap day. The assumption that every day within the four-year time frame is equally likely for birth to occur seems reasonable in a general sense, although some external factors could cause slight deviations. Thus, the given probability appears to be about right.

Step by step solution

01

Compute the Probability

The reasoning behind the probability calculation would be considering the chances of being born on February 29, which happens only once in 1461 days due to the cycle of leap years. Hence the probability for one person to be born on this day is \(1/1461\). Now, for a person born on such a day to have a child also born on February 29, the probability will be \(1/1461 * 1/1461\), which is roughly \(0.00000047\).
02

Reasoning if the Birth is Equally Likely to Occur any Day

The assumption that a birth is equally likely to occur on any of the 1461 days in a four-year period appears to be broadly reasonable. It presumes that there are no factors that materially sway births to occur on particular days over others, which is a generally consistent assumption, barring any specific cultural or personal beliefs or practices that might target certain days for childbirths.
03

Assessment of the Probability

Based on the reasoning above, it can be inferred that the probability of \(0.00000047\) might be about right. The assumption that birth can occur on any day seems reasonable, and albeit certain external factors can slightly influence this, the assumption and hence the calculated probability seem to be a fair estimation.

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 Statistics
Leap years have long fascinated statisticians and mathematicians due to their unique inclusion of an extra day, February 29th, added nearly every four years to help synchronize the calendar year with the astronomical year. The statistical odds of any event occurring on this day, such as a person being born, are reduced because of its infrequency—happening only once in 1,461 days, since a leap year comes after a cycle of three regular years.

To calculate the probability of a leap day baby becoming a leap year mother, one starts by understanding the individual probability, which is \(1/1461\). However, when considering the independent probability of two such rare events happening in sequence—such as a leap year baby then giving birth on a leap day—the calculations become \(1/1461 * 1/1461\), leading to a much smaller number, significantly illustrating the rarity of the event reported.
Birth Probability
Determining the probability of births on any given day involves understanding the conditions under which babies are born. Assuming that there is an equal chance of birth on each day might overlook factors like planned C-sections, induced labor, natural birth timing tendencies, and cultural factors affecting the timing of births.

This assumption however simplifies the complex variables involved in human birth. Under this assumption, the probability of anyone being born on February 29th is as likely as being born on any other day, making it \(1/1461\), due to the fact that February 29th appears only once in the four-year cycle. When discussing the birth probability of a leap year baby to a leap year mom, the improbability is squared, reflecting the rarity of two successive leap day births.
Statistical Assumptions
In every statistical calculation, assumptions play a crucial role in simplifying real-world complexities into calculable probabilities. The assumption that the spokesperson from the hospital made—that a birth is equally likely to happen on any day within a four-year period—is a simplification necessary to perform the calculation without accounting for the myriad factors that can affect birth timing.

While this may broadly hold true and allow for the calculation of a simplified probability, it is important to consider the potential impact of any deviation from this assumption in real life. Factors like leap year superstitions, seasonal birth rates, or medical interventions could skew the actual probability. In this light, the statistical assumption acts as a baseline, recognizing that in practice, the specific probability may be slightly more or less than the calculated estimate of \(0.00000047\).

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 new model of laptop computer can be ordered with one of three screen sizes ( 10 inches, 12 inches, 15 inches) and one of four hard drive sizes \((50 \mathrm{~GB}, 100 \mathrm{~GB}\), \(150 \mathrm{~GB}\), and \(200 \mathrm{~GB}\) ). Consider the chance experiment in which a laptop order is selected and the screen size and hard drive size are recorded. a. Display possible outcomes using a tree diagram. b. Let \(A\) be the event that the order is for a laptop with a screen size of 12 inches or smaller. Let \(B\) be the event that the order is for a laptop with a hard drive size of at most \(100 \mathrm{~GB}\). What outcomes are in \(A^{C}\) ? In \(A \cup B ?\) In \(A \cap B\) ? c. Let \(C\) denote the event that the order is for a laptop with a \(200 \mathrm{~GB}\) hard drive. Are \(A\) and \(C\) disjoint events? Are \(B\) and \(C\) disjoint?

Let \(F\) denote the event that a randomly selected registered voter in a certain city has signed a petition to recall the mayor. Also, let \(E\) denote the event that the randomly selected registered voter actually votes in the recall election. Describe the event \(E \cap F\) in words. If \(P(F)=.10\) and \(P(E \mid F)=.80\), determine \(P(E \cap F)\).

Consider the chance experiment in which both tennis racket head size and grip size are noted for a randomly selected customer at a particular store. The six possible outcomes (simple events) and their probabilities are displayed in the following table: $$ \begin{array}{llll} && {\text { Grip Size }} \\ \text { Head size } & 4_{8}^{\frac{3}{8}} \text { in. } & 4 \frac{1}{2} \text { in. } & 4 \frac{5}{8} \text { in. } \\ \hline \text { Midsize } & O_{1}(.10) & O_{2}(.20) & O_{3}(.15) \\ \text { Oversize } & O_{4}(.20) & O_{5}(.15) & O_{6}(.20) \\ \hline \end{array} $$ a. The probability that grip size is \(4 \frac{1}{2}\) inches (event \(A\) ) is \(P(A)=P\left(O_{2}\right.\) or \(\left.O_{5}\right)=.20+.15=.35\). How would you interpret this probability? b. Use the result of Part (a) to calculate the probability that grip size is not \(4 \frac{1}{2}\) inches c. What is the probability that the racket purchased has an oversize head (event \(B\) ), and how would you interpret this probability? d. What is the probability that grip size is at least \(4 \frac{1}{2}\) inches?

The article "Chances Are You Know Someone with a Tattoo, and He's Not a Sailor" (Associated Press, June 11, 2006) included results from a survey of adults aged 18 to 50 . The accompanying data are consistent with summary values given in the article. $$ \begin{array}{l|cc} & \text { At Least One Tattoo } & \text { No Tattoo } \\ \hline \text { Age 18-29 } & 18 & 32 \\ \text { Age 30-50 } & 6 & 44 \\ \hline \end{array} $$ Assuming these data are representative of adult Americans and that an adult American is selected at random, use the given information to estimate the following probabilities. a. \(P(\) tattoo \()\) b. \(P(\) tattoo \(\mid\) age \(18-29)\) c. \(P(\) tattoo \(\mid\) age \(30-50\) ) d. \(P(\) age \(18-29 \mid\) tattoo \()\)

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?

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.