/*! 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} Q121SE The article 鈥淭hree Sisters Giv... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Q121SE (page 196)

The article 鈥淭hree Sisters Give Birth on the Same Day鈥� (Chance, Spring \({\rm{2001, 23 - 25}}\)) used the fact that three Utah sisters had all given birth on March \({\rm{11, 1998}}\) as a basis for posing some interesting questions regarding birth coincidences.

a. Disregarding leap year and assuming that the other\({\rm{365}}\) days are equally likely, what is the probability that three randomly selected births all occur on March \({\rm{11}}\)? Be sure to indicate what, if any, extra assumptions you are making.

b. With the assumptions used in part (a), what is the probability that three randomly selected births all occur on the same day?

c. The author suggested that, based on extensive data, the length of gestation (time between conception and birth) could be modeled as having a normal distribution with mean value \({\rm{280}}\) days and standard deviation \(\,{\rm{19}}{\rm{.88}}\) days. The due dates for the three Utah sisters were March\({\rm{15}}\), April \({\rm{1}}\), and April \({\rm{4}}\), respectively. Assuming that all three due dates are at the mean of the distribution, what is the probability that all births occurred on March \({\rm{11}}\)? (Hint: The deviation of birth date from due date is normally distributed with mean \({\rm{0}}\).)

d. Explain how you would use the information in part (c) to calculate the probability of a common birth date.

Short Answer

Expert verified

(a) Probability that three randomly selected births all occur on march \({\rm{11}}\) is

\({\rm{P(}}\)March \({\rm{11}}\) thrice\({\rm{) = }}\frac{{\rm{1}}}{{{\rm{36}}{{\rm{5}}^{\rm{3}}}}}\).

(b)Probability that three randomly selected births all occur on the same day is

\({\rm{P(}}\)same day thrice\({\rm{) = }}\frac{{\rm{1}}}{{{\rm{36}}{{\rm{5}}^{\rm{2}}}}}\).

(c) Probability that all births occurred on March \({\rm{11}}\) is

\({\rm{P(}}\)All births on March \({\rm{11}}\)\({\rm{) = 2}}{\rm{.167368 \times 1}}{{\rm{0}}^{{\rm{ - 6}}}}{\rm{ = 0}}{\rm{.000002167368}}\).

(d) the average birth date is within \({\rm{39}}{\rm{.76}}\) days of \({\rm{280}}\) days.

Step by step solution

01

Definition of Probability

Probability refers to the likelihood of a random event's outcome. This word refers to determining the likelihood of a given occurrence occurring.

02

Calculating probability for three selected births occurs on march \({\rm{11}}\)

Independent event multiplication rule:

\({\rm{P(A\;and\;B) = P(A) \times P(B)}}\)

(a) Each of the \({\rm{365}}\) days has an equal chance of occurring, hence March \({\rm{11}}\) is one of the \({\rm{365}}\) equally likely days.

The number of positive outcomes divided by the total number of possible outcomes is the probability:

\({\rm{P(\;March 11\;) = }}\frac{{{\rm{\# \;of favorable outcomes\;}}}}{{{\rm{\# \;of possible outcomes\;}}}}{\rm{ = }}\frac{{\rm{1}}}{{{\rm{365}}}}\)

For independent events, use the multiplication rule:

\({\rm{P(\;March\;11\;thrice\;) = P(\;March\;11) \times P(\;March\;11) \times P(\;March\;11)}}\)

\({\rm{ = }}\frac{{\rm{1}}}{{{\rm{365}}}}\frac{{\rm{1}}}{{{\rm{365}}}}\frac{{\rm{1}}}{{{\rm{365}}}}{\rm{ = }}\frac{{\rm{1}}}{{{\rm{36}}{{\rm{5}}^{\rm{3}}}}}{\rm{ = }}\frac{{\rm{1}}}{{{\rm{48627125}}}} \approx {\rm{2}}{\rm{.0565 \times 1}}{{\rm{0}}^{{\rm{ - 8}}}}\)

03

Calculating probability for three selected births occurs on same day

(b) Because there are \({\rm{365}}\)days in a year, there are \({\rm{365}}\)alternative dates for giving birth:

\({\rm{P(\;same day thrice\;) = 365P(\;March\;11\;thrice\;) = }}\frac{{{\rm{365}}}}{{{\rm{36}}{{\rm{5}}^{\rm{3}}}}}{\rm{ = }}\frac{{\rm{1}}}{{{\rm{36}}{{\rm{5}}^{\rm{2}}}}}\)

\({\rm{ = }}\frac{{\rm{1}}}{{{\rm{133225}}}} \approx {\rm{7}}{\rm{.5061 \times 1}}{{\rm{0}}^{{\rm{ - 6}}}}\)

04

Calculating probability for all births occurs on march \({\rm{11}}\)

(c) Given: The birth date deviation follows a normal distribution.

\({\rm{\mu = 0}}\)

\({\rm{\sigma = 19}}{\rm{.88\;days\;}}\)

The \({\rm{1}}{{\rm{5}}^{{\rm{th}}}}\) of March is four days following the \({\rm{1}}{{\rm{1}}^{{\rm{th}}}}\) of March. The \({{\rm{1}}^{{\rm{st}}}}\) of April is \({\rm{21}}\) days after the \({\rm{1}}{{\rm{1}}^{{\rm{th}}}}\) of March. March \({\rm{11}}\) is \({\rm{24}}\) days after April \({\rm{4}}\).

The standardized score is calculated by dividing the value \({\rm{x}}\) by the mean and then by the standard deviation.

\({\rm{z = }}\frac{{{\rm{x - \mu }}}}{{\rm{\sigma }}}{\rm{ = }}\frac{{{\rm{3}}{\rm{.5 - 0}}}}{{{\rm{19}}{\rm{.88}}}} \approx {\rm{0}}{\rm{.18}}\)

\({\rm{z = }}\frac{{{\rm{x - \mu }}}}{{\rm{\sigma }}}{\rm{ = }}\frac{{{\rm{4}}{\rm{.5 - 0}}}}{{{\rm{19}}{\rm{.88}}}} \approx {\rm{0}}{\rm{.23}}\)

\({\rm{z = }}\frac{{{\rm{x - \mu }}}}{{\rm{\sigma }}}{\rm{ = }}\frac{{{\rm{20}}{\rm{.5 - 0}}}}{{{\rm{19}}{\rm{.88}}}} \approx {\rm{1}}{\rm{.03}}\)

\({\rm{z = }}\frac{{{\rm{x - \mu }}}}{{\rm{\sigma }}}{\rm{ = }}\frac{{{\rm{21}}{\rm{.5 - 0}}}}{{{\rm{19}}{\rm{.88}}}} \approx {\rm{1}}{\rm{.08}}\)

\({\rm{z = }}\frac{{{\rm{x - \mu }}}}{{\rm{\sigma }}}{\rm{ = }}\frac{{{\rm{23}}{\rm{.5 - 0}}}}{{{\rm{19}}{\rm{.88}}}} \approx {\rm{1}}{\rm{.18}}\)

\({\rm{z = }}\frac{{{\rm{x - \mu }}}}{{\rm{\sigma }}}{\rm{ = }}\frac{{{\rm{24}}{\rm{.5 - 0}}}}{{{\rm{19}}{\rm{.88}}}} \approx {\rm{1}}{\rm{.23}}\)

Using the normal probability table in the appendix (which provides the probabilities to the left of the z-scores), calculate the relevant probability:

\(\begin{aligned}{{}{}}{{\rm{P(X = 4) = P(0}}{\rm{.18 < Z < 0}}{\rm{.23) = P(Z < 0}}{\rm{.23) - P(Z < 0}}{\rm{.18) = 0}}{\rm{.5910 - 0}}{\rm{.5714 = 0}}{\rm{.0196}}}\\{{\rm{P(X = 21) = P(1}}{\rm{.03 < Z < 1}}{\rm{.08) = P(Z < 1}}{\rm{.08) - P(Z < 1}}{\rm{.03) = 0}}{\rm{.8599 - 0}}{\rm{.8485 = 0}}{\rm{.0114}}}\\{{\rm{P(X = 24) = P(1}}{\rm{.18 < Z < 1}}{\rm{.23) = P(Z < 1}}{\rm{.23) - P(Z < 1}}{\rm{.18) = 0}}{\rm{.8907 - 0}}{\rm{.8810 = 0}}{\rm{.0097}}}\end{aligned}\)

05

Applying multiplication rule

Apply the following multiplication rule:

\({\rm{P(\;All births on March 11\;) = P(X = 4) \times P(X = 21) \times P(X = 24) = 0}}{\rm{.0196 \times 0}}{\rm{.0114 \times 0}}{\rm{.0097}}\)

\({\rm{ = 2}}{\rm{.167368 \times 1}}{{\rm{0}}^{{\rm{ - 6}}}}{\rm{ = 0}}{\rm{.000002167368}}\)

06

Determining probability of a common birth date

(d) The mean of common birth days is usually within two standard deviations:

\({\rm{\mu \pm 2\sigma = 0 \pm 2(19}}{\rm{.88) = \pm 39}}{\rm{.76}}\)

As a result, the average birth date is within \({\rm{39}}{\rm{.76}}\) days of \({\rm{280}}\) days.

Note: If they mean "birth dates shared by multiple people," you'll need to figure out the odds of each person giving birth on a specific day. Then you may use the multiplication rule to figure out how likely it is that everyone will give birth on that certain day. The chance of a common birth date is calculated by adding the odds for all conceivable "in common" birth days.

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 91影视!

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

Based on an analysis of sample data, the article 鈥淧edestrians鈥� Crossing Behaviours and Safety at Unmarked Roadways in China鈥� (Accident Analysis and Prevention, \({\rm{2011: 1927 - 1936}}\) proposed the pdf \({\rm{f(x) = }}{\rm{.15}}{{\rm{e}}^{{\rm{ - }}{\rm{.15(x - 1)}}}}\) when \({\rm{x}} \ge {\rm{1}}\) as a model for the distribution of \({\rm{X = }}\)time (sec) spent at the median line.

a. What is the probability that waiting time is at most \({\rm{5}}\) sec? More than \({\rm{5}}\) sec?

b. What is the probability that waiting time is between \({\rm{2}}\) and \({\rm{5}}\) sec?

A family of pdf鈥檚 that has been used to approximate the distribution of income, city population size, and size of firms is the Pareto family. The family has two parameters, \({\rm{k}}\) and \({\rm{\theta }}\), both\({\rm{ > 0}}\), and the pdf is

\({\rm{f(x;\theta ) = \{ }}\begin{array}{*{20}{c}}{\frac{{{\rm{k}} \cdot {{\rm{\theta }}^{\rm{k}}}}}{{{{\rm{x}}^{{\rm{k + 1}}}}}}}&{{\rm{x}} \ge {\rm{\theta }}}\\{\rm{0}}&{{\rm{x < \theta }}}\end{array}\)

a. Sketch the graph of \({\rm{f(x;\theta )}}\).

b. Verify that the total area under the graph equals \({\rm{1}}\).

c. If the rv \({\rm{X}}\) has pdf \({\rm{f(x;\theta )}}\), for any fixed \({\rm{b > \theta }}\), obtain an expression for \({\rm{P(X}} \le {\rm{b)}}\).

d. For \({\rm{\theta < a < b}}\) obtain an expression for the probability \({\rm{P(a}} \le {\rm{X}} \le {\rm{b)}}\).

The defect length of a corrosion defect in a pressurized steel pipe is normally distributed with mean value \({\bf{30}}{\rm{ }}{\bf{mm}}\) and standard deviation \({\bf{7}}.{\bf{8}}{\rm{ }}{\bf{mm}}\) (suggested in the article 鈥淩eliability Evaluation of Corroding Pipelines Considering Multiple Failure Modes and Time Dependent Internal Pressure鈥� (J. of Infrastructure Systems, \({\bf{2011}}:{\rm{ }}{\bf{216}}--{\bf{224}})).\)

a. What is the probability that defect length is at most \({\bf{20}}{\rm{ }}{\bf{mm}}\)? Less than 20 mm?

b. What is the \({\bf{75th}}\) percentile of the defect length distribution鈥攖hat is, the value that separates the smallest \({\bf{75}}\% \)of all lengths from the largest \({\bf{25}}\% \)?

c. What is the \({\bf{15th}}\) percentile of the defect length distribution?

d. What values separate the middle \({\bf{80}}\% \) of the defect length distribution from the smallest \({\bf{10}}\% \)and the largest \({\bf{10}}\% \)?

The article "Error Distribution in Navigation"\({\rm{(J}}\). of the Institute of Navigation, \({\rm{1971: 429 442}}\)) suggests that the frequency distribution of positive errors (magnitudes of errors) is well approximated by an exponential distribution. Let \({\rm{X = }}\) the lateral position error (nautical miles), which can be either negative or positive. Suppose the pdf of \({\rm{X}}\) is

\(f(x) = (.1){e^{ - .2k1}} - 楼 < x < 楼\)

a. Sketch a graph of \({\rm{f(x)}}\) and verify that \({\rm{f(x)}}\) is a legitimate pdf (show that it integrates to\({\rm{1}}\)).

b. Obtain the cdf of \({\rm{X}}\) and sketch it.

c. Compute \(P(X拢0),P(X拢2),P( - 1拢X拢2)\), and the probability that an error of more than \({\rm{2}}\) miles is made.

The reaction time (in seconds) to a certain stimulus is a continuous random variable with pdf

\(f(x)= \left\{ {\begin{array}{*{20}{c}}{\frac{3}{2} \times \frac{1}{{{x^2}}}}&{1拢x拢3} \\0&{{\text{ }}otherwise{\text{ }}}\end{array}} \right.\)

a. Obtain the cdf.

b. What is the probability that reaction time is at most\({\rm{2}}{\rm{.5sec}}\)? Between \({\rm{1}}{\rm{.5}}\) and\({\rm{2}}{\rm{.5sec}}\)?

c. Compute the expected reaction time.

d. Compute the standard deviation of reaction time.

e. If an individual takes more than \({\rm{1}}{\rm{.5sec}}\) to react, a light comes on and stays on either until one further second has elapsed or until the person reacts (whichever happens first). Determine the expected amount of time that the light remains lit. (Hint: Let \({\rm{h(X) = }}\) the time that the light is on as a function of reaction time\({\rm{X}}\).)
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Calculus

Read Explanation

Decision Maths

Read Explanation

Pure Maths

Read Explanation

Mechanics Maths

Read Explanation

Discrete Mathematics

Read Explanation
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.