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Chapter 5: Problem 19

The National Center for Health Statistics (www.cdc .gov/nchs/data/nvsr/nvsr64/nvsr64_12.pdf, retrieved April 25,2017 ) gave the following information on births in the United States in 2014 : $$ \begin{array}{|lr|} \hline \text { Type of Birth } & \text { Number of Births } \\ \hline \text { Single birth } & 3,848,214 \\ \text { Twins } & 135,336 \\ \text { Triplets } & 4,233 \\ \text { Quadruplets } & 246 \\ \text { Quintuplets or higher } & 47 \\ \hline \end{array} $$ Use this information to estimate the probability that a randomly selected pregnant woman who gave birth in 2014 a. delivered twins b. delivered quadruplets c. gave birth to more than a single child

Short Answer

Expert verified
The estimated probabilities for a randomly selected pregnant woman who gave birth in 2014 are as follows: a. Probability of delivering twins: \(\approx 0.0339\), b. Probability of delivering quadruplets: \(\approx 0.0000616\), and c. Probability of giving birth to more than a single child: \(\approx 0.0351\).

Step by step solution

01

Add up the total number of births in 2014

To find the total number of births in 2014, we need to add up the number of each type of birth: Total births = Single births + Twins + Triplets + Quadruplets + Quintuplets or higher Total births = 3,848,214 + 135,336 + 4,233 + 246 + 47 = 3,988,076 births
02

Estimate the probability of delivering twins

We will now calculate the probability of a woman giving birth to twins by dividing the number of twins by the total number of births: Probability(Twins) = \(\frac{\text{Number of Twins births}}{\text{Total number of births}}\) = \(\frac{135,336}{3,988,076} \approx 0.0339\)
03

Estimate the probability of delivering quadruplets

Similarly, we can calculate the probability of a woman giving birth to quadruplets by dividing the number of quadruplets by the total number of births: Probability(Quadruplets) = \(\frac{\text{Number of Quadruplets births}}{\text{Total number of births}}\) = \(\frac{246}{3,988,076} \approx 0.0000616\)
04

Estimate the probability of giving birth to more than a single child

To calculate the probability of giving birth to more than a single child, we need to add up the number of all births that are not single births and then divide the sum by the total number of births: Number of multiple births (Twins + Triplets + Quadruplets + Quintuplets or higher) = 135,336 + 4,233 + 246 + 47 = 139,862 Probability(More than one child) = \(\frac{\text{Number of multiple births}}{\text{Total number of births}}\) = \(\frac{139,862}{3,988,076} \approx 0.0351\) To summarize, the estimated probabilities are: a. Probability of delivering twins: \(\approx 0.0339\) b. Probability of delivering quadruplets: \(\approx 0.0000616\) c. Probability of giving birth to more than a single child: \(\approx 0.0351\)

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Key Concepts

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

Birth Statistics
Understanding birth statistics is crucial for grasping broader patterns in health, society, and even policy-making. In our context, birth statistics refer to the numerical data related to the occurrence of births, categorized by types, such as single births, twins, triplets, and so on. This data can reveal trends such as the average family size, the rate of multiple births, and possible factors influencing these trends such as genetics, maternal age, and fertility treatments.

Interpreting this data requires a clear understanding of the categories reported and the context in which the data was collected. For instance, the statistics indicating the number of twins or triplets in a specific year can help us identify the probability of such occurrences in the population, as seen in the exercise.
Probability Calculation
Probability calculation is a fundamental concept in statistics that aids in determining how likely it is that an event will occur. It's expressed as a number between 0 and 1, where 0 indicates impossibility and 1 indicates certainty. To calculate the probability of an event, one can use the formula:
\[ P(E) = \frac{\text{Number of favorable outcomes}}{\text{Total number of possible outcomes}} \]
For instance, based on birth statistics, if you want to find out the likelihood of a woman delivering twins, you'd divide the number of twin births by the total number of births recorded.

Applying this formula to complex scenarios requires a meticulous approach to ensure each step, from data collection to calculation, is accurately handled. Such mathematical rigor ensures that the probabilities derived offer a reliable picture of the event occurring.
Data Interpretation
Data interpretation refers to the process of making sense of numerical data to discern patterns, make predictions, and inform decisions. It is the practice of critically analyzing data to deduce information that isn't immediately obvious. Interpreting data correctly is essential because it influences conclusions and can have significant implications in various fields such as medicine, economics, and public health.

In statistics, representation of data in tables or charts helps to organize information effectively, enabling easier interpretation and comparison of different data points. For instance, in our exercise, the data is neatly tabulated, making it easier to see the number of multiple birth types and to perform probability calculations. This clarity aids in understanding the likelihood of different types of births and supports conclusions drawn from the observed phenomena. Moreover, careful data interpretation allows for more informed predictions about future trends and potential anomalies.

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Most popular questions from this chapter

Lyme disease is transmitted by infected ticks. Several tests are available for people with symptoms of Lyme disease. One of these tests is the EIA/IFA test. The paper "Lyme Disease Testing by Large Commercial Laboratories in the United States" (Clinical Infectious Disease [2014]: \(676-681\) ) found that \(11.4 \%\) of those tested actually had Lyme disease. Consider the following events: \(+\) represents a positive result on the blood test \- represents a negative result on the blood test \(L\) represents the event that the patient actually has Lyme disease \(L^{C}\) represents the event that the patient actually does not have Lyme disease The following probabilities are based on percentages given in the paper: $$ \begin{array}{r} P(L)=0.114 \\ P\left(L^{C}\right)=0.886 \end{array} $$ $$ \begin{array}{c} P(+\mid L)=0.933 \\ P(-\mid L)=0.067 \\ P\left(+\mid L^{C}\right)=0.039 \\ P\left(-\mid L^{C}\right)=0.961 \end{array} $$ a. For each of the given probabilities, write a sentence giving an interpretation of the probability in the context of this problem. b. Use the given probabilities to construct a hypothetical 1000 table with columns corresponding to whether or not a person has Lyme disease and rows corresponding to whether the blood test is positive or negative. c. Notice the form of the known conditional probabilities; for example, \(P(+\mid L)\) is the probability of a positive test given that a person selected at random from the population actually has Lyme disease. Of more interest is the probability that a person has Lyme disease, given that the test result is positive. Use information from the table constructed in Part (b) to calculate this probability.

Suppose that an individual is randomly selected from the population of all adult males living in the United States. Let \(A\) be the event that the selected individual is over 6 feet in height, and let \(B\) be the event that the selected individual is a professional basketball player. Which do you think is greater, \(P(A \mid B)\) or \(P(B \mid A) ?\) Why?

According to The Chronicle for Higher Education Almanac (2016), there were 1,003,329 Associate degrees awarded by U.S. community colleges in the \(2013-2014\) academic year. A total of 613,034 of these degrees were awarded to women. a. If a person who received an Associate degree in \(2013-\) 2014 is selected at random, what is the probability that the selected person will be female? b. What is the probability that the selected person will be male?

Six people hope to be selected as a contestant on a TV game show. Two of these people are younger than 25 years old. Two of these six will be chosen at random to be on the show. a. What is the sample space for the chance experiment of selecting two of these people at random? (Hint: You can think of the people as being labeled \(\mathrm{A}, \mathrm{B}, \mathrm{C}, \mathrm{D}, \mathrm{E},\) and \(\mathrm{F}\). One possible selection of two people is \(\mathrm{A}\) and \(\mathrm{B}\). There are 14 other possible selections to consider.) b. Are the outcomes in the sample space equally likely? c. What is the probability that both the chosen contestants are younger than \(25 ?\) d. What is the probability that both the chosen contestants are not younger than \(25 ?\) e. What is the probability that one is younger than 25 and the other is not?

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 of the possible 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?
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