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Chapter 6: Problem 20

Refer to the following information on births in the United States over a given period of time: $$ \begin{array}{lr} \text { Type of Birth } & \text { Number of Births } \\ \hline \text { Single birth } & 41,500,000 \\ \text { Twins } & 500,000 \\ \text { Triplets } & 5,000 \\ \text { Quadruplets } & 100 \\ \hline \end{array} $$ Use this information to approximate the probability that a randomly selected pregnant woman who reaches full term a. Delivers twins b. Delivers quadruplets c. Gives birth to more than a single child

Short Answer

Expert verified
a. The probability of delivering twins is roughly \(1.19\% \) \n b. The probability of delivering quadruplets is roughly \(0.00024\% \)\n c. The probability of giving birth to more than a single child is around \(1.20\% \)

Step by step solution

01

Calculate Total Births

First, calculate the total number of births by adding together the number of single birth, twins, triplets and quadruplets. This can be calculated as \(41,500,000 + 500,000 +5,000 + 100 = 42,005,100 \)
02

Probability of Delivering Twins

Probability is determined as favorable outcomes divided by total outcomes. In this case, for a woman to deliver twins, the favorable outcomes will be the number of twin births and the total outcome will be the total number of births. Thus, the probability of delivering twins can be calculated as \(500,000/42,005,100 = 0.0119\) or about \(1.19\% \)
03

Probability of Delivering Quadruplets

In a similar way, to calculate the probability of delivering quadruplets, divide the number of quadruplet births (favorable outcomes) by the total outcomes (total births). The probability is then \(100/42,005,100 = 0.0000024\), or roughly \(0.00024\%\)
04

Probability of not delivering a single child

This is the probability that a mother gives birth to more than one child. So, the favorable outcomes are the total births of twins, triplets and quadruplets, and all these are divided by the total number of births. Sum of twins, triplets and quadruplets equals \(500,000 + 5,000 + 100 = 505,100\). Thus, the probability is calculated as \(505,100/42,005,100 = 0.0120\), or around \(1.20\%\)

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Key Concepts

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

Statistical Analysis
Statistical analysis is a powerful tool used to interpret and understand data. In the context of birth statistics, it allows us to analyze the frequency and probability of different types of births. To perform a basic statistical analysis, we start by gathering data. In this exercise, the data includes numbers of single births, twins, triplets, and quadruplets.
Next, we calculate the total number of births, which involves summing up all the different types of births. This gives us a comprehensive view of the dataset.
Analyzing this data involves understanding the distribution of various birth outcomes, which aids in making informed predictions and decisions based on historical data.
  • Gather data effectively.
  • Sum data for total calculations.
  • Analyze distributions.
With these steps, statistical analysis helps uncover patterns and likelihoods in birth statistics.
Birth Statistics
Birth statistics provide insights into the patterns of multiple births, like twins or triplets, in the population. By examining historical birth data, researchers and statisticians can determine the frequency of each type of birth.
This exercise shows that out of about 42 million births, over 41 million are single births, while twins, triplets, and quadruplets occur much less frequently. Birth statistics are crucial for understanding demographics and planning healthcare resources.
Data from birth statistics can inform policies and medical research, ensuring that healthcare systems can manage different birth outcomes effectively.
  • Focus on multiple birth frequencies.
  • Understand healthcare implications.
  • Plan based on demographic insights.
By studying birth statistics, we gain valuable information about population growth and development.
Probability Calculations
Probability calculations help predict the likelihood of various outcomes based on existing data. In this exercise, we are interested in the probability of a woman giving birth to twins, quadruplets, or more than one child.
The basic formula for probability is the number of favorable outcomes divided by the total number of possible outcomes. For example, the probability of delivering twins is computed by dividing the number of twin births by the total births.
Probability helps us understand and quantify the chance of multiple births, which can be beneficial for expectant mothers and healthcare providers to prepare adequately.
  • Use the probability formula effectively.
  • Calculate for various birth scenarios.
  • Prepare for potential outcomes.
Probability calculations provide a numerical perspective, helping to gauge the rarity or commonality of different birth phenomena.

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Most popular questions from this chapter

The following case study was reported in the article 鈥淧arking Tidkets and Missing Women," which appeared in an early edition of the book Statistics: A Guide to the Unknown. 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 \((1 / 12)(1 / 12)=\) \(1 / 144\). a. What reasoning did the expert use to arrive at the probability of \(1 / 144\) ? b. Can you spot the error in the reasoning that leads to the stated probability of \(1 / 144 ?\) What effect does this error have on the probability of occurrence? Do you think that \(1 / 144\) is larger or smaller than the correct probability of occurrence?

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. \(P(D \mid Y)=.001, P\left(D \mid Y^{C}\right)=.004, P(Y)=.7\) b. \(P(D \mid Y)=.001, P\left(D \mid Y^{C}\right)=.001, P(Y)=.7\) c. \(P(D \mid Y)=.004, P\left(D \mid Y^{C}\right)=.004, P(Y)=.7\) d. \(P(D \mid Y)=.001, P\left(D \mid Y^{C}\right)=.004, P(Y)=.4\) e. \(P(D \mid Y)=.001, P\left(D \mid Y^{C}\right)=.001, P(Y)=.4\) f. \(P(D \mid Y)=.004, P\left(D \mid Y^{C}\right)=.004, P(Y)=.4\)

The National Public Radio show Car Talk has a feature called "The Puzzler." Listeners are asked to send in answers to some puzzling questions-usually about cars but sometimes about probability (which, of course, must account for the incredible popularity of the program!). Suppose that for a car question, 800 answers are submitted, of which 50 are correct. a. Suppose that the hosts randomly select two answers from those submitted with replacement. Calculate the probability that both selected answers are correct. (For purposes of this problem, keep at least five digits to the right of the decimal.) b. Suppose now that the hosts select the answers at random but without replacement. Use conditional probability to evaluate the probability that both answers selected are correct. How does this probability compare to the one computed in Part (a)?

An article in the New york Times (March 2. 1994) reported that people who suffer cardiac arrest in New York City have only a 1 in 100 chance of survival. Using probability notation, an equivalent statement would be \(P(\) survival \()=.01\) for people who suffer a cardiac arrest in New York City. (The article attributed this poor survival rate to factors common in large cities: traffic congestion and the difficulty of finding victims in large buildings.) a. Give a relative frequency interpretation of the given probability. b. The research that was the basis for the New York Times article was a study of 2329 consecutive cardiac arrests in New York City. To justify the " 1 in 100 chance of survival" statement, how many of the 2329 cardiac arrest sufferers do you think survived? Explain.

A shipment of 5000 printed circuit boards contains 40 that are defective. Two boards will be chosen at random, without replacement. Consider the two events \(E_{1}=\) event that the first board selected is defective and \(E_{2}=\) event that the second board selected is defective. a. Are \(E_{1}\) and \(E_{2}\) dependent events? Explain in words. b. Let \(n o t E_{1}\) be the event that the first board selected is not defective (the event \(E_{1}^{C}\) ). What is \(P\left(\right.\) not \(E_{1}\) )? c. How do the two probabilities \(P\left(E_{2} \mid E_{1}\right)\) and \(P\left(E_{2} \mid\right.\) not \(\left.E_{1}\right)\) compare? d. Based on your answer to Part (c), would it be reasonable to view \(E_{1}\) and \(E_{2}\) as approximately independent?
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