/*! 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 72 Consider babies born in the "nor... [FREE SOLUTION] | 91Ó°ÊÓ

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Consider babies born in the "normal" range of \(37-43\) weeks gestational age. The paper referenced in Example \(7.27\) ("Fetal Growth Parameters and Birth Weight. Their Relationship to Neonatal Body Composition." Ultrasound in Obstetrics and Gynecology [2009]: \(441-446\) ) suggests that a normal distribution with mean \(\mu=3500\) grams and standard deviation \(\sigma=\) 600 grams is a reasonable model for the probability distribution of the continuous numerical variable \(x=\) birth weight of a randomly selected full-term baby. a. What is the probability that the birth weight of a randomly selected full- term baby exceeds \(4000 \mathrm{~g}\) ? is between 3000 and \(4000 \mathrm{~g} ?\) b. What is the probability that the birth weight of a randomly selected full- term baby is cither less than \(2000 \mathrm{~g}\) or greater than \(5000 \mathrm{~g}\) ? c. What is the probability that the birth weight of a randomly selected full- term baby exceeds 7 pounds? (Hint: \(1 \mathrm{lb}=453.59 \mathrm{~g} .\) ) d. How would you characterize the most extreme \(0.1 \%\) of all full-term baby birth weights? e. If \(x\) is a random variable with a normal distribution and \(a\) is a numerical constant \((a \neq 0)\), then \(y=a x\) also has a normal distribution. Use this formula to determine the distribution of full-term baby birth weight expressed in pounds (shape, mean, and standard deviation), and then recalculate the probability from Part (c). How does this compare to your previous answer?

Short Answer

Expert verified
a. Probability of a baby's weight exceeding 4000 grams is approximately 0.2033. The probability of a baby's weight being between 3000 and 4000 grams is approximately 0.4332. b. The combined probability of a baby's weight being less than 2000 grams or greater than 5000 grams is approximately 0.0566. c. The probability of a baby's weight exceeding 7 pounds (or approximately 3175.13 grams) is about 0.3694. d. The most extreme 0.1% of all weights might be considered those over approximately 5021.38 grams or under about 1378.62 grams. e. When converting weights from grams to pounds, the mean (mu) is approximately 7.7 pounds and the standard deviation (sigma) is approximately 1.32 pounds. The recalculated probability of a weight exceeding 7 pounds is essentially the same as the result of step 4.

Step by step solution

01

Find probability of birth weight exceeding 4000 grams

A Z-score should be calculated using the formula \(Z = (X - \mu) / \sigma = (4000 - 3500) / 600 \approx 0.833\) . Using a standard normal distribution (Z-table) to find the area to the right of this Z-score, gives the required probability.
02

Find probability of a birth weight being between 3000 and 4000 grams

First, calculate the Z-scores for 3000 and 4000 grams using \(Z = (X - \mu) / \sigma\) . Then use the standard normal distribution (Z-table) to correspond these Z-scores with the probabilities and find the difference between the two to obtain the probability of a birth weight being between 3000 and 4000 grams.
03

Compare probabilities of a birth weight less than 2000 grams or greater than 5000 grams

Similarly to steps 1 and 2, calculate the Z-scores for 2000 and 5000 grams and find the associated probabilities on the Z-table. Since these are two separate cases, add the two resulting probabilities.
04

Calculate probability of a birth weight exceeding 7 pounds

First, convert pounds into grams. 7 pounds is equivalent to approximately 3175.13 grams. Now calculate the Z-score for 3175.13, then find the proportion of the standard normal curve to the right of this Z-score – this is the required probability.
05

Determine distribution of birth weights in pounds

Apply \(\mu = E(ax) = aE(x)\) and \(\sigma = \sqrt{Var(ax)} = |a|\sqrt{Var(x)}\) to transform the birth weights from grams to pounds and recalculate the probability for a weight exceeding 7 pounds. Compare this probability with the one calculated in step 4.

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Key Concepts

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

Probability Distribution
A probability distribution is like a map that represents all possible values a random variable can take and the likelihood of each value occurring. When we're talking about birth weights of babies born in the normal range of 37-43 weeks gestational age, we're looking at a continuous probability distribution. Specifically, it's a normal distribution with a mean and a standard deviation.
The mean is the average value, in this case, 3500 grams. The standard deviation, 600 grams, measures how spread out the numbers are around the mean. For a normal distribution like this, the famous "bell curve" depicts how birth weights cluster around the mean. Most babies will have weights near 3500 grams, but the curve tails off on either side, showing fewer occurrences as we move towards the less common weights.
To find the probability of a specific birth weight range, like exceeding 4000 grams, we'd use this distribution. This involves calculating a Z-score, which tells us how many standard deviations away from the mean our value is. Knowing this helps us find probabilities associated with different weights by looking them up on a standard normal distribution table.
Z-Score Calculation
Z-score calculation is a handy way to compare a specific weight to the overall distribution. In simpler terms, the Z-score tells us how unusual or typical a particular birth weight is in comparison to the average. The formula for calculating a Z-score is:
  • \[ Z = \frac{X - \mu}{\sigma} \]

Here, \(X\) is the weight of the baby, \(\mu\) is the mean birth weight (3500 grams in our problem), and \(\sigma\) is the standard deviation (600 grams here).
Let's say we want to find out how common it is for a baby to weigh more than 4000 grams. We plug these values into the formula, and the result is our Z-score. A Z-score of 0 would mean the weight is exactly at the mean. Positive or negative Z-scores tell us how far from the mean a particular weight is, in terms of standard deviations. And by checking this score against a Z-table, we can figure out the probability or proportion of birth weights that are above, below, or between certain values.
Gestational Age
Gestational age refers to the length of time a baby spends developing in the womb, usually measured in weeks. Full-term gestation typically spans from 37 to 43 weeks. This time frame is crucial because the baby's development during this period can significantly impact their birth weight. Generally, a longer gestational age allows for more growth, potentially resulting in a higher birth weight.
However, even within this 'normal' gestational age range, there's variability. That's why statistics like the mean and standard deviation are important when talking about birth weights. They help us understand and predict the typical birth weight while acknowledging that each baby's development is unique. Understanding gestational age is essential for expecting parents, as it offers insight into what might be a healthy birth weight for their baby.
Birth Weight Analysis
Birth weight analysis involves examining the distribution and probabilities of various birth weights to understand patterns and predict outcomes. This type of analysis can inform healthcare decisions and parental expectations.
Looking at a normal distribution of birth weights, we examine extreme observations and ordinary occurrences. For instance, calculating the likelihood of a birth weight less than 2000 grams or greater than 5000 grams allows us to assess risks associated with significantly underweight or overweight newborns.
These probabilities hold clinical importance. Healthcare professionals use such analyses to determine if interventions might be needed to support the health of the mother and child. Furthermore, birth weight can serve as an indicator of a baby's immediate health needs or potential future health issues. Understanding both common and rare birth weights through rigorous analysis helps in crafting better healthcare strategies and ensuring the well-being of newborns.

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Most popular questions from this chapter

Classify each of the following random variables as either discrete or continuous: a. The fuel efficiency \((\mathrm{mpg})\) of an automobile b. The amount of rainfall at a particular location during the next year c. The distance that a person throws a baseball d. The number of questions asked during a l-hour lecture c. The tension (in pounds per square inch) at which a tennis racket is strung f. The amount of water used by a household during a given month g. The number of traffic citations issued by the highway patrol in a particular county on a given day

The article "FBI Says Fewer than 25 Failed Polygraph Test" (San Luis Obispo Tribune, July 29,2001 ) states that false-positives in polygraph tests (i.e., tests in which an individual fails even though he or she is telling the truth) are relatively common and occur about \(15 \%\) of the time. Suppose that such a test is given to 10 trustworthy individuals. a. What is the probability that all 10 pass? b. What is the probability that more than two fail, even though all are trustworthy? c. The article indicated that 500 FBI agents were required to take a polygraph test. Consider the random variable \(x=\) number of the 500 tested who fail. If all 500 agents tested are trustworthy, what are the mean and standard deviation of \(x ?\) d. The headline indicates that fewer than 25 of the 500 agents tested failed the test. Is this a surprising result if all 500 are trustworthy? Answer based on the values of the mean and standard deviation from Part (c).

Consider two binomial experiments. a. The first binomial experiment consists of six trials. How many outcomes have exactly one success, and what are these outcomes? b. The second binomial experiment consists of 20 trials. How many outcomes have exactly 10 successes? exactly 15 successes? exactly five successes?

You are to take a multiple-choice exam consisting of 100 questions with five possible responses to each question. Suppose that you have not studied and so must guess (select one of the five answers in a completely random fashion) on each question. Let \(x\) represent the number of correct responses on the test. a. What kind of probability distribution does \(x\) have? b. What is your expected score on the exam? (Hint: Your expected score is the mean value of the \(x\) distribution.) c. Compute the variance and standard deviation of \(x\). d. Based on your answers to Parts (b) and (c), is it likely that you would score over 50 on this exam? Explain the reasoning behind your answer.

The paper "Risk Behavior, Decision Making, and Music Genre in Adolescent Males" (Marshall University, May 2009 ) examined the effect of type of music playing and performance on a risky, decision-making task. a. Participants in the study responded to a questionnaire that was used to assign a risk behavior score. Risk behavior scores (read from a graph that appeared in the paper) for 15 participants follow. Use these data to construct a normal probability plot (the normal scores for a sample of size 15 appear in the previous exercise). \(\begin{array}{llllllll}102 & 105 & 113 & 120 & 125 & 127 & 134 & 135\end{array}\) \(\begin{array}{lllllll}139 & 141 & 144 & 145 & 149 & 150 & 160\end{array}\) b. Participants also completed a positive and negative affect scale (PANAS) designed to measure emotional response to music. PANAS values (read from a graph that appeared in the paper) for 15 participants follow. Use these data to construct a normal probability plot (the normal scores for a sample of size 15 appear in the previous exercise). \(\begin{array}{llllllll}36 & 40 & 45 & 47 & 48 & 49 & 50 & 52\end{array}\) \(\begin{array}{lllllll}53 & 54 & 56 & 59 & 61 & 62 & 70\end{array}\) c. The author of the paper states that he believes that it is reasonable to consider both risk behavior scores and PANAS scores to be approximately normally distributed. Do the normal probability plots from Parts (a) and (b) support this conclusion? Explain.

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