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Chapter 6: Problem 110

The paper referenced in the previous exercise also included data on left atrial diameter for children who were considered overweight. For these children, left atrial diameter was approximately normally distributed with a mean of \(28 \mathrm{~mm}\) and a standard deviation of \(4.7 \mathrm{~mm}\). a. Approximately what proportion of overweight children have left atrial diameters less than \(25 \mathrm{~mm}\) ? b. Approximately what proportion of overweight children have left atrial diameters greater than \(32 \mathrm{~mm} ?\) c. Approximately what proportion of overweight children have left atrial diameters between 25 and \(30 \mathrm{~mm}\) ? d. What proportion of overweight children have left atrial diameters greater than the mean for healthy children?

Short Answer

Expert verified
a. Using the z-score formula, we find \(z_1 = -0.64\). The proportion of overweight children with left atrial diameters less than 25 mm is approximately 0.2611. b. Calculating the z-score for 32 mm, we get \(z_2 = 0.85\). The proportion of overweight children with left atrial diameters greater than 32 mm is approximately 0.1977. c. We have already calculated the z-scores for 25 mm and 30 mm as \(z_3 = -0.64\) and \(z_4 = 0.43\). The proportion of overweight children with left atrial diameters between 25 and 30 mm is approximately 0.3347. d. Assuming the mean left atrial diameter for healthy children is 29 mm, the z-score is \(z_5 = 0.21\). The proportion of overweight children with left atrial diameters greater than the mean for healthy children is approximately 0.4170.

Step by step solution

01

Identifying the given parameters

For overweight children, we know that the mean left atrial diameter (碌) is 28 mm, and the standard deviation (蟽) is 4.7 mm. We will use these parameters to calculate the z-scores for the specified left atrial diameter values.
02

Formula for z-score

The z-score formula is given by \(z = \frac{X - \mu}{\sigma}\), where X denotes the given left atrial diameter value, 碌 is the mean, and 蟽 is the standard deviation.
03

Calculating z-scores for given diameters and finding proportions

a. For a left atrial diameter less than 25 mm, calculate the z-score as follows: \(z_1 = \frac{25 - 28}{4.7}\) Using the z-score table or calculator, find the proportion of values below this z-score. This will give us the proportion of overweight children with left atrial diameters less than 25 mm. b. For a left atrial diameter greater than 32 mm, calculate the z-score as follows: \(z_2 = \frac{32 - 28}{4.7}\) Using the z-score table or calculator, find the proportion of values above this z-score. This will give us the proportion of overweight children with left atrial diameters greater than 32 mm. c. For left atrial diameters between 25 and 30 mm, calculate the z-scores corresponding to these diameters as follows: \(z_3 = \frac{25 - 28}{4.7}\) \(z_4 = \frac{30 - 28}{4.7}\) Using the z-score table or calculator, find the proportion of values between these z-scores. This will give us the proportion of overweight children with left atrial diameters between 25 and 30 mm. d. To answer this question, we first need to know the mean left atrial diameter for healthy children. Let us assume it is 29 mm (let's denote it as X). Calculate the z-score corresponding to the mean left atrial diameter for healthy children as follows: \(z_5 = \frac{X - 28}{4.7}\) Using the z-score table or calculator, find the proportion of values above this z-score. This will give us the proportion of overweight children with left atrial diameters greater than the mean left atrial diameter for healthy children.

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Key Concepts

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

Standard Deviation
Standard deviation is a measure of the amount of variation or dispersion in a set of values. It tells us how much the values in a dataset deviate from the mean. In the context of the original exercise, the standard deviation of left atrial diameters for overweight children is given as 4.7 mm. This means that most of the children's left atrial diameters are within 4.7 mm of the mean. The smaller the standard deviation, the closer the data points are to the mean. Conversely, a larger standard deviation indicates the data points are more spread out. To calculate standard deviation, use the formula:
\[\sigma = \sqrt{\frac{\sum (X_i - \mu)^2}{N}}\]where \(X_i\) represents each data point, \(\mu\) is the mean, and \(N\) is the number of data points.
Understanding the standard deviation helps us determine the variability of the dataset, which is crucial when analyzing how typical or exceptional a particular data point is in the context of the data.
Mean
The mean, often referred to as the average, is a measure of central tendency. It is calculated by adding all the data points together and then dividing by the total number of points. In our exercise, the mean left atrial diameter for overweight children is 28 mm. It acts as a central point around which the rest of the data is distributed. The mean is central to calculations for both standard deviation and z-scores.
To find the mean of a dataset, use the following formula:
\[\mu = \frac{\sum X_i}{N}\]where \(\sum X_i\) is the sum of all data points and \(N\) is the number of points in the dataset.
The mean provides a quick summary of the dataset, giving a good starting point for further statistical analysis.
Z-Score
A z-score is a statistical measurement that describes a value's relation to the mean of a group of values. It tells us how many standard deviations an element is from the mean. In this exercise, z-scores help us determine the proportion of the population that falls below, above, or between certain measurements of left atrial diameters. The formula for calculating a z-score is:
\[z = \frac{X - \mu}{\sigma}\]where \(X\) is the value in question, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.
Calculating the z-score allows us to standardize scores on a single scale, which aids in comparing different datasets or different segments of the same dataset. By referring to z-score tables, we can find the probability or proportion of values within given ranges.
Proportion Calculation
Proportion calculation is used to determine the fraction of the dataset that meets a specific condition. In the context of the exercise, we determine the proportion of overweight children with certain left atrial diameter measurements. This involves calculating and interpreting z-scores. Once the z-scores are calculated, a standard normal distribution table, or software tools, can be used to find the corresponding proportions.
  • For a measurement less than a certain value, the z-score provides the cumulative probability.
  • For values greater than a given measurement, subtract the cumulative probability from one.
  • For measurements between two values, calculate the difference in cumulative probabilities of respective z-scores.
These calculations help in understanding the distribution and prevalence of certain characteristics within the population, providing a clear statistical grounding for decision making and analysis.

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Most popular questions from this chapter

6.27 The article "Probabilistic Risk Assessment of Infrastructure Networks Subjected to Hurricanes" (12th International Conference on Applications of Statistics and Probability in Civil Engineering, 2015) suggests a uniform distribution as a model for the actual landfall position of the eye of a hurricane. Consider the random variable \(x=\) distance of actual landfall from predicted landfall. Suppose that a uniform distribution on the interval that ranges from \(0 \mathrm{~km}\) to \(400 \mathrm{~km}\) is a reasonable model for \(x\). a. Draw the density curve for \(x\). b. What is the height of the density curve? c. What is the probability that \(x\) is at most \(100 ?\) d. What is the probability that \(x\) is between 200 and \(300 ?\) Between 50 and \(150 ?\) Why are these two probabilities equal?

A machine producing vitamin E capsules operates so that the actual amount of vitamin \(\mathrm{E}\) in each capsule is normally distributed with a mean of \(5 \mathrm{mg}\) and a standard deviation of \(0.05 \mathrm{mg}\). What is the probability that a randomly selected capsule contains less than \(4.9 \mathrm{mg}\) of vitamin \(\mathrm{E}\) ? At least \(5.2 \mathrm{mg}\) of vitamin \(\mathrm{E}\) ?

Consider the random variable \(y=\) the number of broken eggs in a randomly selected carton of one dozen eggs. Suppose the probability distribution of \(y\) is as follows: \(\begin{array}{cccccc}y & 0 & 1 & 2 & 3 & 4 \\ p(y) & 0.65 & 0.20 & 0.10 & 0.04 & ?\end{array}\) a. Only \(y\) values of \(0,1,2,3,\) and 4 have probabilities greater than \(0 .\) What is \(p(4) ?\) b. How would you interpret \(p(1)=0.20 ?\) c. Calculate \(P(y \leq 2)\), the probability that the carton contains at most two broken eggs, and interpret this probability. d. Calculate \(P(y<2),\) the probability that the carton contains fewer than two broken eggs. Why is this smaller than the probability in Part (c)? e. What is the probability that the carton contains exactly 10 unbroken eggs? f. What is the probability that at least 10 eggs are unbroken?

Suppose that fuel efficiency (miles per gallon, mpg) for a particular car model under specified conditions is normally distributed with a mean value of \(30.0 \mathrm{mpg}\) and a standard deviation of \(1.2 \mathrm{mpg}\). a. What is the probability that the fuel efficiency for a randomly selected car of this model is between 29 and \(31 \mathrm{mpg}\) ? b. Would it surprise you to find that the efficiency of a randomly selected car of this model is less than \(25 \mathrm{mpg} ?\) c. If three cars of this model are randomly selected, what is the probability that each of the three have efficiencies exceeding \(32 \mathrm{mpg}\) ? d. Find a number \(x^{*}\) such that \(95 \%\) of all cars of this model have efficiencies exceeding \(x^{*}\) (i.e., \(P\left(x>x^{*}\right)=0.95\) ).

Suppose that your statistics professor tells you that the scores on a midterm exam were approximately normally distributed with a mean of 78 and a standard deviation of 7 . The top \(15 \%\) of all scores have been designated A's. Your score is \(89 .\) Did you earn an \(\mathrm{A}\) ? Explain.
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