/*! 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 17 The mean weight gain for women d... [FREE SOLUTION] | 91Ó°ÊÓ

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The mean weight gain for women during a full-term pregnancy is \(30.2\) pounds. The standard deviation of weight gain for this group is \(9.9\) pounds, and the shape of the distribution of weight gains is symmetric and unimodal. (Source: BMJ 2016; 352 doi: https://doi.org/10.1136/bmj.i555) a. State the weight gain for women one standard deviation below the mean and for one standard deviation above the mean. b. Is a weight gain of 35 pounds more or less than one standard deviation from the mean?

Short Answer

Expert verified
a. The weight gain for women one standard deviation below the mean is 20.3 pounds and for one standard deviation above the mean is 40.1 pounds. \nb. A weight gain of 35 pounds is less than one standard deviation from the mean.

Step by step solution

01

Calculate weight gain one standard deviation below and above the mean

The mean weight gain is 30.2 pounds and the standard deviation is 9.9 pounds. To find the weight gain one standard deviation below the mean, subtract the standard deviation from the mean. This gives \(30.2 - 9.9 = 20.3\) pounds. To find the weight gain one standard deviation above the mean, add the standard deviation to the mean. This gives \(30.2 + 9.9 = 40.1\) pounds.
02

Compare 35 pounds with weight gain one standard deviation above the mean

In order to find whether the weight gain of 35 pounds is more or less than one standard deviation from the mean, compare this value with the weight gain for one standard deviation above the mean, which is 40.1 pounds. Since 35 pounds is less than 40.1 pounds, it is therefore less than one standard deviation from the mean.

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Key Concepts

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

Mean Weight Gain
Understanding the concept of mean weight gain during pregnancy is crucial for healthcare professionals and expectant mothers alike. Let's take the finding from the BMJ study, which indicates that the average weight gain is 30.2 pounds. The mean, or average, represents the central point of a data set. In this case, it's the average amount of weight gained by women during a full-term pregnancy. To calculate it, you would sum all individual weight gains and divide by the number of women in the study.

To provide a clearer picture, imagine a group of women whose weight gains are plotted on a graph. The mean weight gain of 30.2 pounds would be at the center of this plot. It's a helpful indicator that gives us a starting point for comparing individual cases. For instance, we can quickly identify if a weight gain of 35 pounds is above or below average. By understanding the mean, health professionals can better advise expectant mothers regarding nutrition and health during pregnancy.
Symmetric Distribution
A symmetric distribution is quite intuitive once you get the hang of it. Imagine you're looking at the silhouette of a mountain, mirrored perfectly on either side of its peak. That's the essence of a symmetric distribution. In the context of weight gain during pregnancy from the BMJ study, a symmetric distribution suggests that most women's weight gains are distributed evenly around the mean (in our case, 30.2 pounds).

This symmetry means that if you were to select a woman at random who gained less than the mean, another woman who gained more by the same amount would likely exist. The symmetry also has implications for predicting outcomes. For instance, knowing that the distribution is symmetric enables us to say with some certainty that deviations from the mean on the lower and upper ends are equally probable. For expectant mothers, this might convey reassurance: substantial deviations in weight gain, either above or below the mean, are part of a natural pattern.
Unimodal Distribution
Unimodal distribution may sound complex, but it's a concept that can be dissected easily. 'Uni' means one, and 'modal' refers to modes — which in statistics is another term for peaks in a distribution. So, a unimodal distribution has a single peak. In our case, it indicates that there's one most common weight gain value among women during pregnancy, and this value occurs near the mean of 30.2 pounds mentioned in the BMJ study.

Visualizing a unimodal distribution, imagine a hill with one clear peak, as opposed to a range with multiple peaks. It simplifies understanding of data by highlighting the most frequent outcome. For healthcare purposes, knowing that the distribution of weight gain is unimodal allows experts to focus on what the most typical pregnancy weight gain looks like, rather than being concerned with many different peaks of common weight gains. This singular focus can guide recommendations and expectations for both practitioners and patients.

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Most popular questions from this chapter

In 2017 a pollution index was calculated for a sample of cities in the western states using data on air and water pollution. Assume the distribution of pollution indices is unimodal and symmetric. The mean of the distribution was \(43.0\) points with a standard deviation of \(11.3\) points. (Source: numbeo.com) a. What percentage of western cities would you expect to have a pollution index between \(31.7\) and \(54.3\) points? b. What percentage of western cities would you expect to have a pollution index between \(20.4\) and \(65.6\) ? c. The pollution index for San Jose in 2017 was \(51.9\) points. Based on this distribution, was this unusually high? Explain.

Quantitative SAT scores have a mean of 500 and a standard deviation of 100, while ACT scores have a mean of 21 and a standard deviation of \(5 .\) Assuming both types of scores have distributions that are unimodal and symmetric, which is more unusual: a quantitative SAT score of 750 or an ACT score of 28 ? Show your work.

This list represents the number of children for the first six "first ladies" of the United States. (Source: 2009 World Almanac and Book of Facts) $$ \begin{array}{ll} \text { Martha Washington } & 0 \\ \text { Abigail Adams } & 5 \\ \hline \text { Martha Jefferson } & 6 \\ \text { Dolley Madison } & 0 \\ \text { Elizabeth Monroe } & 2 \\ \hline \text { Louisa Adams } & 4 \end{array} $$ a. Find the mean number of children, rounding to the nearest tenth. Interpret the mean in this context. b. According to eh.net/encyclopedia, women living around 1800 tended to have between 7 and 8 children. How does the mean of these first ladies compare to that? c. Which of the first ladies listed here had the number of children that is farthest from the mean and therefore contributes most to the standard deviation? d. Find the standard deviation, rounding to the nearest tenth.

The dotplot shows the distribution of the heights (in feet) of a sample of roller coasters. The five-number summary of the data is given in the following table. Sketch a boxplot of the data. Explain how you determined the length of the whiskers. $$ \begin{array}{lclcl} \text { Minimum } & \text { Q1 } & \text { Median } & \text { Q3 } & \text { Maximum } \\ \hline 2.438 & 8.526 & 18.288 & 33.223 & 128.016 \end{array} $$

Note: Reported interquartile ranges will vary depending on technology. Name two measures of the variation of a distribution, and state the conditions under which each measure is preferred for measuring the variability of a single data set.

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