/*! 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 119 According to the \(95 \%\) rule,... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

According to the \(95 \%\) rule, the largest value in a sample from a distribution which is approximately symmetric and bell-shaped should be between 2 and 3 standard deviations above the mean, while the smallest value should be between 2 and 3 standard deviations below the mean. Thus the range should be roughly 4 to 6 times the standard deviation. As a rough rule of thumb, we can get a quick estimate of the standard deviation for a bell-shaped distribution by dividing the range by \(5 .\) Check how well this quick estimate works in the following situations. (a) Pulse rates from the StudentSurvey dataset discussed in Example 2.17 on page \(77 .\) The five number summary of pulse rates is \((35,62,70,\) 78,130) and the standard deviation is \(s=12.2\) bpm. Find the rough estimate using all the data, and then excluding the two outliers at 120 and \(130,\) which leaves the maximum at 96 . (b) Number of hours a week spent exercising from the StudentSurvey dataset discussed in Example 2.21 on page 81 . The five number summary of this dataset is (0,5,8,12,40) and the standard deviation is \(s=5.741\) hours. (c) Longevity of mammals from the MammalLongevity dataset discussed in Example 2.22 on page 82 . The five number summary of the longevity values is (1,8,12,16,40) and the standard deviation is \(s=7.24\) years.

Short Answer

Expert verified
Rough standard deviation estimates: For Pulse Rates (using all data): 19 bpm, (excluding outliers): 12.2 bpm. For Exercise hours: 8 hours. For Mammal longevity: 7.8 years.

Step by step solution

01

Calculate rough estimate for Pulse Rates

First, consider the pulse rates from the StudentSurvey dataset. The five number summary given is (35,62,70,78,130). The range of this dataset is \(130 - 35 = 95\). Now, to get a quick estimation of the standard deviation, divide the range by 5: \(95 / 5 = 19\) bpm. This is the rough estimate using all the data. For the second part, excluding the outliers 120 and 130 and taking the maximum as 96, the range would be \(96 - 35 = 61\). So, the rough standard deviation estimate would be \(61 / 5 = 12.2\) bpm.
02

Calculate rough estimate for exercise hours

In this step, consider the number of hours a week spent exercising from the StudentSurvey dataset. The five number summary of this dataset is (0,5,8,12,40). The range here is \(40 - 0 = 40\). Dividing the range by 5 gives us the rough estimate of standard deviation which is \(40 / 5 = 8\) hours.
03

Calculate rough estimate for mammal longevity

Lastly, evaluate the MammalLongevity dataset. The five number summary of the longevity values is (1,8,12,16,40). The range here is \(40 - 1 = 39\). Divide this range by 5 to get the rough estimate of the standard deviation, which is \(39 / 5 = 7.8\) years.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Standard Deviation
Standard deviation is a measure of how spread out the numbers are in a data set. Think of it as an average distance from the mean (average) value. For example, in a group of students, if we measure the heights and calculate the mean height, standard deviation tells us, on average, how far away each student's height is from the mean height.

When we have a set of numbers, like pulse rates, exercise hours, or lifespan of mammals, standard deviation gives us a precise numerical value that helps quantify their variability. It's calculated using a specific formula, but for a quick estimate, the range rule of thumb comes in handy—especially when dealing with symmetrical, bell-shaped distributions. In real-life situations, you’ll often find that data won’t be perfect, and outliers can skew your standard deviation, revealing the potential for extreme values in a dataset.
Normal Distribution
A normal distribution, often depicted as a bell-shaped curve, represents a distribution where most of the data points are clustered around the mean and less are found as you move away from the center. It's symmetric, meaning the left and right sides are mirror images of each other.

Characteristics such as pulse rates, hours spent exercising, or longevity of mammals often follow this pattern when enough random samples are collected. The amazing thing about normal distributions is that they enable us to predict probabilities and make inferences about the entire population based on sample data. This is why knowing if a dataset closely follows a normal distribution is essential in statistics.
Range Rule of Thumb
The range rule of thumb is like a quick math trick to estimate the standard deviation without doing any complex calculations. As we saw in the exercise solutions, you simply take the range of your data (the difference between the largest and smallest values) and divide it by 5 for a dataset that follows a normal distribution.

Why 5, though?

It’s based on the empirical rule that in a normal distribution, approximately 95% of the data falls within 2 standard deviations of the mean—thus the range is often around 4 standard deviations. Dividing by a number a bit larger than 4 (like 5) accounts for slight asymmetry and provides a buffer, making the rule applicable to a broader range of bell-shaped distributions.
Data Variability
Data variability, or dispersion, tells us how much the data points in a set differ from each other and from the mean value. It's not just one number—different statistics measure it, including range, variance, and standard deviation. High variability means that the data points are spread out over a wider range of values. Low variability means that the data points are more clustered closely to the mean.

Understanding variability is crucial because it affects the conclusions we can draw from data. For instance, two sets of mammal longevity could have the same mean life span, but one could have high variability with some mammals living much longer or shorter than others, while the other set could have low variability with most mammals living close to the average age.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Two variables are defined, a regression equation is given, and one data point is given. (a) Find the predicted value for the data point and compute the residual. (b) Interpret the slope in context. (c) Interpret the intercept in context, and if the intercept makes no sense in this context, explain why. Weight \(=\) maximum weight capable of bench pressing (pounds), Training \(=\) number of hours spent lifting weights a week \(\widehat{\text { Weigh }} t=95+11.7\) (Training); data point is an individual who trains 5 hours a week and can bench 150 pounds

Do Movies with Larger Budgets Get Higher Audience Ratings? The dataset HollywoodMovies2011 is introduced on page \(93,\) and includes many variables for movies that were produced in Hollywood in 2011, including Budget and AudienceScore. (a) Use technology to create a scatterplot to show the relationship between the budget of a movie, in millions of dollars, and the audience score. We want to see if the budget has an effect on the audience score. (b) Is there a linear relationship? How strong is it? Give your answer in the context of movies. (c) There is an outlier with a very large budget. What is the audience rating for this movie and what movie is it? There is another data value with a budget of about 125 million dollars and an audience score over 90 . To what movie does that dot correspond? (d) Use technology to find the correlation between these two variables.

Exercises 2.145 and 2.146 examine issues of location and spread for boxplots. In each case, draw sideby-side boxplots of the datasets on the same scale. There are many possible answers. One dataset has median 25, interquartile range 20 , and range 30 . The other dataset has median \(75,\) interquartile range 20 , and range 30 .

Scientists are working to train dogs to smell cancer, including early stage cancer that might not be detected with other means. In previous studies, dogs have been able to distinguish the smell of bladder cancer, lung cancer, and breast cancer. Now, it appears that a dog in Japan has been trained to smell bowel cancer. \({ }^{12}\) Researchers collected breath and stool samples from patients with bowel cancer as well as from healthy people. The dog was given five samples in each test, one from a patient with cancer and four from healthy volunteers. The dog correctly selected the cancer sample in 33 out of 36 breath tests and in 37 out of 38 stool tests. (a) The cases in this study are the individual tests. What are the variables? (b) Make a two-way table displaying the results of the study. Include the totals. (c) What proportion of the breath samples did the dog get correct? What proportion of the stool samples did the dog get correct? (d) Of all the tests the \(\operatorname{dog}\) got correct, what proportion were stool tests?

Laptop Computers and Sperm Count Stu dies have shown that heating the scrotum by jus \(1^{\circ} \mathrm{C}\) can reduce sperm count and sperm quality so men concerned about fertility are cautioned to avoid too much time in the hot tub or sauna. A new study \(^{41}\) suggests that men also keep their lap top computers off their laps. The study measurec scrotal temperature in 29 healthy male volunteer as they sat with legs together and a laptop compute on the lap. Temperature increase in the left scrotun over a 60 -minute session is given as \(2.31 \pm 0.96\) anc a note tells us that "Temperatures are given as \({ }^{\circ} \mathrm{C}\) values are shown as mean \(\pm \mathrm{SD} . "\) The abbreviatior SD stands for standard deviation. (Men who sit witl their legs together without a laptop computer do not show an increase in temperature.) (a) If we assume that the distribution of the temper ature increases for the 29 men is symmetric anc bell-shaped, find an interval that we expect to contain about \(95 \%\) of the temperature increases (b) Find and interpret the \(z\) -score for one of the men, who had a temperature increase of \(4.9^{\circ}\).

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.