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Is your breathing rate normal? Actually, there is no standard breathing rate for humans. It can vary from as low as 4 breaths per minute to as high as 70 or 75 for a person engaged in strenuous exercise, Suppose that the resting breathing rates for college-age students have a relative frequency distribution that is mound-shaped, with a mean equal to 12 and a standard deviation of 2.3 breaths per minute.What fraction of all students would have breathing rates in the following intervals? a. 9.7 to 14.3 breaths per minute b. 7.4 to 16.6 breaths per minute c. More than 18.9 or less than 5.1 breaths per minute

Short Answer

Expert verified
Answer: - Approximately 68.26% of students have a resting breathing rate within 9.7 to 14.3 breaths per minute. - Approximately 95.44% of students have a resting breathing rate within 7.4 to 16.6 breaths per minute. - Approximately 0.26% of students have a resting breathing rate more than 18.9 or less than 5.1 breaths per minute.

Step by step solution

01

Understand the problem and the given data

The exercise provides us with a distribution of resting breathing rates that is assumed to be approximately normal. We are given a mean (µ) of 12 breaths per minute, a standard deviation (σ) of 2.3 breaths per minute, and interval limits for each part. We will use the standard normal (z) table to find the desired fraction of students for each interval.
02

Calculate z-scores for the interval limits

For each interval, we will first calculate the respective z-scores. The z-score is a measure of how many standard deviations an observation is from the mean. The formula used to calculate the z-score for a value x is: $$ z = \frac{(x - \mu)}{\sigma} $$
03

Part a: 9.7 to 14.3 breaths per minute

Find the z-scores for 9.7 and 14.3 breaths per minute: $$ z_1 = \frac{(9.7 - 12)}{2.3} = -1 $$ $$ z_2 = \frac{(14.3 - 12)}{2.3} = 1 $$
04

Part b: 7.4 to 16.6 breaths per minute

Find the z-scores for 7.4 and 16.6 breaths per minute: $$ z_1 = \frac{(7.4 - 12)}{2.3} = -2 $$ $$ z_2 = \frac{(16.6 - 12)}{2.3} = 2 $$
05

Part c: More than 18.9 or less than 5.1 breaths per minute

Find the z-scores for 18.9 and 5.1 breaths per minute: $$ z_1 = \frac{(18.9 - 12)}{2.3} = 3 $$ $$ z_2 = \frac{(5.1 - 12)}{2.3} = -3 $$
06

Find the fractions of students with the calculated z-scores

Using a standard normal (z) table or calculator, find the values that correspond to the calculated z-scores, and then find the desired fractions.
07

Part a: 9.7 to 14.3 breaths per minute

Find the probabilities corresponding to the z-scores -1 and 1: $$ P(-1) = 0.1587 $$ $$ P(1) = 0.8413 $$ The fraction of students for this interval is the difference between these probabilities: $$ P(a) = P(1) - P(-1) = 0.8413 - 0.1587 = 0.6826 \approx 68.26\% $$
08

Part b: 7.4 to 16.6 breaths per minute

Find the probabilities corresponding to the z-scores -2 and 2: $$ P(-2) = 0.0228 $$ $$ P(2) = 0.9772 $$ The fraction of students for this interval is the difference between these probabilities: $$ P(b) = P(2) - P(-2) = 0.9772 - 0.0228 = 0.9544 \approx 95.44\% $$
09

Part c: More than 18.9 or less than 5.1 breaths per minute

Find the probabilities corresponding to the z-scores -3 and 3: $$ P(-3) = 0.0013 $$ $$ P(3) = 0.9987 $$ The fraction of students for this interval is the sum of the probabilities in the tails: $$ P(c) = P(-3) + (1 - P(3)) = 0.0013 + (1 - 0.9987) = 0.0013 + 0.0013 = 0.0026 \approx 0.26\% $$
10

Interpret the results

From our calculations, the results are as follows: - For the interval of 9.7 to 14.3 breaths per minute, approximately 68.26% of students have a resting breathing rate within this range. - For the interval of 7.4 to 16.6 breaths per minute, approximately 95.44% of students have a resting breathing rate within this range. - For the intervals with the breathing rate more than 18.9 or less than 5.1 breaths per minute, approximately 0.26% of students have a resting breathing rate within these ranges.

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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 crucial concept in statistics that measures how spread out numbers are in a data set. In simpler terms, it tells us how much the individual data points differ from the mean. A low standard deviation means that most of the numbers are close to the mean, while a high one indicates that the numbers are more spread out.
For example, if we're looking at breathing rates among college-age students, and the standard deviation is 2.3 breaths per minute, it means most students' breathing rates hover around the mean of 12, but there can still be some variability. The standard deviation helps us understand this variation and play a key role in calculating other statistics like z-scores.
Mean
The mean is the average of a set of numbers, representing the central value. To find the mean, add up all the numbers and divide by the amount of numbers there are. In the case of breathing rates, the mean is given as 12 breaths per minute.
This tells us that most college students have a resting breathing rate near this value. It acts as a central point from which we measure the dispersion of the data, making it vital for understanding how data tends to cluster around this central point.
Z-scores
A z-score measures how many standard deviations a data point is away from the mean. The z-score formula is \( z = \frac{(x - \mu)}{\sigma} \), where \( x \) is the data point, \( \mu \) is the mean, and \( \sigma \) is the standard deviation.
Z-scores allow us to standardize different data points for comparison. If we have a student with a breathing rate of 14.3 breaths per minute, we can compute its z-score to see how uncommon this rate is relative to the average. Negative z-scores indicate values below the mean, while positive scores reflect values above the mean. It's a way to determine how extreme or typical a data point is, based on the distribution's standard deviation.
Probability Calculations
The probability calculations in a normal distribution context tell us about the likelihood of a data point occurring within a certain range. We rely on the z-table when performing these calculations. The z-table provides the probability of a z-score being below a certain value.
For instance, by converting the interval limits to z-scores, we can find out how likely it is for a student's breathing rate to fall within that interval. For the interval 9.7 to 14.3 breaths per minute, using z-scores that we calculated, we find a probability of approximately 68.26%.
This showcases how these probability calculations offer insights into the percentage of occurrences, helping us to understand, interpret, and forecast data patterns effectively.

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Most popular questions from this chapter

As professional sports teams become a more and more lucrative business for their owners, the salaries paid to the players have also increased. In fact, sports superstars are paid astronomical salaries for their talents. If you were asked by a sports management firm to describe the distribution of players" salaries in several different categories of professional sports, what measure of center would you choose? Why?

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High school students in an International Baccalaureate (IB) program are placed in accelerated or advanced courses and must take IB examinations in each of six subject areas at the end of their junior or senior year. Students are scored on a scale of \(1-7,\) with \(1-2\) being poor, 3 mediocre, 4 average, and \(5-7\) excellent. During its first year of operation at John W. North High School in Riverside, California, 17 juniors attempted the IB economics exam, with these results: $$ \begin{array}{cc} \text { Exam Grade } & \text { Number of Students } \\ \hline 7 & 1 \\ 6 & 4 \\ 5 & 4 \\ 4 & 4 \\ 3 & 4 \end{array} $$ Calculate the mean and standard deviation for these scores.

In the seasons that followed his 2001 record-breaking season, Barry Bonds hit \(46,45,45,5,26,\) and 28 homers, respectively, until he retired from major league baseball in 2007 (www.ESPN,com). \(^{16}\) Two box plots, one of Bond's homers through \(2001,\) and a second including the years \(2002-2007\) follow. The statistics used to construct these box plots are given in the table. \begin{tabular}{lccccccc} Years & Min & \(a_{1}\) & Median & \(a_{3}\) & IQR & Max & \(n\) \\ \hline 2001 & 16 & 25.00 & 34.00 & 41.50 & 16.5 & 73 & 16 \\ 2007 & 5 & 25.00 & a. Calculate the upper fences for both of these box plots. b. Can you explain why the record number of homers is an outlier in the 2001 box plot, but not in the 2007 box plot?34.00 & 45.00 & 20.0 & 73 & 22 \end{tabular}

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