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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: a. Approximately 68.27% of students have a breathing rate between 9.7 and 14.3 breaths per minute. b. Approximately 95.45% of students have a breathing rate between 7.4 and 16.6 breaths per minute. c. Approximately 0.27% of students have a breathing rate more than 18.9 or less than 5.1 breaths per minute.

Step by step solution

01

Convert the intervals into z-scores

For each of the intervals, we'll convert the values into z-scores using the formula: `z = (X - μ) / σ` where z is the z-score, X is the value of the interval, μ is the mean, and σ is the standard deviation. For part a (9.7 to 14.3 breaths per minute), X1 = 9.7 and X2 = 14.3, μ = 12, and σ = 2.3. So, we calculate the z-scores for both values: `z1 = (9.7 - 12) / 2.3 = -1` `z2 = (14.3 - 12) / 2.3 = 1` For part b (7.4 to 16.6 breaths per minute), X1 = 7.4 and X2 = 16.6, μ = 12, and σ = 2.3. So, we calculate their z-scores: `z1 = (7.4 - 12) / 2.3 = -2` `z2 = (16.6 - 12) / 2.3 = 2` For part c (more than 18.9 or less than 5.1 breaths per minute), X1 = 5.1 and X2 = 18.9, μ = 12, and σ = 2.3. So, we calculate their z-scores: `z1 = (5.1 - 12) / 2.3 = -3` `z2 = (18.9 - 12) / 2.3 = 3`
02

Compute the area under the normal curve for each interval

Using a z-table, find the area between each pair of z-scores, which represents the fraction of students whose breathing rate falls within those intervals. For interval a (z1 = -1 to z2 = 1), we can find the area from the z-table, or we can recall that this is a standard result: approximately 68.27% of the area under a normal curve falls between z-scores of -1 and 1. For interval b (z1 = -2 to z2 = 2), we can similarly find the area from the z-table, or recall the standard result that approximately 95.45% of the area under a normal curve falls between z-scores of -2 and 2. For interval c (z1 < -3 and z2 > 3), first find the area between z-scores of -3 and 3 from the z-table, which is approximately 99.73%. Then, notice that this situation is a complement of the interval between -3 and 3, so we subtract this area from 100%: `100% - 99.73% = 0.27%`.
03

Interpret the results

Now we have the fractions of students who fall within each interval: a. Approximately 68.27% of students have a breathing rate between 9.7 and 14.3 breaths per minute. b. Approximately 95.45% of students have a breathing rate between 7.4 and 16.6 breaths per minute. c. Approximately 0.27% of students have a breathing rate more than 18.9 or less than 5.1 breaths per minute.

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Key Concepts

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

Z-scores
Understanding Z-scores is crucial when working with normal distributions. A Z-score tells us how many standard deviations an individual data point is from the mean. The formula for calculating a Z-score is:
  • \( z = \frac{X - \mu}{\sigma} \)
Where:
  • \( X \) is the data value we are considering,
  • \( \mu \) is the mean of the distribution,
  • \( \sigma \) is the standard deviation.

Let's take an example: we have a student's breathing rate at 9.7 breaths per minute, a mean \( \mu = 12 \), and a standard deviation \( \sigma = 2.3 \). The Z-score calculation would be:
  • \( z = \frac{9.7 - 12}{2.3} = -1 \)
This tells us that a 9.7 rate is 1 standard deviation below the mean. Similarly, a Z-score of 1 would mean 1 standard deviation above the mean.
Interpreting these scores with a Z-table or standard results helps quantify what percentage of the population falls within certain intervals.
Standard Deviation
Standard deviation is a measure of spread within a set of data. It tells us how much the values in our data set typically vary from the mean. In the world of normal distributions, it's a crucial measure. It helps us understand how data disperses itself around the mean.
The formula for standard deviation in a sample size is:
  • \( \sigma = \sqrt{\frac{\sum (X_i - \mu)^2}{N}} \)
Where:
  • \( X_i \) are the data points,
  • \( \mu \) is the mean,
  • \( N \) is the number of data points.

For a normal distribution like the breathing rates of students, the standard deviation helps us determine what is generally considered a 'normal' range.
  • Within 1 standard deviation (both directions from the mean), we find about 68% of data points.
  • Within 2 standard deviations, around 95% of data points.
  • And within 3 standard deviations, roughly 99.7%.
This gives us a tool to understand variability and to identify when something is more unusual. If a student's breathing rate is more than 3 standard deviations away, it's quite rare in this distribution.
Mean
The mean is a measure of central tendency. It gives us the average value in a dataset, serving as a useful baseline in normal distributions. Mathematically, the mean is calculated by summing all the numbers in a dataset and then dividing by the number of values in that dataset. Here's the formula:
  • \( \mu = \frac{\sum X_i}{N} \)
Where:
  • \( X_i \) are the individual data points,
  • \( N \) is the total number of data points.

In the context of our exercise on breathing rates, the mean is 12 breaths per minute. This number serves as the 'center' of our distribution, with the standard deviation telling us how much the other data points deviate from this center. Knowing the mean, along with the standard deviation, allows us to place any data point in context with the rest of the data. For example, it tells us that 68% of all values will be between approximately 9.7 and 14.3 breaths per minute, centered on this mean.
The mean is a useful and quick reference to determine where most of the observations lie. It's especially powerful in conjunction with other statistics like the standard deviation, to paint a full picture of our data distribution.

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Most popular questions from this chapter

The miles per gallon (mpg) for each of 20 medium-sized cars selected from a production line during the month of March follow. $$ \begin{array}{llll} 23.1 & 21.3 & 23.6 & 23.7 \\ 20.2 & 24.4 & 25.3 & 27.0 \\ 24.7 & 22.7 & 26.2 & 23.2 \\ 25.9 & 24.7 & 24.4 & 24.2 \\ 24.9 & 22.2 & 22.9 & 24.6 \end{array} $$ a. What are the maximum and minimum miles per gallon? What is the range? b. Construct a relative frequency histogram for these data. How would you describe the shape of the distribution? c. Find the mean and the standard deviation. d. Arrange the data from smallest to largest. Find the \(z\) -scores for the largest and smallest observations. Would you consider them to be outliers? Why or why not? e. What is the median? f. Find the lower and upper quartiles.

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