Question

Your pulse rate is a measure of the number of heartbeats per minute. It can be measured in several places on your body, where an artery passes close to the skin. Once you find the pulse, count the number of beats per minute, or, count for 30 seconds and multiply by two. What's a normal pulse rate? That depends on a variety of factors. Pulse rates between 60 and 100 beats per minute are considered normal for children over 10 and adults. \({ }^{4}\) Suppose that these pulse rates are approximately normally distributed with a mean of 78 and a standard deviation of 12 . a. What proportion of adults will have pulse rates between 60 and \(100 ?\) b. What is the 95 th percentile for the pulse rates of adults? c. Would a pulse rate of 110 be considered unusual? Explain.

Short answer

Answer: The proportion of adults with pulse rates between 60 and 100 is approximately 0.8996 or 89.96%. Question: What is the 95th percentile of pulse rates for adults? Answer: The 95th percentile of the pulse rates for adults is approximately 97.74 beats per minute. Question: Is a pulse rate of 110 considered unusual? Answer: Yes, a pulse rate of 110 is considered unusual because its corresponding z-score is 2.67, which is outside the typical range (-2 to 2) that covers approximately 95% of the data.

Step-by-step solution

Find the Z-scores

To convert the pulse rates (60 and 100) into their corresponding z-scores, we use the formula: z = (pulse_rate - mean) / standard_deviation For 60: z1 = (60 - 78) / 12 z1 = -1.5 For 100: z2 = (100 - 78) / 12 z2 = 1.83

Find the proportion between the Z-scores

To find the proportion of adults with pulse rates between 60 and 100, we need to find the area under the standard normal curve between z1 and z2. Using a standard normal distribution table, we can find the probabilities corresponding to z1 and z2: P(z1) = 0.0668 P(z2) = 0.9664 Now, subtract P(z1) from P(z2) to find the proportion of adults with pulse rates between 60 and 100: Proportion = P(z2) - P(z1) Proportion = 0.9664 - 0.0668 Proportion = 0.8996 #b. The 95th percentile of pulse rates for adults#

Find the Z-score corresponding to 95th percentile

To find the z-score corresponding to the 95th percentile, we can look up the standard normal distribution table and find the value of z for which the probability is 0.95. z = 1.645

Convert Z-score to pulse rate

Now, we convert the z-score back to the original pulse rate using the formula: pulse_rate = (z * standard_deviation) + mean pulse_rate = (1.645 * 12) + 78 pulse_rate ≈ 97.74 The 95th percentile of the pulse rates for adults is approximately 97.74 beats per minute. #c. Determine if a pulse rate of 110 is considered unusual#

Find the Z-score for pulse rate 110

To check if a pulse rate of 110 is considered unusual, we first convert it into a z-score, using the same formula as before: z = (110 - 78) / 12 z = 2.67

Check if Z-score is unusual

Generally, a z-score above 2 or below -2 is considered unusual because it covers approximately 95% of the data. Since the z-score for a pulse rate of 110 is 2.67, it is considered unusual.

Key concepts

Z-scores

Understanding Z-scores is fundamental in the field of statistics when dealing with the normal distribution. A Z-score, sometimes called a standard score, represents how many standard deviations a data point is from the mean. The formula to calculate a Z-score is:
\[ z = \frac{(X - \mu)}{\sigma} \]
where \(X\) is the value of the data point, \(\mu\) is the mean, and \(\sigma\) is the standard deviation. Z-scores are dimensionless, which allows for comparison between different sets of data. For instance, if a pulse rate of 60 has a Z-score of -1.5, it is 1.5 standard deviations below the mean pulse rate. A Z-score can tell whether a data point is typical for a given set of data or if it's unusual. Generally, a Z-score greater than 2 or less than -2 is considered unusual, as data within those ranges are typically found in the extremes of the distribution curve.

Standard Normal Curve

The standard normal curve, or bell curve, is a depiction of a normal distribution with a mean of 0 and a standard deviation of 1. It is vital in the field of statistics because it allows for the analysis of variables that follow a normal distribution. The area under the curve represents probabilities and corresponds to the percentage of data within certain ranges. For example, the area between the Z-scores of -1.5 and 1.83 represents the proportion of adult pulse rates between 60 and 100 per minute. Analysts use the standard normal curve to determine areas under the curve, which equate to percentiles. This helps in assessing whether a data point falls within a typical range or if it's part of the outliers.

Percentiles in Statistics

Percentiles in statistics are measures that divide a set of data into 100 equal parts. They are used to compare scores across a wide array of datasets. In the context of pulse rates, if an individual's pulse rate is at the 95th percentile, their pulse rate is faster than 95% of the population's rates. To determine the 95th percentile of pulse rates for adults, you can use the Z-score associated with 95% (which is approximately 1.645) and convert it using the mean and standard deviation of the data. This process translates the standard score back to the specific measurement scale of interest. This percentile helps health professionals and researchers gauge how an individual's pulse rate stacks up against the broader population.

Unusual Events in Statistics

In statistics, unusual events refer to occurrences that are rare within a set of data. They often lie on the outer edges of the normal distribution curve. Typically, data points that have Z-scores less than -2 or greater than 2 are considered unusual because they fall outside the range where about 95% of the values lie. The pulse rate of 110 discussed in the problem, with a Z-score of 2.67, is thus unusual. Recognizing unusual events is crucial for outlier detection which can be important in fields ranging from medical diagnostics to quality control in manufacturing. Identifying such events helps professionals to inquire further to determine if there is a special cause for the abnormal reading.