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Chapter 3: Problem 23

Find the mean and median for each of the two samples, then compare the two sets of results. Listed below are pulse rates (beats per minute) from samples of adult males and females (from Data Set 1 "Body Data"in Appendix B). Does there appear to be a difference? \(\begin{array}{llllllllllllllll}\text { Male: } & 86 & 72 & 64 & 72 & 72 & 54 & 66 & 56 & 80 & 72 & 64 & 64 & 96 & 58 & 66 \\ \text { Female: } & 64 & 84 & 82 & 70 & 74 & 86 & 90 & 88 & 90 & 90 & 94 & 68 & 90 & 82 & 80\end{array}\)

Short Answer

Expert verified
Mean(Male) = 69.47, Median(Male) = 66Mean(Female) = 81.47, Median(Female) = 86Female pulse rates are higher than male pulse rates.

Step by step solution

01

Organize the Data

List out the pulse rates for both males and females separately:Male: 86, 72, 64, 72, 72, 54, 66, 56, 80, 72, 64, 64, 96, 58, 66Female: 64, 84, 82, 70, 74, 86, 90, 88, 90, 90, 94, 68, 90, 82, 80
02

Calculate the Mean Pulse Rate for Males

First sum the pulse rates for males: 86 + 72 + 64 + 72 + 72 + 54 + 66 + 56 + 80 + 72 + 64 + 64 + 96 + 58 + 66 = 1042Then divide by the number of male pulse rates (15):Mean (Male) = \( \frac{1042}{15} \approx 69.47 \)
03

Calculate the Mean Pulse Rate for Females

First sum the pulse rates for females:64 + 84 + 82 + 70 + 74 + 86 + 90 + 88 + 90 + 90 + 94 + 68 + 90 + 82 + 80 = 1222Then divide by the number of female pulse rates (15):Mean (Female) = \( \frac{1222}{15} \approx 81.47 \)
04

Determine the Median Pulse Rate for Males

Sort the male pulse rates in ascending order:54, 56, 58, 64, 64, 64, 66, 66, 72, 72, 72, 80, 86, 96Since there are 15 numbers, the median will be the 8th value in the sorted list:Median (Male) = 66
05

Determine the Median Pulse Rate for Females

Sort the female pulse rates in ascending order:64, 68, 70, 74, 80, 82, 82, 84, 86, 88, 90, 90, 90, 90, 94Since there are 15 numbers, the median will be the 8th value in the sorted list:Median (Female) = 86
06

Compare the Mean and Median of the Two Samples

Comparing the two sets of results:Mean (Male) = 69.47, Median (Male) = 66Mean (Female) = 81.47, Median (Female) = 86Both the mean and median are higher for females than for males, indicating that female pulse rates are generally higher than male pulse rates.

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Key Concepts

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

mean calculation
The mean, or average, is a measure of central tendency that sums up all the values in a set and divides by the number of values. In this exercise, we calculated the mean pulse rate for both males and females.
To find the mean pulse rate for males, we added all their pulse rates to get a total of 1042 and then divided by the number of male participants (15). This yields a mean of approximately 69.47.
Similarly, for females, we summed their pulse rates to get 1222 and divided by the number of female participants (15), resulting in a mean of approximately 81.47.
These values help us understand the typical pulse rate within each group more concretely.
median calculation
The median is another measure of central tendency. It represents the middle value in a sorted list of numbers.
For the male pulse rates, after arranging them in ascending order, the list is: 54, 56, 58, 64, 64, 64, 66, 66, 72, 72, 72, 80, 86, 96. The median is the 8th value, which is 66.
For the female pulse rates, the sorted list is: 64, 68, 70, 74, 80, 82, 82, 84, 86, 88, 90, 90, 90, 90, 94. The median here is the 8th value, which is 86.
The median provides a better measure when there are outliers in the data as it isn't affected by extremely high or low values.
pulse rates
Pulse rate is the number of heartbeats per minute, an essential indicator of cardiac health.
In this exercise, we examined the pulse rates from samples of adult males and females. Male pulse rates were generally between 54 and 96, while female pulse rates ranged from 64 to 94.
Understanding pulse rate distributions can help in identifying normal and abnormal heart function, potentially indicating underlying health issues if the pulse rate is significantly high or low relative to the general population.
comparison of samples
When comparing pulse rates between the two samples, we used both mean and median to summarize each group's central tendency.
The mean pulse rate for males was approximately 69.47, while for females, it was approximately 81.47. This indicates that, on average, females had higher pulse rates.
The median pulse rate for males was 66, compared to 86 for females, further supporting the observation that females generally exhibited higher pulse rates.
Such comparisons are crucial in statistics as they reveal trends and differences across different groups, helping make informed conclusions.

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Most popular questions from this chapter

Find the range, variance, and standard deviation for the given sample data. Include appropriate units (such as "minutes") in your results. (The same data were used in Section 3-1, where we found measures of center. Here we find measures of variation.) Then answer the given questions. Listed below are the numbers of Atlantic hurricanes that occurred in each year. The data are listed in order by year, starting with the year 2000 . What important feature of the data is not revealed by any of the measures of variation? $$ \begin{array}{llllllllllllll} 8 & 9 & 8 & 7 & 9 & 15 & 5 & 6 & 8 & 4 & 12 & 7 & 8 & 2 \end{array} $$

Find the mean and median for each of the two samples, then compare the two sets of results. Waiting times (in seconds) of customers at the Madison Savings Bank are recorded with two configurations: single customer line; individual customer lines. Carefully examine the data to determine whether there is a difference between the two data sets that is not apparent from a comparison of the measures of center. If so, what is it? \(\begin{array}{lllllllllll}\text { Single Line } & 390 & 396 & 402 & 408 & 426 & 438 & 444 & 462 & 462 & 462 \\ \text { Individual Lines } & 252 & 324 & 348 & 372 & 402 & 462 & 462 & 510 & 558 & 600\end{array}\)

Use the following cell phone airport data speeds (Mbps) from Sprint. Find the percentile corresponding to the given data speed. $$\begin{array}{llllllllll} 0.2 & 0.3 & 0.3 & 0.3 & 0.3 & 0.3 & 0.3 & 0.4 & 0.4 & 0.4 \\ 0.5 & 0.5 & 0.5 & 0.5 & 0.5 & 0.6 & 0.6 & 0.7 & 0.8 & 1.0 \\ 1.1 & 1.1 & 1.2 & 1.2 & 1.6 & 1.6 & 2.1 & 2.1 & 2.3 & 2.4 \\ 2.5 & 2.7 & 2.7 & 2.7 & 3.2 & 3.4 & 3.6 & 3.8 & 4.0 & 4.0 \\ 5.0 & 5.6 & 8.2 & 9.6 & 10.6 & 13.0 & 14.1 & 15.1 & 15.2 & 30.4 \end{array}$$ \(9.6 \mathrm{Mbps}\)

Watch out for these little buggers. Each of these exercises involves some feature that is somewhat tricky. Find the (a) mean, \((b)\) median, (c) mode, (d) midrange, and then answer the given question. Listed below are the measured radiation absorption rates (in \(\mathrm{W} / \mathrm{kg}\) ) corresponding to these cell phones: iPhone 5S, BlackBerry Z30, Sanyo Vero, Optimus V, Droid Razr, Nokia N97, Samsung Vibrant, Sony Z750a, Kyocera Kona, LG G2, and Virgin Mobile Supreme. The data are from the Federal Communications Commission (FCC). The media often report about the dangers of cell phone radiation as a cause of cancer. The FCC has a standard that a cell phone absorption rate must be \(1.6 \mathrm{~W} / \mathrm{kg}\) or less. If you are planning to purchase a cell phone, are any of the measures of center the most important statistic? Is there another statistic that is most relevant? If so, which one? $$ \begin{array}{lllllllllll} 1.18 & 1.41 & 1.49 & 1.04 & 1.45 & 0.74 & 0.89 & 1.42 & 1.45 & 0.51 & 1.38 \end{array} $$

Because the mean is very sensitive to extreme values, we say that it is not a resistant measure of center. By deleting some low values and high values, the trimmed mean is more resistant. To find the \(10 \%\) trimmed mean for a data set, first arrange the data in order. then delete the bottom \(10 \%\) of the values and delete the top \(10 \%\) of the values, then calculate the mean of the remaining values. Use the axial loads (pounds) of aluminum cans listed below (from Data Set 30 "Aluminum Cans" in Appendix B) for cans that are \(0.0111\) in. thick. An axial load is the force at which the top of a can collapses. Identify any outliers, then compare the median, mean, \(10 \%\) trimmed mean, and \(20 \%\) trimmed mean. \(\begin{array}{llllllllllll}247 & 260 & 268 & 273 & 276 & 279 & 281 & 283 & 284 & 285 & 286 & 288\end{array}\) \(\begin{array}{llllllll}289 & 291 & 293 & 295 & 296 & 299 & 310 & 504\end{array}\)
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