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River Lengths (Example 2) The table shows the lengths (in miles) of major rivers in North America. (Source: World Almanac and Book of Facts 2017) $$ \begin{array}{|lc|} \hline \text { River } & \text { Length (in miles) } \\ \hline \text { Arkansas } & 1459 \\ \hline \text { Colorado } & 1450 \\ \hline \text { Mackenzie } & 2635 \\ \hline \text { Mississippi-Missouri-Red Rock } & 3710 \\ \hline \text { Rio Grande } & 1900 \\ \hline \end{array} $$ a. Find and interpret (report in context) the mean, rounding to the nearest tenth mile. Be sure to include units for your answer. b. Find the standard deviation, rounding to the nearest tenth mile. Be sure to include units for your answer. Which river contributes most to the size of the standard deviation? Explain. c. If the St. Lawrence River (length 800 miles) were included in the data set, explain how the mean and standard deviation from parts (a) and (b) would be affected? Then recalculate these values including the St. Lawrence River to see if your prediction was correct.

Short Answer

Expert verified
a. The mean length of the rivers is 2230.8 miles. b. The standard deviation is 904.83 miles with the Mississippi-Missouri-Red Rock River contributing most due to its length being significantly longer than the other river's lengths. c. When the St. Lawrence River is included, the mean reduces to 1992.33 miles and the standard deviation is inflated to 965.01 miles.

Step by step solution

01

Calculate the Mean

Add together the lengths of all the rivers (1459+1450+2635+3710+1900) and then divide by the number of rivers, which is 5. This will give the mean value.
02

Calculate the Standard Deviation

To find the standard deviation, each river's length needs to be subtracted from the mean, squared, and then summed up. This sum, divided by the number of rivers and then taking the square root, gives the standard deviation. The river that has a value furthest away from the mean will contribute the most to the standard deviation.
03

Predict Changes

To predict how the mean and standard deviation will change when the St. Lawrence River is added, it can be assumed that the mean will decrease because the length of the St. Lawrence River is less than the initial mean. The standard deviation might increase, because a smaller value increases the distance between individual measurements and the mean.
04

Recalculate Mean and Standard Deviation

Recalculate the mean and standard deviation including the St. Lawrence River in the same way as in Step 1 and Step 2, but now divide by 6 rivers.

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Key Concepts

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

Mean Calculation
Understanding the mean, or average, of a data set is fundamental in statistics. When calculating the mean river length, we simply add up the lengths of all specified rivers and then divide by the number of rivers. In the exercise example, we add the lengths of the Arkansas, Colorado, Mackenzie, Mississippi-Missouri-Red Rock, and Rio Grande, then divide by 5. The mean is a measure of the central tendency of the data, offering a quick glimpse at what a typical value might be.

For example, the mean calculated from the provided lengths is \( (1459 + 1450 + 2635 + 3710 + 1900) / 5 \), which gives us the average length. The mean helps us understand the general size scope of rivers in North America, although it does not tell us about the distribution or variability of these lengths.
Standard Deviation
Standard deviation is a measure of the amount of variation or dispersion in a set of values. A low standard deviation means that most numbers are close to the mean, while a high standard deviation means that the values are spread out over a wider range. In our example, after finding the mean river length, we calculate the standard deviation by subtracting each river's length from the mean, squaring the result, summing these squares, dividing by the number of rivers, and finally taking the square root of that result. This gives us an understanding of how much the river lengths differ from the average length.

For instance, the Mississippi-Missouri-Red Rock, which has the greatest length, is likely to be the river that contributes the most to the standard deviation, because its length is much further from the mean compared to the other rivers. Recognizing which data points contribute most to the standard deviation is valuable for interpreting the variability of the dataset.
Data Interpretation
Interpreting data goes beyond just calculating statistical measures; it involves making sense of those numbers in context. For example, the mean gives us an indication of the middle ground of river lengths, while the standard deviation tells us how much the rivers' lengths vary from this central value. In our exercise, interpreting the mean and standard deviation in the context of North American rivers helps us understand the geographical landscape in terms of river sizes.

Adding the St. Lawrence River to the original data set—which is considerably shorter than the mean—would result in a lower overall mean. This introduces a new perspective on the diversity of river sizes. Meanwhile, the standard deviation would likely increase, indicating a wider variation in river lengths as we add more data points, particularly ones that diverge significantly from the original mean. This is an essential skill, as it enables students to draw conclusions and make predictions based on statistical analysis.

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Most popular questions from this chapter

The top seven movies based on DC comic book characters for the U.S. box office as of fall 2017 are shown in the following table, rounded to the nearest hundred million. (Source: ultimatemovieranking.com) a. Find and interpret the median in context. b. Find and interpret the IQR in context. c. Find the range of the data. Explain why the IQR is preferred over the range as a measure of variability. $$ \begin{array}{|lc|} \hline \text { Movie } & \begin{array}{c} \text { Adjusted Domestic } \\ \text { Gross (\$ millions) } \end{array} \\ \hline \text { The Dark Knight (2008) } & \$ 643 \\ \text { Batman (1989) } & \$ 547 \\ \hline \text { Superman (1978) } & \$ 543 \\ \hline \text { The Dark Knight Rises (2012) } & \$ 487 \\ \text { Wonder Woman (2017) } & \$ 407 \\ \text { Batman Forever (1995) } & \$ 366 \\ \hline \text { Superman II (1981) } & \$ 346 \\ \hline \end{array} $$

A dieter recorded the number of calories he consumed at lunch for one week. As you can see, a mistake was made on one entry. The calories are listed in increasing order: $$ 331,374,387,392,405,4200 $$ When the error is corrected by removing the extra 0, will the median calories change? Will the mean? Explain without doing any calculations.

Construct two sets of numbers with at least five numbers in each set with the following characteristics: The mean of set \(A\) is smaller than that of set \(B\), but the median of set \(B\) is smaller than that of set A. Report the mean and the median of both sets of data.

Is it possible for a standard deviation to be negative? Explain.

Babies born weighing 2500 grams (about \(5.5\) pounds) or less are called low- birthweight babies, and this condition sometimes indicates health problems for the infant. The mean birth weight for U.S.-born children is about 3462 grams (about \(7.6\) pounds). The mean birth weight for babies born one month early is 2622 grams. Suppose both standard deviations are 500 grams. Also assume that the distribution of birth weights is roughly unimodal and symmetric. (Source: www.babycenter.com) a. Find the standardized score \((z\) -score), relative to all U.S. births, for a baby with a birth weight of 2500 grams. b. Find the standardized score for a birth weight of 2500 grams for a child born one month early, using 2622 as the mean. c. For which group is a birth weight of 2500 grams more common? Explain what that implies. Unusual z-scores are far from \(0 .\)

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