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A sample of 12 drivers shows the time that they spent (in minutes) stopped in rush-hour traffic on a specific snowy day last winter. Find the range, variance, and standard deviation for the data. \(\begin{array}{lll}52 & 56 & 53 \\ 61 & 49 & 51\end{array}\) \(\begin{array}{lll}53 & 58 & 53 \\ 60 & 71 & 58\end{array}\)

Short Answer

Expert verified
The range is 22, variance is approximately 32.57, and standard deviation is approximately 5.71.

Step by step solution

01

List the Data

First, list all the time values of the drivers: 52, 56, 53, 61, 49, 51, 53, 58, 53, 60, 71, 58. Double-check to ensure all values are included.
02

Calculate the Range

The range is the difference between the highest and lowest values in the dataset. Identify the maximum value, 71, and the minimum value, 49. Calculate the range: \( 71 - 49 = 22 \).
03

Calculate the Mean

Calculate the mean by adding all the data values and dividing by the number of values. The sum is \( 52 + 56 + 53 + 61 + 49 + 51 + 53 + 58 + 53 + 60 + 71 + 58 = 675 \). Divide by 12 (number of values): \( \frac{675}{12} = 56.25 \).
04

Find Deviations from the Mean

Subtract the mean from each data point to find deviations: \(52 - 56.25 = -4.25,\ 56 - 56.25 = -0.25,\ 53 - 56.25 = -3.25,\ 61 - 56.25 = 4.75,\ 49 - 56.25 = -7.25,\ 51 - 56.25 = -5.25,\ 53 - 56.25 = -3.25,\ 58 - 56.25 = 1.75,\ 53 - 56.25 = -3.25,\ 60 - 56.25 = 3.75,\ 71 - 56.25 = 14.75,\ 58 - 56.25 = 1.75\).
05

Calculate Squared Deviations

Square each deviation calculated in Step 4: \((-4.25)^2 = 18.06,\ (-0.25)^2 = 0.06,\ (-3.25)^2 = 10.56,\ (4.75)^2 = 22.56,\ (-7.25)^2 = 52.56,\ (-5.25)^2 = 27.56,\ (-3.25)^2 = 10.56,\ (1.75)^2 = 3.06,\ (-3.25)^2 = 10.56,\ (3.75)^2 = 14.06,\ (14.75)^2 = 217.56,\ (1.75)^2 = 3.06\).
06

Sum the Squared Deviations

Add the squared deviations: \(18.06 + 0.06 + 10.56 + 22.56 + 52.56 + 27.56 + 10.56 + 3.06 + 10.56 + 14.06 + 217.56 + 3.06 = 390.8 \).
07

Calculate Variance

Divide the sum of squared deviations by the number of data points to find the variance: \( \frac{390.8}{12} \approx 32.57 \).
08

Calculate Standard Deviation

The standard deviation is the square root of the variance: \( \sqrt{32.57} \approx 5.71 \).

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Key Concepts

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

Range Calculation
In the world of descriptive statistics, the range offers a straightforward glimpse into how spread out a set of numbers are. It is one of the simplest measures of variability in a dataset.

To calculate the range, the first step is identifying the maximum and minimum values in your dataset. For the example provided, these are 71 minutes and 49 minutes respectively. Subtracting the smallest number from the largest one gives you the range:

- Maximum value = 71
- Minimum value = 49
- Range = Maximum value - Minimum value = 71 - 49 = 22

This information tells us that the longest time a driver was stopped was 22 minutes more than the shortest time. The range provides a quick way to see the extent of traffic hold-ups experienced by this group on that snowy day.
Variance Calculation
Variance takes range a step further by delving into how much each number in the dataset varies from the mean. It gives you more insight than the range by considering each data point's spread around the average.

First, find the mean (average) of the data by adding up all the values and dividing by the count of numbers:
  • Mean = Total sum of values / Number of values = 675 / 12 = 56.25 minutes
The next step involves calculating the deviation of each number from the mean. Deviations are simply the differences found by subtracting the mean from each data point.

Next, you square each deviation to get rid of negative numbers and emphasize larger discrepancies:

- Example for a few squared deviations:
  • Deviation for 52 minutes: \((52 - 56.25)^2 = 18.06\)
  • Deviation for 61 minutes: \((61 - 56.25)^2 = 22.56\)

After squaring, sum up all these squared deviations to get 390.8. Finally, divide by the number of observations (12 in this case) to find the variance:

Variance = Sum of squared deviations / Number of data points = 390.8 / 12 = 32.57

Variance, therefore, shows how the data points differ from the mean, highlighting the average of the squared differences from the mean. A higher variance means more spread out data.
Standard Deviation Calculation
Standard deviation is the natural follow-up to variance and provides a measure of how spread out the numbers in a dataset are in the same unit as the data itself.

After calculating the variance (32.57 in this instance), the standard deviation is simply the square root of this figure. This is a key step because it converts squared units back into the original units of the data (in this example, minutes).

- Standard Deviation = \( \sqrt{32.57} \approx 5.71 \)

The standard deviation of approximately 5.71 minutes provides a clear view of variability within the dataset. It tells us that on average, each data point (the time drivers spent in traffic) is about 5.71 minutes away from the mean time of 56.25 minutes.

In practical terms, this means there's a moderate spread around the typical time spent in rush-hour traffic, giving us a more complete picture than the range or variance alone.

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Most popular questions from this chapter

The data shown are the total compensation (in millions of dollars) for the 50 top-paid CEOs for a recent year. Compare the averages, and state which one you think is the best measure. $$\begin{array}{lllll}17.5 & 18.0 & 36.8 & 31.7 & 31.7 \\\17.3 & 24.3 & 47.7 & 38.5 & 17.0 \\\23.7 & 16.5 & 25.1 & 17.4 & 18.0 \\\37.6 & 19.7 & 21.4 & 28.6 & 21.6 \\\19.3 & 20.0 & 16.9 & 25.2 & 19.8 \\\25.0 & 17.2 & 20.4 & 20.1 & 29.1 \\\19.1 & 25.2 & 23.2 & 25.9 & 24.0 \\\41.7 & 24.0 & 16.8 & 26.8 & 31.4 \\\16.9 & 17.2 & 24.1 & 35.2 & 19.1 \\\22.9 & 18.2 & 25.4 & 35.4 & 25.5\end{array}$$

Forbes magazine prints an annual Top-Earning Nonliving Celebrities list (based on royalties and estate earnings). Find the mean, median, mode, and midrange for the data. Comment on the skewness. Figures represent millions of dollars. $$\begin{array}{llll}\text { Kurt Cobain } & 50 & \text { Ray Charles } & 10 \\\\\text { Elvis Presley } & 42 & \text { Marilyn Monroe } & 8 \\\\\text { Charles M. Schulz } & 35 & \text { Johnny Cash } & 8 \\\\\text { John Lennon } & 24 & \text { J.R.R. Tolkien } & 7 \\\\\text { Albert Einstein } & 20 & \text { George Harrison } & 7 \\ \text { Andy Warhol } & 19 & \text { Bob Marley } & 7 \\\\\text { Theodore Geisel } & 10 & &\end{array}$$

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Americans spend an average of 3 hours per day online. If the standard deviation is 32 minutes, find the range in which at least \(88.89 \%\) of the data will lie. Use Chebyshev's theorem.

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