Question

Find the range and use it to approximate the value of s. Then calculate the actual value of \(s\). Is the actual value close to the estimate? The length of time it takes for a driver to respond to a particular emergency situation was recorded for 10 drivers. The times (in seconds) were \(0.5,0.8,1.1,0.7,0.6,0.9,0.7,0.8,0.7,0.8 .\)

Short answer

Answer: The range of the given response times is 0.6 seconds. The actual value of s (standard deviation) is approximately 0.0766 seconds. The initial estimate obtained using the range for s was between 0.15 seconds and 0.3 seconds, which does not accurately represent the actual value.

Step-by-step solution

Calculate the Range

To find the range, first list the given response times in ascending order and then subtract the smallest time from the largest time: \(0.5, 0.6, 0.7, 0.7, 0.7, 0.8, 0.8, 0.8, 0.9, 1.1\) Range = Largest Time - Smallest Time = \(1.1 - 0.5 = 0.6\) seconds.

Approximate the value of s using the Range

To approximate the value of s (assuming s represents standard deviation) using the range, we often use the following inequality for the Standard Deviation: \(Range/4 \leq s \leq Range/2\) So we have: \(0.6/4 \leq s \leq 0.6/2\) \(0.15 \leq s \leq 0.3\) Hence, the approximate value of s lies between 0.15 seconds and 0.3 seconds.

Calculate the actual value of s (Standard Deviation)

To calculate the actual value of the standard deviation, first, find the mean of the given response times: Mean = \(\frac{0.5+0.6+0.7+0.7+0.7+0.8+0.8+0.8+0.9+1.1}{10} = \frac{7.7}{10} = 0.77\) seconds. Next, find the deviations of each time from the mean and square them: \((0.27)^2\), \((0.17)^2\), \((0.07)^2\), \((0.07)^2\), \((0.07)^2\), \((0.03)^2\), \((0.03)^2\), \((0.03)^2\), \((0.13)^2\), \((0.33)^2\) Now, find the average of these squared deviations: Variance = \(\frac{\Sigma(X_i -\bar{X})^2}{n}\) Variance = \(\frac{(0.27)^2 +(0.17)^2 +(0.07)^2 +(0.07)^2 +(0.07)^2 +(0.03)^2 +(0.03)^2 +(0.03)^2 +(0.13)^2 +(0.33)^2}{10}\) Variance ≈ 0.00587 Finally, find the square root of the variance to get the actual value of s: \(s=\sqrt{Variance}=\sqrt{0.00587} \approx 0.0766\) seconds

Compare the Actual Value to the Estimate

Our initial estimate for s using the range was between 0.15 seconds and 0.3 seconds. The actual value of s is approximately 0.0766 seconds. The actual value of s is not close to the estimate obtained from the range since it is outside the interval [\(0.15 , 0.30\)]. In conclusion, the range is not an accurate method to approximate the standard deviation, and the actual value is not close to this estimate. The actual value of s (approximately 0.0766 seconds) represents how much the response times deviate from their mean value (0.77 seconds).

Key concepts

Range

The range is a simple measure of how varied or spread out a dataset is. For any given set of numbers, the range is calculated by subtracting the smallest number from the largest number. In the context of our driver response times, which are listed in ascending order as \(0.5, 0.6, 0.7, 0.7, 0.7, 0.8, 0.8, 0.8, 0.9, 1.1\), the range is \(1.1 - 0.5 = 0.6\) seconds.
This tells us that the driver's response times span a duration of 0.6 seconds.
While the range is easy to compute, it is not very informative about the distribution of numbers between the smallest and largest values.

Mean

The mean, or average, is a measure of the central tendency of a dataset. It gives us an idea of the typical value within the data set by adding up all the values and dividing by the number of values.
In our driver's response times example, the mean is calculated as follows:

• Add all the response times: \(0.5 + 0.6 + 0.7 + 0.7 + 0.7 + 0.8 + 0.8 + 0.8 + 0.9 + 1.1 = 7.7\)
• Divide by the number of observations: \(7.7/10\)

Thus, the mean response time is \(0.77\) seconds.
The mean provides a single value that summarizes all the data, but it doesn’t give information on how the individual responses are distributed around this average.

Variance

Variance is a statistical measure that provides insight into the spread of a dataset. It looks at how far each data point is from the mean and how these differences are spread out in the dataset.
To calculate the variance for our drivers' response times, you need to:

• Find the deviation of each time from the mean \( (0.77) \).
• Square each of these deviations to eliminate negative values.
• Take the average of these squared deviations.

The formula for variance is \[Variance = \frac{\sum (X_i - \bar{X})^2}{n}\]where \( X_i \) represents each observation, and \(n\) is the number of observations.
After calculating, the variance in our example is approximately \(0.00587\). This small variance indicates that the response times do not vary widely from the mean, suggesting consistency among the drivers.

Approximation Method

When estimating the standard deviation (\(s\)), without performing detailed calculations, a quick approximation method uses the range.
This method involves dividing the range by a number between 4 and 2. The approximation represented as follows:

• \(\text{Range/4} \leq s \leq \text{Range/2}\)

This inequality gives a rough estimate of where the standard deviation might lie. In our driver's example, with a range of 0.6, the approximation is \(0.15 \leq s \leq 0.3\).
However, observation shows this method does not always accurately predict the actual variance. In our example, the actual standard deviation calculated was \(0.0766\), outside the estimated range.
This indicates that while this method can give a quick reference, it can lack precision, hence detailed calculations should be preferred for more accurate data analyses.