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The following data represent the amount of time (in minutes) a random sample of eight students took to complete the online portion of an exam in Sullivan's Statistics course. Compute the range, sample variance, and sample standard deviation time. $$ 60.5,128.0,84.6,122.3,78.9,94.7,85.9,89.9 $$

Short Answer

Expert verified
Range: 67.5, Variance: 495.0034571, Standard Deviation: 22.248

Step by step solution

01

- Arrange the Data

First, arrange the data in ascending order: 60.5, 78.9, 84.6, 85.9, 89.9, 94.7, 122.3, 128.0
02

- Compute the Range

The range is the difference between the maximum and minimum values. Here, the range is: \( 128.0 - 60.5 = 67.5 \)
03

- Compute the Mean

Calculate the mean (average) of the data: \[ \text{Mean} = \frac{60.5 + 78.9 + 84.6 + 85.9 + 89.9 + 94.7 + 122.3 + 128.0}{8} = \frac{745.8}{8} = 93.225 \]
04

- Compute Each Deviation from the Mean

Subtract the mean from each data point and square the result: \((60.5 - 93.225)^2 = 1067.700625,\ (78.9 - 93.225)^2 = 203.715625,\ (84.6 - 93.225)^2 = 74.592025,\ (85.9 - 93.225)^2 = 52.802025,\ (89.9 - 93.225)^2 = 11.012025,\ (94.7 - 93.225)^2 = 2.175625,\ (122.3 - 93.225)^2 = 852.225625,\ (128.0 - 93.225)^2 = 1201.800625 \)
05

- Compute the Sum of Squared Deviations

Sum the squared deviations: \[ 1067.700625 + 203.715625 + 74.592025 + 52.802025 + 11.012025 + 2.175625 + 852.225625 + 1201.800625 = 3465.0242 \]
06

- Calculate the Sample Variance

Divide the sum of squared deviations by the degrees of freedom (n-1): \[ s^2 = \frac{3465.0242}{8-1} = \frac{3465.0242}{7} = 495.0034571 \]
07

- Calculate the Sample Standard Deviation

Take the square root of the variance to get the standard deviation: \[ s = \sqrt{495.0034571} \approx 22.248 \]

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Key Concepts

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

Range
The range is a simple descriptive statistic that indicates the spread or dispersion of a data set. It tells us the difference between the highest and lowest values. To compute the range, you subtract the smallest number from the largest number in your data set.

In our example with the times: 60.5, 128.0, 84.6, 122.3, 78.9, 94.7, 85.9, 89.9 minutes, we arranged the data in ascending order first: 60.5, 78.9, 84.6, 85.9, 89.9, 94.7, 122.3, 128.0.

The range is calculated as:
\[\text{Range} = 128.0 - 60.5 = 67.5 \]

This result means that the amount of time taken by students varies by 67.5 minutes between the fastest and slowest students.
Sample Variance
Sample variance measures the average of the squared differences from the mean. It gives us an idea about the spread of data points. Here's how to compute it step-by-step.

1. **Calculate the mean (average) of data**: Add up all values and divide by the number of values.
In our case:
\[\text{Mean} = \frac{60.5 + 78.9 + 84.6 + 85.9 + 89.9 + 94.7 + 122.3 + 128.0}{8} = \frac{745.8}{8} = 93.225 \]
2. **Find the deviations from the mean**: Subtract the mean from each data point and square these deviations.
For example, \((60.5 - 93.225)^2 = 1067.700625\)
Continue this for all values.
3. **Calculate the sum of squared deviations**: Add all squared deviations:
\[\text{Sum} = 1067.700625 + 203.715625 + 74.592025 + 52.802025 + 11.012025 + 2.175625 + 852.225625 + 1201.800625 = 3465.0242 \]
4. **Divide by the degrees of freedom (number of values minus one)**: This adjusts for the fact we are using a sample, not the entire population.
\[\text{Sample Variance } (s^2) = \frac{3465.0242}{8-1} = \frac{3465.0242}{7} = 495.0034571 \]

So, our sample variance is approximately 495.0034571, indicating how spread out the exam completion times are from the mean.
Sample Standard Deviation
The sample standard deviation is a measure that indicates the typical distance of data points from the mean of the sample. It is simply the square root of the sample variance. This value gives us a sense of the average spread of data points.

To compute the sample standard deviation:
1. **Start with the sample variance**: We already calculated our sample variance to be 495.0034571.
2. **Take the square root of the variance**: This step is crucial as it transforms our variance into the same units as the original data.
\[\text{Sample Standard Deviation} (s) = \sqrt{495.0034571} \]
\[\text{Sample Standard Deviation} (s) \approx 22.248 \]

The sample standard deviation is approximately 22.248 minutes. This number indicates that, on average, the exam completion times vary by about 22.248 minutes from the mean time of 93.225 minutes.

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