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x\( and \)s\( by Hand. Radon is a naturally occurring gas and is the second leading cause of lung cancer in the United States. \)\frac{12}{}\( It comes from the natural breakdown of uranium in the soil and enters buildings through cracks and other holes in foundations. Radon is found throughout the United States, but levels vary considerably from state to state. Several methods can reduce the levels of radon in a home, and the Environmental Protection Agency recommends using one of them if the measured level in a home is above 4 picocuries per liter. Four readings from Franklin County, Ohio, where the county average is \)8.2\( picocuries per liter, were \)3.8,1.9,12.1\(, and \)14.4\(. a. Find the mean step-by-step. That is, find the sum of the four observations and divide by 4 . b. Find the standard deviation step-by-step. That is, find the deviation of each observation from the mean, square the deviations, and obtain the variance and the standard deviation. Example 2.7 (page 57 ) shows the method. c. Now enter the data into your calculator and use the mean and standard deviation buttons to obtain \)x\( and \)s$. Do the results agree with your hand calculations?

Step by step solution

01

Compute the Mean

To find the mean of the four observations, sum the values and divide by the number of observations. The readings are 3.8, 1.9, 12.1, and 14.4. First, find the sum:\[ 3.8 + 1.9 + 12.1 + 14.4 = 32.2 \] Then, divide the sum by 4 to find the mean:\[ \bar{x} = \frac{32.2}{4} = 8.05 \]

02

Find Deviations from the Mean

Compute the deviation of each observation from the mean calculated in Step 1. The mean is 8.05.- Deviation of 3.8: \( 3.8 - 8.05 = -4.25 \)- Deviation of 1.9: \( 1.9 - 8.05 = -6.15 \)- Deviation of 12.1: \( 12.1 - 8.05 = 4.05 \)- Deviation of 14.4: \( 14.4 - 8.05 = 6.35 \)

03

Square the Deviations

Square each deviation obtained in Step 2 to eliminate negative values.- \((-4.25)^2 = 18.0625\)- \((-6.15)^2 = 37.8225\)- \((4.05)^2 = 16.4025\)- \((6.35)^2 = 40.3225\)

04

Calculate the Variance

Find the variance by averaging the squared deviations found in Step 3.\[ \text{Variance} = \frac{18.0625 + 37.8225 + 16.4025 + 40.3225}{4} = \frac{112.61}{4} = 28.1525 \]

05

Determine the Standard Deviation

The standard deviation is the square root of the variance:\[ s = \sqrt{28.1525} \approx 5.305 \]

06

Calculator Verification

Input the values 3.8, 1.9, 12.1, and 14.4 into a calculator's statistical function and check the mean and standard deviation outputs.The hand calculations should produce: Mean \( \bar{x} = 8.05 \) and Standard Deviation \( s \approx 5.305 \). Verify these values match the calculator results.

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Key Concepts

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

Mean Calculation

The mean is one of the most common measures of central tendency. It gives an average value which represents the center of a dataset. To calculate the mean, also known as the arithmetic average, you add up all the values in the set and then divide by the number of values.
In our example, we have four radon readings: 3.8, 1.9, 12.1, and 14.4 picocuries per liter. The mean of these readings is calculated by first summing them up:

• Sum: 3.8 + 1.9 + 12.1 + 14.4 = 32.2

Then, you divide this total by the number of observations (4 in this case):

• Mean: \( \bar{x} = \frac{32.2}{4} = 8.05 \)

The mean is significant because it provides a single value that summarizes the entire data set, offering a quick glimpse into the distribution of the data.
It is important to remember that the mean can be affected by extremely high or low values, known as outliers. These values can skew the mean, making it not always an ideal representation of the central tendency for skewed distributions.

Standard Deviation

The standard deviation is a measure that indicates the amount of variation or dispersion in a set of values. A low standard deviation suggests that the values are close to the mean, while a high standard deviation indicates that the values are spread out over a wider range.
To compute the standard deviation, follow these steps:
- First, calculate the deviation of each observation from the mean:

• Deviation of 3.8: \( 3.8 - 8.05 = -4.25 \)
• Deviation of 1.9: \( 1.9 - 8.05 = -6.15 \)
• Deviation of 12.1: \( 12.1 - 8.05 = 4.05 \)
• Deviation of 14.4: \( 14.4 - 8.05 = 6.35 \)

- Next, square each of these deviations to make all values positive. This step helps handle negative deviations:

• \((-4.25)^2 = 18.0625\)
• \((-6.15)^2 = 37.8225\)
• \((4.05)^2 = 16.4025\)
• \((6.35)^2 = 40.3225\)

- Lastly, take the square root of the average of these squared deviations (this average is known as variance) to get the standard deviation:

• Standard Deviation: \( s = \sqrt{28.1525} \approx 5.305 \)

The standard deviation is indispensable in statistics because it helps quantify the uncertainty or risk associated with a set of data. Additionally, it greatly enhances interpretation by providing context about how much variation exists among data points.

Variance Calculation

Variance is a fundamental concept in statistics and helps quantify the degree of spread in a data set. Essentially, it is the average of the squared deviations from the mean. Variance offers insight into the level of dispersion, similar to standard deviation, but expressed in squared units.
To calculate the variance, you perform the following steps:

• First, square the deviations of each observation from the mean as calculated previously.

This results from our example being:

• 18.0625, 37.8225, 16.4025, 40.3225

Then, average these squared deviations:

• Variance: \( \frac{18.0625 + 37.8225 + 16.4025 + 40.3225}{4} = 28.1525 \)

Variance is particularly useful when comparing variability between two different data sets, as it clearly depicts how data points are spread out around the mean. However, note that because variance is in squared units, it may not always be as directly interpretable as the standard deviation. This is why often, standard deviation plays a complementary role, providing a clearer metric for assessing spread in the original units of the data.