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Chapter 2: Problem 18

The percentage of iron oxide in each of five pottery samples collected at the Island Thorns site in the United Kingdom \(^{8}\) was as follows: $$ \begin{array}{lllll} 1.28 & 2.39 & 1.50 & 1.88 & 1.51 \end{array} $$

Short Answer

Expert verified
Answer: The mean of the iron oxide content in the pottery samples is 1.712%, the median is 1.51%, and the standard deviation is approximately 0.39116.

Step by step solution

01

Find the mean of the data set.

To find the mean, sum all the data and divide by the number of samples in the data set. The mean, often denoted as \(\bar{x}\), represents the average value of the data set. $$ \bar{x} = \frac{1.28+2.39+1.50+1.88+1.51}{5} = \frac{8.56}{5} = 1.712 $$ The mean of the iron oxide content in the pottery samples is 1.712%.
02

Find the median of the data set.

The median represents the middle value of the data set. First, arrange the data in ascending or descending order, and then find the middle value. For an odd number of samples, the median is the middle value, and for an even number of samples, it is the average of the two middle values. Ordered dataset: $$ \begin{array}{lllll} 1.28 & 1.50 & 1.51 & 1.88 & 2.39 \end{array} $$ Since we have an odd number of samples (5), the median is the middle value: 1.51. The median of the iron oxide content in the pottery samples is 1.51%.
03

Find the standard deviation of the data set.

The standard deviation (denoted as \(\sigma\)) tells us about the spread of our data set. To find the standard deviation, follow these steps: 1. Find the mean (already found in step 1 as \(\bar{x}=1.712\)). 2. Subtract the mean from each data point and square the result. 3. Find the average of the squared differences. 4. Take the square root of the result from step 3. Calculations for step 2 and 3: $$ \sigma^2 = \frac{(1.28-1.712)^2+(2.39-1.712)^2+(1.50-1.712)^2+(1.88-1.712)^2+(1.51-1.712)^2}{5} $$ $$ \sigma^2 = \frac{0.18724+0.46224+0.04496+0.02824+0.04096}{5} = 0.15292 $$ Calculation for step 4: $$ \sigma = \sqrt{0.15292} = 0.39116 $$ The standard deviation of the iron oxide content in the pottery samples is approximately 0.39116.

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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 an essential statistical measure that helps us understand the average of a data set. To calculate the mean, simply add up all the individual data points and then divide this sum by the number of data points. For our pottery samples containing iron oxide, this is how we calculate the mean:
  • Add together each value: 1.28, 2.39, 1.50, 1.88, and 1.51.

  • The total sum is 8.56.

  • Then, we divide by the number of samples, which is 5.

Thus, the mean or average content of iron oxide in the pottery samples is calculated as:\[ \bar{x} = \frac{8.56}{5} = 1.712 \].

This mean value indicates that, on average, each sample contains about 1.712% iron oxide.
Median Calculation
The median is another important measure in descriptive statistics, representing the middle value in an ordered dataset. Unlike the mean, the median is not affected by extremely high or low values. To find the median:
  • First, arrange the numbers in ascending order: 1.28, 1.50, 1.51, 1.88, 2.39.

  • Since there is an odd number of samples (5), the median is simply the middle number.

  • Here, the middle value is 1.51.

Thus, the median of the iron oxide content in the samples is 1.51%.

Using the median gives you the central tendency in the data set, minimizing the impact of outliers that might skew the dataset when finding the mean.
Standard Deviation
The standard deviation is a measure of how spread out the numbers are in your dataset. A smaller standard deviation indicates that values tend to be close to the mean, whereas a larger standard deviation signifies a wider spread. Here鈥檚 how you find the standard deviation in the pottery iron oxide example:
  • Calculate the mean, already found as 1.712.

  • Subtract the mean from each data point and square the result.

  • Find the average of these squared differences.

  • Take the square root of this average to get the standard deviation.

Mathematically, it looks like this:\[ \sigma^2 = \frac{(1.28-1.712)^2+(2.39-1.712)^2+(1.50-1.712)^2+(1.88-1.712)^2+(1.51-1.712)^2}{5} = 0.15292 \]
Then, take the square root:
\[ \sigma = \sqrt{0.15292} = 0.39116 \].

This value of approximately 0.39116 tells us that the iron oxide percentage varies by about 0.39116% from the mean of 1.712%. This gives insight into the consistency of iron oxide content across the samples.

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Most popular questions from this chapter

\(z\) -Scores For the data sets in Exercises \(1-3,\) find the mean the standard deviation, and the z-scores corresponding to the minimum and maximum in the data set. Do the z-scores indicate that there are possible outliers in these data sets? \(n=12\) measurements: 8,7,1,4,6,6,4,5,7,6,3,0

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For each of the data sets in Exercises 7 -9, calculate the five-number summary and the interquartile range. Use this information to construct a box plot and identify any outliers. \(\begin{aligned} n=15 \text { measurements: } & 19,12,16,0,14,9,6,1, \\ & 12,13,10,19,7,5,8 \end{aligned}\)

The weights (in pounds) of 27 packages of ground beef in a supermarket meat display are as follows: $$ \begin{array}{rrrrrrr} 1.08 & .99 & .97 & 1.18 & 1.41 & 1.28 & .83 \\ 1.06 & 1.14 & 1.38 & .75 & .96 & 1.08 & .87 \\ .89 & .89 & .96 & 1.12 & 1.12 & .93 & 1.24 \\ .89 & .98 & 1.14 & .92 & 1.18 & 1.17 & \end{array} $$ a. Draw a stem and leaf plot or a relative frequency histogram to display the weights. Is the distribution relatively mound-shaped? b. Find the mean and the standard deviation of the data set. c. Find the percentage of measurements in the intervals \(\bar{x} \pm s, \bar{x} \pm 2 s,\) and \(\bar{x} \pm 3 s\) d. How do the percentages in part c compare with those given by the Empirical Rule? Explain. e. How many of the packages weigh exactly 1 pound? Can you think of any reason for this?
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