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Chapter 3: Problem 9

The following data set belongs to a population: 5 \(\begin{array}{llllll}2 & 0 & -9 & 16 & 10 & 7\end{array}\)

Short Answer

Expert verified
The mean is \(26/6\approx 4.33\), the median is \(4.5\), there is no mode and the standard deviation can be calculated as described above.

Step by step solution

01

Calculating the mean

The mean is the sum of all data values divided by the count of values. Calculate the sum of all data values \(2 + 0 + -9 + 16 + 10 + 7\) and divide it by 6 which is the count of data values.
02

Calculating the median

First arrange the numbers in increasing order: \(-9, 0, 2, 7, 10, 16\). The median is the middle number in a set of ordered data values. Because the count of data values is even (6), the median is the arithmetic mean of the two middlemost numbers. In this case it is the arithmetic mean of 2 and 7.
03

Finding the mode

The mode is the value that appears most frequently in a data set. In this case, all numbers appear exactly once, so the data set has no mode.
04

Calculating the standard deviation

First find the mean of the data set.\nThen, subtract the mean from each data point and square the result. Sum up these squared results. Divide this sum by the count of data points (6 in this case) to get the variance. Finally, take the square root of the variance to find the standard deviation.

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Key Concepts

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

Mean Calculation
The mean provides us with the average value of a data set. This is a simple yet powerful concept in descriptive statistics. To find the mean, we need to sum all the data values and then divide this sum by the number of data points.

Let's consider the data set: 2, 0, -9, 16, 10, and 7. Here, we begin by calculating the sum of these numbers:
  • 2 + 0 + (-9) + 16 + 10 + 7 = 26
Then, we count the items in our data set, giving us 6 data points.

Finally, divide the total sum by the number of points:
  • Mean = \( \frac{26}{6} \approx 4.33 \)
The mean for this data set is approximately 4.33. This tells us, on average, where the data values tend to cluster.
Median Determination
The median is the value that lies in the middle of a data set when the numbers are in order. It is known as a measure of central tendency and helps us understand the distribution of the data values.

For our data set: -9, 0, 2, 7, 10, and 16, we must first arrange the numbers in increasing order:
  • -9, 0, 2, 7, 10, 16
As there are 6 numbers, an even count, the median will be the average of the two central numbers. These are 2 and 7.

To find their mean:
  • Median = \( \frac{2 + 7}{2} = 4.5 \)
Thus, 4.5 is the median, indicating that half of the data points are below this value and half are above.
Mode Identification
The mode is the value that appears most frequently in a data set. It helps identify the most common value but can be quite tricky if no value repeats.

In our example data set: 2, 0, -9, 16, 10, and 7, we need to check the frequency of each number.
  • Each number appears exactly once.
In this special case, the data set has no mode because no number repeats. Sometimes data sets can have one mode, more than one mode (bimodal or multimodal), or no mode at all. Lack of a mode indicates there is no common trend or value.
Standard Deviation Calculation
Standard deviation is a measure of data dispersion or how spread out the numbers in a data set are around the mean. A low standard deviation means the values are close to the mean, while a high one indicates wide variation.

To calculate standard deviation, follow these steps:
  • First, find the mean of the data set. We've already calculated it as 4.33.
  • Subtract the mean from each number and square each result:
    • (2 - 4.33)虏 = 5.4289
    • (0 - 4.33)虏 = 18.7489
    • (-9 - 4.33)虏 = 175.8289
    • (16 - 4.33)虏 = 136.2889
    • (10 - 4.33)虏 = 32.4489
    • (7 - 4.33)虏 = 7.1289
  • Sum these squared differences:
    • Total = 5.4289 + 18.7489 + 175.8289 + 136.2889 + 32.4489 + 7.1289 = 375.8724
  • Divide by the number of data points (6) to get the variance:
    • Variance = \( \frac{375.8724}{6} \approx 62.6454 \)
  • Finally, take the square root of the variance:
    • Standard Deviation = \( \sqrt{62.6454} \approx 7.914 \)
The standard deviation for this data set is approximately 7.914, showing moderate spread around the mean.

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Most popular questions from this chapter

The following stem-and-leaf diagram gives the distances (in thousands of miles) driven during the past year by a sample of drivers in a city. $$ \begin{array}{l|llllll} 0 & 3 & 6 & 9 & & & \\ 1 & 2 & 8 & 5 & 1 & 0 & 5 \\ 2 & 5 & 1 & 6 & & & \\ 3 & 8 & & & & & \\ 4 & 1 & & & & & \\ 5 & & & & & & \\ 6 & 2 & & & & \end{array} $$ a. Compute the sample mean, median, and mode for the data on distances driven. b. Compute the range, variance, and standard deviation for these data. c. Compute the first and third quartiles. d. Compute the interquartile range. Describe what properties the interquartile range has. When would the IQR be preferable to using the standard deviation when measuring variation?

The following data give the 2010 gross domestic product (in billions of dollars) for all 50 states. The data are entered in alphabetical order by state (Source: Bureau of Economic Analysis). \(\begin{array}{rrrrrrrrrr}173 & 49 & 254 & 103 & 1901 & 258 & 237 & 62 & 748 & 403 \\ 67 & 55 & 652 & 276 & 143 & 127 & 163 & 219 & 52 & 295 \\ 379 & 384 & 270 & 97 & 244 & 36 & 90 & 126 & 60 & 487 \\ 80 & 1160 & 425 & 35 & 478 & 148 & 174 & 570 & 49 & 164 \\\ 40 & 255 & 1207 & 115 & 26 & 424 & 340 & 65 & 248 & 39\end{array}\) a. Calculate the mean and median for these data. Are these values of the mean and the median sample statistics or population parameters? Explain. b. Do these data have a mode? Explain.

The mean age of six persons is 46 years. The ages of five of these six persons are \(57,39,44,51\), and 37 years, respectively. Find the age of the sixth person.

The following table gives the standard deductions and personal exemptions for persons filing with "single" status on their 2011 state income taxes in a random sample of 9 states. Calculate the mean and median for the data on standard deductions for these states. $$ \begin{array}{lcc} \hline \text { State } & \begin{array}{c} \text { Standard Deduction } \\ \text { (in dollars) } \end{array} & \begin{array}{c} \text { Personal Exemption } \\ \text { (in dollars) } \end{array} \\ \hline \text { Delaware } & 3250 & 110 \\ \text { Hawaii } & 2000 & 1040 \\ \text { Kentucky } & 2190 & 20 \\ \text { Minnesota } & 5450 & 3500 \\ \text { North Dakota } & 5700 & 3650 \\ \text { Oregon } & 1945 & 176 \\ \text { Rhode Island } & 5700 & 3650 \\ \text { Vermont } & 5700 & 3650 \\ \text { Virginia } & 3000 & 930 \\ \hline \end{array} $$

The following data give the numbers of text messages sent by a high school student on 40 randomly selected days during 2012: \(\begin{array}{llllllllll}32 & 33 & 33 & 34 & 35 & 36 & 37 & 37 & 37 & 37 \\\ 38 & 39 & 40 & 41 & 41 & 42 & 42 & 42 & 43 & 44 \\ 44 & 45 & 45 & 45 & 47 & 47 & 47 & 47 & 47 & 48 \\ 48 & 49 & 50 & 50 & 51 & 52 & 53 & 54 & 59 & 61\end{array}\) a. Calculate the values of the three quartiles and the interquartile range. Where does the value 49 fall in relation to these quartiles? b. Determine the approximate value of the 91 st percentile. Give a brief interpretation of this percentile. c. For what percentage of the days was the number of text messages sent 40 or higher? Answer by finding the percentile rank of 40 .
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