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Chapter 3: Problem 11

Of the 25 brightest stars, the distances from earth (in light-years) for those with distances less than 100 light-years are found below. Find the mean, median, mode, and midrange for the data. $$\begin{array}{lllllll}8.6 & 36.7 & 42.2 & 16.8 & 33.7 & 77.5 & 87.9 \\\4.4 & 25.3 & 11.4 & 65.1 & 25.1 & 51.5 &\end{array}$$

Short Answer

Expert verified
Mean: 37.4, Median: 33.7, Mode: None, Midrange: 46.15

Step by step solution

01

Arrange the Data

First, arrange the data set in increasing order to facilitate calculation of the median and mode. The ordered set is: 4.4, 8.6, 11.4, 16.8, 25.1, 25.3, 33.7, 36.7, 42.2, 51.5, 65.1, 77.5, 87.9.
02

Calculate the Mean

The mean is the sum of all the data points divided by the number of points. Add all the distances: \[ 4.4 + 8.6 + 11.4 + 16.8 + 25.1 + 25.3 + 33.7 + 36.7 + 42.2 + 51.5 + 65.1 + 77.5 + 87.9 = 486.2 \]Then divide by the total number of data points, which is 13: \[ \text{Mean} = \frac{486.2}{13} \approx 37.4 \text{ light-years} \]
03

Determine the Median

The median is the middle value of a data set. With 13 values, the 7th value is the median. By observation, the 7th value in the ordered set is 33.7 light-years. Therefore, the median is 33.7.
04

Identify the Mode

The mode is the value that appears most frequently in the data set. In this data set, no number repeats. Therefore, there is no mode.
05

Calculate the Midrange

The midrange is the average of the minimum and maximum values. The minimum value is 4.4, and the maximum value is 87.9:\[ \text{Midrange} = \frac{4.4 + 87.9}{2} = \frac{92.3}{2} = 46.15 \text{ light-years} \]

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Key Concepts

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

Mean Calculation
Mean calculation is a fundamental concept in descriptive statistics, serving as a measure of central tendency that represents the average of a data set. To find the mean, you begin by summing all the individual data points together. In this case, you would add up the light-year distances of the bright stars:
  • 4.4
  • 8.6
  • 11.4
  • 16.8
  • 25.1
  • 25.3
  • 33.7
  • 36.7
  • 42.2
  • 51.5
  • 65.1
  • 77.5
  • 87.9
The total sum of these numbers is 486.2. Next, you divide this sum by the total number of data points, which in our example is 13. This division yields the mean: \[ \text{Mean} = \frac{486.2}{13} \approx 37.4 \text{ light-years} \]This value tells us that, on average, the stars are approximately 37.4 light-years away from Earth.
Median Determination
To determine the median, which is another measure of central tendency, we first need to arrange the data in ascending order. This assists in easily locating the middle value. In our exercise, the ordered distances are:
4.4, 8.6, 11.4, 16.8, 25.1, 25.3, 33.7, 36.7, 42.2, 51.5, 65.1, 77.5, 87.9. Since there are 13 data values 鈥� an odd number鈥� the median is simply the 7th value in this ordered list.
In this case, that value is 33.7 light-years, making it the median distance.
  • The median provides the middle point of the data, showing that half of the star distances are less than 33.7 light-years and the other half are greater.
  • This can sometimes be more informative than the mean if the dataset contains outliers, as it is not affected by extremely high or low values.
Mode Identification
The mode is a simple measure of central tendency that identifies the value that occurs most frequently in a dataset. However, in some datasets 鈥� like the one we have here 鈥� it is possible for there to be no mode. In our dataset of star distances in light-years, each number appears exactly once. As a result:
  • No value is repeated.
  • The dataset is considered to be "mode-less" or having "no mode" because no value satisfies the condition of appearing more frequently than others.
Even though a mode is a useful measure in data where repetition occurs, it's perfectly normal and acceptable for some datasets to have no mode at all.
Midrange Calculation
Calculating the midrange involves determining the arithmetic mean of the smallest and largest values in your dataset. This calculation provides a simple measure of central tendency that gives a sense of the center or balance of data from its extremes. In this example, the smallest distance is 4.4 light-years, and the largest is 87.9 light-years. The midrange calculation is done as follows:\[ \text{Midrange} = \frac{4.4 + 87.9}{2} = \frac{92.3}{2} = 46.15 \text{ light-years} \]This indicates that, from a simplistic extreme-based sense of balance, the average point between the closest and farthest stars is 46.15 light-years.
  • The midrange is easy to calculate and provides a rapid sense of the dataset's central point relative to its range.
  • However, unlike the mean, it can be heavily influenced by outliers since it only considers the two extreme values in the dataset.

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Most popular questions from this chapter

Harmonic Mean The harmonic mean (HM) is defined as the number of values divided by the sum of the reciprocals of each value. The formula is $$\mathrm{HM}=\frac{n}{\Sigma(1 / X)}$$ For example, the harmonic mean of \(1,4,5,\) and 2 is $$\mathrm{HM}=\frac{4}{1 / 1+1 / 4+1 / 5+1 / 2} \approx 2.051$$ This mean is useful for finding the average speed. Suppose a person drove 100 miles at 40 miles per hour and returned driving 50 miles per hour. The average miles per hour is not 45 miles per hour, which is found by adding 40 and 50 and dividing by 2 . The average is found as shown. Since Time \(=\) distance \(\div\) rate then Time \(1=\frac{100}{40}=2.5\) hours to make the trip Time \(2=\frac{100}{50}=2\) hours to return Hence, the total time is 4.5 hours, and the total miles driven are \(200 .\) Now, the average speed is $$\text { Rate }=\frac{\text { distance }}{\text { time }}=\frac{200}{4.5} \approx 44.444 \text { miles per hour }$$ This value can also be found by using the harmonic mean formula $$\mathrm{HM}=\frac{2}{1 / 40+1 / 50} \approx 44.444$$ Using the harmonic mean, find each of these. a. A salesperson drives 300 miles round trip at 30 miles per hour going to Chicago and 45 miles per hour returning home. Find the average miles per hour. b. A bus driver drives the 50 miles to West Chester at 40 miles per hour and returns driving 25 miles per hour. Find the average miles per hour. c. A carpenter buys \(\$ 500\) worth of nails at \(\$ 50\) per pound and \(\$ 500\) worth of nails at \(\$ 10\) per pound. Find the average cost of 1 pound of nails.

The number of incidents in which police were needed for a sample of 9 schools in Allegheny County is \(7,37,3,8,48,11,6,0,10 .\) Find the first and third quartiles for the data.

Which is a better relative position, a score of 83 on a geography test that has a mean of 72 and a standard deviation of \(6,\) or a score of 61 on an accounting test that has a mean of 55 and a standard deviation of \(3.5 ?\)

Using the weighted mean, find the average number of grams of fat per ounce of meat or fish that a person would consume over a 5 -day period if he ate these: $$\begin{array}{lc}\text { Meat or fish } & \text { Fat (g/oz) } \\\\\hline 3 \text { oz fried shrimp } & 3.33 \\\3 \text { oz veal cutlet (broiled) } & 3.00 \\\2 \text { oz roast beef (lean) } & 2.50 \\\2.5 \text { oz fried chicken drumstick } & 4.40 \\\4 \text { oz tuna (canned in oil) } & 1.75\end{array}$$

The geometric mean (GM) is defined as the \(n\) th root of the product of \(n\) values. The formula is $$\mathrm{GM}=\sqrt[n]{\left(X_{1}\right)\left(X_{2}\right)\left(X_{3}\right) \cdots\left(X_{n}\right)}$$ The geometric mean of 4 and 16 is $$\mathrm{GM}=\sqrt{(4)(16)}=\sqrt{64}=8$$ The geometric mean of \(1,3,\) and 9 is $$\mathrm{GM}=\sqrt[3]{(1)(3)(9)}=\sqrt[3]{27}=3$$ The geometric mean is useful in finding the average of percentages, ratios, indexes, or growth rates. For example, if a person receives a \(20 \%\) raise after 1 year of service and a \(10 \%\) raise after the second year of service, the average percentage raise per year is not 15 but \(14.89 \%,\) as shown. $$\mathrm{GM}=\sqrt{(1.2)(1.1)} \approx 1.1489$$ Or $$\mathrm{GM}=\sqrt{(120)(110)} \approx 114.89 \%$$ His salary is \(120 \%\) at the end of the first year and \(110 \%\) at the end of the second year. This is equivalent to an average of \(14.89 \%\), since \(114.89 \%-100 \%=\) \(14.89 \% .\) This answer can also be shown by assuming that the person makes \(\$ 10,000\) to start and receives two raises of \(20 \%\) and \(10 \%\). $$\begin{array}{l}\text { Raise } 1=10,000 \cdot 20 \%=\$ 2000 \\\\\text { Raise } 2=12,000 \cdot 10 \%=\$ 1200\end{array}$$ Find the geometric mean of each of these. a. The growth rates of the Living Life Insurance Corporation for the past 3 years were \(35,24,\) and \(18 \%\). b. A person received these percentage raises in salary over a 4-year period: \(8,6,4,\) and \(5 \%\). c. A stock increased each year for 5 years at these percentages: \(10,8,12,9,\) and \(3 \%\). d. The price increases, in percentages, for the cost of food in a specific geographic region for the past 3 years were \(1,3,\) and \(5.5 \% .\)
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