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Chapter 3: Problem 28

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.

Short Answer

Expert verified
The first quartile is 4.5, and the third quartile is 24.

Step by step solution

01

Order the Data

Begin by organizing the data in ascending order. The original data set is \(\{7, 37, 3, 8, 48, 11, 6, 0, 10\}\). When ordered, it becomes \(\{0, 3, 6, 7, 8, 10, 11, 37, 48\}\).
02

Identify Quartile Positions

To find the quartiles, we first determine their positions. With 9 data points: \(Q_1 = (\frac{1(9+1)}{4}) = 2.5\) and \(Q_3 = (\frac{3(9+1)}{4}) = 7.5\).
03

Calculate the First Quartile (Q1)

Since \(Q_1\) is at position 2.5, we find the average of the values at positions 2 and 3 in the ordered list: \(\frac{3 + 6}{2} = 4.5\). Thus, \(Q_1 = 4.5\).
04

Calculate the Third Quartile (Q3)

For \(Q_3\), positioned at 7.5, average the values at positions 7 and 8 in the ordered list: \(\frac{11 + 37}{2} = 24\). Thus, \(Q_3 = 24\).

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Key Concepts

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

Data Ordering
Before diving into calculating quartiles, we need to sort our data. Why? Ordering data from smallest to largest helps us find positions accurately.
Think of it as arranging books on a shelf from shortest to tallest; it makes finding the right one much easier. In our example, the original data \( \{7, 37, 3, 8, 48, 11, 6, 0, 10\} \) becomes \( \{0, 3, 6, 7, 8, 10, 11, 37, 48\} \) after ordering.
  • Ensure each number is sorted.
  • Double-check for any missed data entries.
  • Revisit sorted list if your results seem odd; you might have missed a step!
By organizing data, placing quartiles becomes straightforward.
Quartile Positions
Quartiles are values that split your data into four equal parts, helping describe the spread and center of your data.
Finding their positions aids in calculating their values. Imagine dividing a pizza into equal slices; quartile positions tell you where to cut.Here's a quick breakdown:- **Position of First Quartile (Q1):** Use the formula \( \frac{1(n+1)}{4} \), where \(n\) is the number of data points. For our nine data points, this becomes \(2.5\).- **Position of Third Quartile (Q3):** Employ \(\frac{3(n+1)}{4}\), leading to \(7.5\) in our example.Knowing these positions tells us where to go next when calculating the quartile values.
First Quartile (Q1)
Now, let's compute the first quartile, or \(Q_1\). The position found was 2.5, indicating that \(Q_1\) lies between the second and third values in our ordered data list.To find \(Q_1\):- Average the numbers at positions 2 and 3: \( \frac{3 + 6}{2} = 4.5 \).Thus, \(Q_1\) equals 4.5 in this data set. It reflects the value below which 25% of data falls, highlighting lower data spread.
Third Quartile (Q3)
Calculating the third quartile, \(Q_3\), is just as straightforward. With \(Q_3\)'s position at 7.5, we look between the seventh and eighth values.Here's how to find \(Q_3\):
  • Identify values at these positions (11 and 37).
  • Average these two: \( \frac{11 + 37}{2} = 24 \).
Thus, \(Q_3 = 24\).
This means 75% of the data values lie below this point, providing insight into the upper half of your data set.
Statistical Analysis
Understanding quartiles is essential in statistical analysis. They provide insights into data variability and distribution. Here's how: - **Data Spread:** Quartiles present a data summary, showing distribution. - **Comparison Aid:** Allows comparison across different datasets. - **Outlier Detection:** Helps isolate unusual data points. By using quartiles, you gain a clearer view of data behavior.
Analysts often present this as a boxplot, visually conveying the data's distribution.
Recognizing where your data stands aids in building informed conclusions.

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Most popular questions from this chapter

The data for a recent year show the taxes (in millions of dollars) received from a random sample of 10 states. Find the first and third quartiles and the IQR. \(\begin{array}{llllllllll}13 & 15 & 32 & 36 & 11 & 24 & 6 & 25 & 11 & 71\end{array}\)

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 \(3.5 ?\)

Find the mean of \(10,20,30,40,\) and \(50 .\) a. Add 10 to each value and find the mean. b. Subtract 10 from each value and find the mean. c. Multiply each value by 10 and find the mean. d. Divide each value by 10 and find the mean. e. Make a general statement about each situation.

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.

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