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Chapter 5: Problem 12

Bill is looking for original taillights for his 1932 Ford. The prices vary depending on the condition. He fi nds these prices: \(\$ 450, \$ 100, \$ 180, \$ 600, \$ 300, \$ 350, \$ 300,\) and \(\$ 400\) a. Find the four quartiles. b. Find the interquartile range. c. Find the boundary for the lower outliers. Are there any lower outliers? d. Find the boundary for the upper outliers. Are there any upper outliers?

Short Answer

Expert verified
The four quartiles are \(\$240\), \(\$325\), \(\$425\) respectively. The interquartile range is \(\$ 185\). There are no lower or upper outliers in the data set as the boundaries for outliers fall outside the range of the dataset.

Step by step solution

01

Sort the data

First, arrange the given data in ascending order. That would be: \( \$ 100, \$ 180, \$ 300, \$ 300, \$ 350, \$ 400, \$ 450, \$ 600\).
02

Find the quartiles

The first quartile (Q1) is the middle number between the smallest number and the median of the data set. The second quartile (Q2) is the median of the data. The third quartile (Q3) is the middle value between the median and the highest value. For this dataset, the Q1 would be \($240\), Q2 or median would be \($325\) and Q3 would be \($425\).
03

Find the interquartile range

The interquartile range is calculated by subtracting Q1 from Q3. So, for this dataset, the IQR would be \(Q3 - Q1 = \$ 425 - \$ 240 = \$ 185\).
04

Find the boundary for the lower outliers

The boundary for the lower outliers can be found by subtracting \(1.5*IQR\) from Q1. So, Lower boundary = \(Q1 - 1.5*IQR = \$ 240 - 1.5*\$ 185 = -\$ 37.5\). Since this value is below zero, and we do not have any negative price, there won't be any lower outliers.
05

Find the boundary for the upper outliers

The boundary for the upper outliers is calculated by adding \(1.5*IQR\) to Q3. So, Upper boundary = \(Q3 + 1.5*IQR = \$ 425 + 1.5*\$ 185 = \$ 702.5\). Any data point greater than this value would be an upper outlier. In our dataset, there are no prices greater than \$702.5, so there are no upper outliers.

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Key Concepts

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

Quartiles
Quartiles are a foundational concept in statistics. They are used to divide a dataset into four equal parts. Think of them as markers that split your data into quarters. Here's a quick breakdown:
  • First Quartile (Q1): This is the value below which 25% of the data falls. In our sorted price list, this is \(\\(240\).
  • Second Quartile (Q2): Also known as the median, it's the midpoint of the dataset. Here, half the prices fall below this value, and half are above. For our example, Q2 is \(\\)325\).
  • Third Quartile (Q3): This marks the value below which 75% of the data rests. In this dataset, our third quartile is \(\$425\).
To find these quartiles, ensure your data is arranged in ascending order. Then, determine the position of the quartiles to easily calculate them.
Interquartile Range
The interquartile range (IQR) measures the variability or spread of the middle 50% of a dataset. It's an essential tool in descriptive statistics.
To compute the IQR, subtract the first quartile from the third quartile. In our case:\[ \text{IQR} = Q3 - Q1 = \\(425 - \\)240 = \\(185 \]This span of \(\\)185\) indicates how spread out the central half of your data is. The IQR is valuable because it provides a good sense of the data's distribution by excluding outliers. It gives a more robust insight into variability than the full range, especially when data contains anomalies.
Upper Outliers
Upper outliers are extremely high values that sit far from the rest of the dataset. Detecting them is crucial, especially when analyzing data for financial decisions, as they can skew results.
To identify upper outliers, calculate an upper boundary using the formula:\[ \text{Upper Boundary} = Q3 + 1.5 \times \text{IQR} \]
In our dataset, this becomes:\[ \\(425 + 1.5 \times \\)185 = \\(702.5 \]Any price above \(\\)702.5\) would be considered an upper outlier. Since none of Bill's prices exceed this value, he does not have any upper outliers.
Lower Outliers
Lower outliers are values that are unusually low compared to the rest of the data. Identifying them helps in maintaining the integrity of your data analysis.
To find the boundary for lower outliers, use the formula:\[ \text{Lower Boundary} = Q1 - 1.5 \times \text{IQR} \]
For Bill's data, this calculates to:\[ \\(240 - 1.5 \times \\)185 = -\$37.5 \]Since negative amounts aren鈥檛 applicable in this context, there are no lower outliers in his dataset. The absence of lower outliers suggests that all price entries fall within a reasonable range for the collected data.

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Most popular questions from this chapter

Five Smithtown High School students are saving up to buy their fi rst cars. They all have after-school jobs, and their weekly salaries are listed in the table. $$ \begin{array}{ll}{\text { Emily }} & {\$ 110} \\ {\text { Sam }} & {\$ 145} \\\ {\text { Danielle }} & {\$ 130} \\ {\text { Katie }} & {\$ 160} \\\ {\text { Stephanie }} & {\$ 400}\end{array} $$ a. What is the mean weekly salary for these students? b. What is the median salary? c. Whose salary would you consider to be an outlier? d. Which number do you think is better representative of the data, the mean or the median? e. Explain your answer to part d.

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Ron鈥檚 car left four skid marks on the road after he slammed his foot on the brake pedal to make an emergency stop. The police measured them to be 55 ft, 55 ft, 62 ft, and 62 ft. What skid distance will be used when calculating the skid speed formula?

The following list of prices is for a used original radio for a 1955 Thunderbird. The prices vary depending on the condition of the radio. \(\$ 210, \$ 210, \$ 320, \$ 200, \$ 300, \$ 10, \$ 340\) \(\$ 300, \$ 245, \$ 325, \$ 700, \$ 250, \$ 240, \$ 200\) a. Find the mean of the radio prices. b. Find the median of the radio prices. c. Find the mode of the radio prices. d. Find the four quartiles. e. Find the interquartile range for this data set. f. Find the boundary for the lower outliers. Are there any lower outliers? g. Find the boundary for the upper outliers. Are there any upper outliers?

The straight line depreciation equation for a car is \(y=-2,680 x+26,800 .\) a. How much is the car worth after 48 months? b. How much is the car worth after 75 months? c. Suppose that \(M\) represents the length of time in months when the car still has value. Write an algebraic expression to represent the value of this car after \(M\) months.
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