Question

the following are distances (in miles) traveled to the workplace by 18 employees of a certain brokerage firm. 28, 18, 32, 20, 29, 15, 18, 2, 10, 27, 14, 14, 11, 5, 9, 37, 31, 34 (send data to calculator) (send data to excel) find ( 10^{\text{th}} ) and ( 75^{\text{th}} ) percentiles for these distances. (if necessary, consult a list of formulas.) (a) the ( 10^{\text{th}} ) percentile: (square) miles (b) the ( 75^{\text{th}} ) percentile: (square) miles

Explanation:

Step1: Sort the data

First, we sort the given data in ascending order: \(2, 5, 9, 10, 11, 14, 14, 15, 18, 18, 20, 27, 28, 29, 31, 32, 34, 37\) (note: there are 18 data points).

Step2: Calculate the position for 10th percentile

The formula for the position of the \(p\)-th percentile is \(i=\frac{p}{100}\times n\), where \(n = 18\) (number of data points) and \(p = 10\).
So, \(i=\frac{10}{100}\times18 = 1.8\). Since \(i\) is not an integer, we round up to the next integer. So the position is \(2\) (because we round up \(1.8\) to \(2\)). Looking at the sorted data, the 2nd value is \(5\)? Wait, no, wait: sorted data is \(2, 5, 9, 10, 11, 14, 14, 15, 18, 18, 20, 27, 28, 29, 31, 32, 34, 37\). Wait, the first value is index 1: \(2\) (index 1), \(5\) (index 2), \(9\) (index 3), etc. Wait, \(i = 1.8\), so we take the value at the ceiling of \(i\), which is \(2\). Wait, but some methods use linear interpolation. Wait, let's check the formula. The percentile formula: if \(i=\frac{p}{100}(n + 1)\) or \(i=\frac{p}{100}n\). Let's confirm the formula. The common formula for percentile is:

For a set of \(n\) ordered data values \(x_1\leq x_2\leq\cdots\leq x_n\), the \(p\)-th percentile is the value such that at least \(p\%\) of the data is less than or equal to it and at least \((100 - p)\%\) is greater than or equal to it.

The position \(i\) is calculated as \(i=\frac{p}{100}\times n\). If \(i\) is an integer, the percentile is the average of \(x_i\) and \(x_{i + 1}\). If \(i\) is not an integer, we round up to the next integer, and the percentile is \(x_{\lceil i ceil}\).

Wait, let's recalculate \(i\) for 10th percentile: \(p = 10\), \(n = 18\). So \(i=\frac{10}{100}\times18=1.8\). Since \(i\) is not an integer, we round up to \(2\). So the 10th percentile is the value at position \(2\), which is \(5\)? Wait, no, wait the sorted data is:

1: 2

2: 5

3: 9

4: 10

5: 11

6: 14

7: 14

8: 15

9: 18

10: 18

11: 20

12: 27

13: 28

14: 29

15: 31

16: 32

17: 34

18: 37

Wait, but let's check another method. The formula \(i=\frac{p}{100}(n - 1)+1\). For \(p = 10\), \(n = 18\): \(i=\frac{10}{100}(17)+1=1.7 + 1=2.7\). Then we take the integer part \(2\) and the decimal part \(0.7\). So the percentile is \(x_2+0.7(x_3 - x_2)\). \(x_2 = 5\), \(x_3 = 9\). So \(5+0.7(9 - 5)=5 + 2.8 = 7.8\). Wait, this is a different result. So maybe the formula is different. Let's check the standard method. The two common methods are:

Method 1: \(i=\frac{p}{100}\times n\). If \(i\) is integer, percentile is \(\frac{x_i + x_{i + 1}}{2}\). If not, round up to next integer, take \(x_{\lceil i ceil}\).

Method 2: \(i=\frac{p}{100}(n + 1)\).

Let's check with an example. Let's see, for \(n = 18\), \(p = 10\):

Method 1: \(i = 0.1\times18 = 1.8\). Round up to \(2\), so percentile is \(x_2 = 5\).

Method 2: \(i = 0.1\times19 = 1.9\). Round up to \(2\), so percentile is \(x_2 = 5\). Wait, no, \(x_2\) is \(5\). But when we use linear interpolation, for \(i = 1.9\), the value is \(x_1+0.9(x_2 - x_1)=2 + 0.9(5 - 2)=2 + 2.7 = 4.7\). Hmm, this is confusing. Maybe the problem expects the method where we use \(i=\frac{p}{100}\times n\), and if \(i\) is not integer, round up. Let's proceed with the first method.

Now for 75th percentile: \(p = 75\), \(n = 18\). So \(i=\frac{75}{100}\times18 = 13.5\). Since \(i\) is not an integer, we take the average of \(x_{13}\) and \(x_{14}\). \(x_{13}=28\), \(x_{14}=29\). So the 75th percentile is \(\frac{28 + 29}{2}=28.5\). Wait, let's check the sorted data:

1: 2

2: 5

3: 9

4: 10

5: 11

6: 14

7: 14

8: 15

9: 18

10: 18

11: 20

12: 27

13: 28

14: 29

15: 31

16: 32

17: 34

18: 37

So \(x_{13}=28\), \(x_{14}=29\). So \(i = 13.5\), so we take the average of \(x_{13}\) and \(x_{14}\), which is \(\frac{28 + 29}{2}=28.5\).

Wait, let's recheck the 10th percentile. Let's use the formula \(i=\frac{p}{100}(n + 1)\). For \(p = 10\), \(n = 18\), \(i = 0.1\times19 = 1.9\). So we take the value at position \(2\) (since \(1.9\) is between \(1\) and \(2\)), and use linear interpolation: \(x_1 + 0.9(x_2 - x_1)=2 + 0.9(5 - 2)=2 + 2.7 = 4.7\). But this is different. However, different textbooks use different methods. Let's check the problem statement: it says "consult a list of formulas". The common formula for percentile is:

The \(p\)-th percentile of a data set is a value such that at least \(p\%\) of the data is less than or equal to it and at least \((100 - p)\%\) is greater than or equal to it. To find the \(p\)-th percentile:

1. Order the data from smallest to largest.

2. Calculate the index \(i=\frac{p}{100}\times n\), where \(n\) is the number of data points.

3. If \(i\) is an integer, the \(p\)-th percentile is the average of the \(i\)-th and \((i + 1)\)-th values.

4. If \(i\) is not an integer, round up to the next integer, and the \(p\)-th percentile is the value at that position.

So applying this:

For 10th percentile:

\(n = 18\), \(p = 10\), \(i=\frac{10}{100}\times18 = 1.8\). Since \(i\) is not an integer, round up to \(2\). So the 10th percentile is the 2nd value in the sorted data, which is \(5\)? Wait, no, the sorted data is:

1: 2

2: 5

3: 9

4: 10

5: 11

6: 14

7: 14

8: 15

9: 18

10: 18

11: 20

12: 27

13: 28

14: 29

15: 31

16: 32

17: 34

18: 37

Wait, the 2nd value is \(5\). But let's check the percentage: how many values are less than or equal to \(5\)? The values \(2\) and \(5\), so 2 values. \(2/18\approx11.11\%\), which is more than \(10\%\). So that works.

For 75th percentile:

\(i=\frac{75}{100}\times18 = 13.5\). Since \(i\) is not an integer, we take the average of the 13th and 14th values. \(x_{13}=28\), \(x_{14}=29\). So the 75th percentile is \(\frac{28 + 29}{2}=28.5\).

Wait, let's verify the percentage for 75th percentile. The number of values less than or equal to \(28.5\): the first 13 values are up to \(28\), and the 14th is \(29\). So 13.5 values? Wait, no, the formula says that at least \(75\%\) of the data is less than or equal to it. \(13.5/18 = 0.75\), which is \(75\%\). So that works.

So:

(a) 10th percentile: position \(i = 1.8\), round up to \(2\), so \(x_2 = 5\)? Wait, but when we calculate \(2/18\approx11.1\%\), which is more than \(10\%\), so that's correct.

Wait, but let's check with another approach. Let's list the data sorted:

1: 2 (2)

2: 5 (5)

3: 9 (9)

4: 10 (10)

5: 11 (11)

6: 14 (14)

7: 14 (14)

8: 15 (15)

9: 18 (18)

10: 18 (18)

11: 20 (20)

12: 27 (27)

13: 28 (28)

14: 29 (29)

15: 31 (31)

16: 32 (32)

17: 34 (34)

18: 37 (37)

For 10th percentile, we need a value where at least \(10\%\) of the data is ≤ to it. \(10\%\) of \(18\) is \(1.8\), so we need the smallest value such that at least \(2\) values (since \(1.8\) rounded up is \(2\)) are ≤ to it. The 2nd value is \(5\), and there are 2 values (2 and 5) ≤ to \(5\), which is \(2/18\approx11.1\%\), which is ≥ \(10\%\). So that's correct.

For 75th percentile, \(75\%\) of \(18\) is \(13.5\), so we need the average of the 13th and 14th values, because \(13.5\) is the midpoint between \(13\) and \(14\). So \(x_{13}=28\), \(x_{14}=29\), average is \(28.5\). And \(13.5\) values are ≤ to \(28.5\), which is \(75\%\) of \(18\), so that's correct.

So:

(a) 10th percentile: \(5\)

(b) 75th percentile: \(28.5\)

Wait, but let's check with the formula \(i=\frac{p}{100}(n + 1)\) for 10th percentile: \(i = 0.1\times19 = 1.9\). So we take the value at position \(2\) (since \(1.9\) is between \(1\) and \(2\)), and use linear interpolation: \(x_1 + 0.9(x_2 - x_1)=2 + 0.9(5 - 2)=2 + 2.7 = 4.7\). But then the percentage of values ≤ \(4.7\) is \(1\) (only \(2\)) or \(2\) (if \(4.7\) is included). Wait, \(2\) is ≤ \(4.7\), \(5\) is > \(4.7\), so only \(1\) value, which is \(1/18\approx5.56\%\), which is less than \(10\%\). So that method is not correct here. So the first method is better.

Therefore, the 10th percentile is \(5\) and the 75th percentile is \(28.5\).

Answer:

(a) The \(10^{\text{th}}\) percentile: \(\boxed{5}\) miles

(b) The \(75^{\text{th}}\) percentile: \(\boxed{28.5}\) miles