Question

the data show the number of vacation days used by a sample of 20 employees in a recent year. use technology to answer parts (a) and (b).
a. find the data sets first, second, and third quartiles.
b. draw a box - and - whisker plot that represents the data set.
0 1 7 4 6 7 6 5 6 1 1 1 10 0 6 5 3 5 2 3
a. find the three quartiles.
$q_1 = \square$
$q_2 = \square$
$q_3 = \square$
(type integers or decimals. do not round.)
b. choose the correct answer plot below. note that different technologies will produce slightly different results.
\\(\bigcirc\\) a.
\\(\bigcirc\\) b.
\\(\bigcirc\\) c.

Explanation:

Step 1: Organize the data

First, we need to sort the data set in ascending order. The given data is: 0, 1, 7, 4, 6, 7, 6, 5, 6, 1, 1, 1, 10, 0, 6, 5, 3, 5, 2, 3. Sorting it gives: 0, 0, 1, 1, 1, 1, 2, 3, 3, 4, 5, 5, 5, 6, 6, 6, 6, 7, 7, 10.

Step 2: Find the median (Q₂)

Since there are \( n = 20 \) data points (even number), the median \( Q_2 \) is the average of the 10th and 11th values. The 10th value is 4 and the 11th value is 5. So, \( Q_2=\frac{4 + 5}{2}=4.5 \).

Step 3: Find the first quartile (Q₁)

The first quartile \( Q_1 \) is the median of the lower half of the data. The lower half consists of the first 10 data points: 0, 0, 1, 1, 1, 1, 2, 3, 3, 4. There are 10 data points (even number), so the median of this subset is the average of the 5th and 6th values. The 5th value is 1 and the 6th value is 1. Thus, \( Q_1=\frac{1+1}{2} = 1 \). Wait, no, wait. Wait, when \( n \) is even for the lower half, let's re - check. Wait, the lower half for \( n = 20 \) is the first 10 observations. The 5th and 6th terms: positions 5 and 6 (1 - based index). The 5th term is 1 (value at index 5: 1) and 6th term is 1 (value at index 6: 1). So \( Q_1=\frac{1 + 1}{2}=1 \)? Wait, no, maybe I made a mistake. Wait, another way: using the formula for quartiles. For a data set with \( n \) observations, the position of \( Q_1 \) is \( \frac{n + 1}{4} \) when using the "inclusive" method or \( \frac{n}{4} \) when using the "exclusive" method. Let's use the "inclusive" method. \( n = 20 \), so the position of \( Q_1 \) is \( \frac{20+ 1}{4}=5.25 \). So we take the 5th value plus 0.25 times the difference between the 6th and 5th values. The 5th value is 1, the 6th value is 1. So \( Q_1=1+0.25\times(1 - 1)=1 \).

Wait, no, maybe the "exclusive" method: split the data into lower half (first 10) and upper half (last 10). The lower half: 0, 0, 1, 1, 1, 1, 2, 3, 3, 4. The median of the lower half: since there are 10 values, the median is the average of the 5th and 6th values. 5th value: 1, 6th value: 1. So \( Q_1 = 1 \).

Step 4: Find the third quartile (Q₃)

The third quartile \( Q_3 \) is the median of the upper half of the data. The upper half consists of the last 10 data points: 5, 5, 5, 6, 6, 6, 6, 7, 7, 10. There are 10 data points (even number), so the median of this subset is the average of the 5th and 6th values. The 5th value is 6 and the 6th value is 6. So \( Q_3=\frac{6 + 6}{2}=6 \). Wait, using the position formula for \( Q_3 \) with inclusive method: \( \frac{3(n + 1)}{4}=\frac{3\times(20 + 1)}{4}=\frac{63}{4}=15.75 \). The 15th value is 6, the 16th value is 6. So \( Q_3=6+0.75\times(6 - 6)=6 \).

Wait, but let's check with a different approach. Let's use the "Tukey's hinges" method. For \( n = 20 \), the median is between the 10th and 11th values (4 and 5), so \( Q_2 = 4.5 \). The lower half is the first 10 values: 0, 0, 1, 1, 1, 1, 2, 3, 3, 4. The median of the lower half (Q₁) is the median of these 10 values. The 5th and 6th values are both 1, so \( Q_1 = 1 \). The upper half is the last 10 values: 5, 5, 5, 6, 6, 6, 6, 7, 7, 10. The median of the upper half (Q₃) is the median of these 10 values. The 5th and 6th values are both 6, so \( Q_3 = 6 \). Wait, but when I check with a calculator or software, maybe I made a mistake. Wait, let's re - sort the data again:

Sorted data: 0, 0, 1, 1, 1, 1, 2, 3, 3, 4, 5, 5, 5, 6, 6, 6, 6, 7, 7, 10.

Number of data points \( n = 20 \).

- **First quartile (Q₁)**: The position of \( Q_1 \) is \( \frac{n}{4}=5 \) (using the "exclusive" method where we don't include the median in the lower/upper half when \( n \) is even). Wait, no, the "exclusive" method for quartiles: when \( n \) is even, the lower half is the first \( \frac{n}{2} \) observations and the upper half is the last \( \frac{n}{2} \) observations. So for \( n = 20 \), lower half: 0, 0, 1, 1, 1, 1, 2, 3, 3, 4 (10 observations), upper half: 5, 5, 5, 6, 6, 6, 6, 7, 7, 10 (10 observations).

For the lower half (10 observations), the median (Q₁) is the average of the 5th and 6th observations. The 5th observation in the lower half is 1 (value at index 5: 1) and the 6th observation is 1 (value at index 6: 1). So \( Q_1=\frac{1 + 1}{2}=1 \).

For the upper half (10 observations), the median (Q₃) is the average of the 5th and 6th observations. The 5th observation in the upper half is 6 (value at index 14 + 1=15? Wait, no, the upper half starts at index 11 (1 - based). The upper half data: 5 (index 11), 5 (12), 5 (13), 6 (14), 6 (15), 6 (16), 6 (17), 7 (18), 7 (19), 10 (20). So the 5th observation in the upper half is at index 11+4 = 15 (value 6) and the 6th is at index 16 (value 6). So \( Q_3=\frac{6+6}{2}=6 \).

The median (Q₂) is the average of the 10th and 11th observations. The 10th observation is 4 (index 10) and the 11th is 5 (index 11). So \( Q_2=\frac{4 + 5}{2}=4.5 \).

Answer:

\( Q_1=\boxed{1} \), \( Q_2=\boxed{4.5} \), \( Q_3=\boxed{6} \)