Question

answer the statistical measures and create a box and whiskers plot for the following set of data. enter all values as whole numbers or decimals. 5,6,8,9,9,9,10,10,11,13,13,16,19 min: q1: med: q3: max: create the box plot by dragging the lines:

Explanation:

Step1: Find Minimum (Min)

The minimum value in the data set is the smallest number. Looking at \(5,6,8,9,9,9,10,10,11,13,13,16,19\), the smallest is \(5\).

Step2: Find Quartile 1 (Q1)

First, the data is already ordered. The median (Med) will split the data into two halves. The number of data points \(n = 13\), so the median is at position \(\frac{n + 1}{2}=\frac{13+1}{2}=7\)th term. The first half (lower half) is \(5,6,8,9,9,9\) (since the 7th term is the median, the lower half has 6 terms). The median of the lower half (Q1) is the average of the 3rd and 4th terms of the lower half. The 3rd term is \(8\), 4th term is \(9\), so \(Q1=\frac{8 + 9}{2}=8.5\)? Wait, no, wait. Wait, \(n = 13\), so the lower half is the first 6 terms: positions 1 - 6: \(5,6,8,9,9,9\). Wait, no, when \(n\) is odd, the median is the middle term, and the lower half is the terms before the median (not including the median). So median is the 7th term (\(10\)? Wait no, wait the data is \(5,6,8,9,9,9,10,10,11,13,13,16,19\). Let's index them: 1:5, 2:6, 3:8, 4:9, 5:9, 6:9, 7:10, 8:10, 9:11, 10:13, 11:13, 12:16, 13:19. So median is the 7th term, which is \(10\). Then the lower half is terms 1 - 6: \(5,6,8,9,9,9\). The median of the lower half (Q1) is the average of the 3rd and 4th terms of this lower half. 3rd term: \(8\), 4th term: \(9\), so \(Q1=\frac{8 + 9}{2}=8.5\)? Wait, no, wait, for an even number of terms in the lower half (6 terms), the median is the average of the 3rd and 4th. Wait, but maybe I made a mistake. Wait, actually, another method: Q1 is the 25th percentile. The formula for the position of Q1 is \(\frac{n + 1}{4}=\frac{13+1}{4}=3.5\)th term. So we take the average of the 3rd and 4th terms. 3rd term: \(8\), 4th term: \(9\), so \(Q1=\frac{8 + 9}{2}=8.5\)? Wait, but let's check again. Wait, the data is \(5,6,8,9,9,9,10,10,11,13,13,16,19\). Let's list the positions:

1:5, 2:6, 3:8, 4:9, 5:9, 6:9, 7:10, 8:10, 9:11, 10:13, 11:13, 12:16, 13:19.

Q1: position \(\frac{13 + 1}{4}=3.5\), so average of 3rd (8) and 4th (9) terms: \(\frac{8+9}{2}=8.5\).

Step3: Find Median (Med)

As \(n = 13\), median is at position \(\frac{13 + 1}{2}=7\)th term. The 7th term is \(10\) (since index 7 is \(10\)).

Step4: Find Quartile 3 (Q3)

The upper half is terms 8 - 13: \(10,11,13,13,16,19\) (wait, no, the median is the 7th term, so upper half is terms 8 - 13: indices 8:10, 9:11, 10:13, 11:13, 12:16, 13:19. The median of the upper half (Q3) is the average of the 3rd and 4th terms of the upper half. The 3rd term of upper half is \(13\), 4th term is \(13\), so \(Q3=\frac{13 + 13}{2}=13\). Alternatively, using the position formula: \(\frac{3(n + 1)}{4}=\frac{3(13 + 1)}{4}=10.5\)th term. So average of 10th (13) and 11th (13) terms: \(\frac{13+13}{2}=13\).

Step5: Find Maximum (Max)

The largest value in the data set is \(19\).

Wait, wait, earlier when I thought Q1, let's re - check. Wait, maybe I made a mistake in the lower half. Wait, the data is \(5,6,8,9,9,9,10,10,11,13,13,16,19\). The median is the 7th term, which is \(10\). The lower half is the first 6 terms: \(5,6,8,9,9,9\). The median of these 6 terms (Q1) is the average of the 3rd and 4th terms. 3rd term: \(8\), 4th term: \(9\), so \(Q1=\frac{8 + 9}{2}=8.5\)? But let's check with another approach. The 25th percentile: for \(n = 13\) data points, the position is \(0.25\times(13 + 1)=3.5\), so we take the 3rd and 4th terms and average them. 3rd term: \(8\), 4th term: \(9\), so \(Q1 = 8.5\).

Wait, but let's list all steps again:

1. Min: The smallest number in the data set \(5,6,8,9,9,9,10,10,11,13,13,16,19\) is \(5\).

2. Q1: The data is ordered. The number of observations \(n = 13\). The position of Q1 is \(n_1=\frac{n + 1}{4}=\frac{13+1}{4}=3.5\). So we take the 3rd and 4th values from the start. The 3rd value is \(8\), the 4th value is \(9\). So \(Q1=\frac{8 + 9}{2}=8.5\).

3. Med (Median): The position of the median is \(n_2=\frac{n + 1}{2}=\frac{13+1}{2}=7\). The 7th value in the ordered data is \(10\).

4. Q3: The position of Q3 is \(n_3=\frac{3(n + 1)}{4}=\frac{3\times14}{4}=10.5\). So we take the 10th and 11th values from the start. The 10th value is \(13\), the 11th value is \(13\). So \(Q3=\frac{13+13}{2}=13\).

5. Max: The largest number in the data set is \(19\).

Answer:

Min: \(5\), Q1: \(8.5\), Med: \(10\), Q3: \(13\), Max: \(19\)

For the box - and - whiskers plot:

• The left whisker starts at the minimum value (\(5\)) and ends at Q1 (\(8.5\)).
• The box starts at Q1 (\(8.5\)), has a line at the median (\(10\)), and ends at Q3 (\(13\)).
• The right whisker starts at Q3 (\(13\)) and ends at the maximum value (\(19\)).