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. 2,3,4,4,5,5,8,10,13,13,14 min: 2 q1: 1 med: q3: max: create the box plot by dragging the lines: box plot image with x - axis from 0 to 20

Explanation:

Step1: Identify the data set

The data set is \(2, 3, 4, 4, 5, 5, 8, 10, 13, 13, 14\). The number of data points \(n = 11\).

Step2: Find the minimum (Min)

The smallest value in the data set is \(2\), so Min \(= 2\) (already given).

Step3: Find the first quartile (Q1)

The first quartile is the median of the lower half of the data. The lower half of the data (excluding the median if \(n\) is odd) is the first \(\frac{n - 1}{2}= 5\) values? Wait, no. For \(n = 11\) (odd), the median is the 6th value. The lower half is the first 5 values: \(2, 3, 4, 4, 5\). The median of this lower half is the 3rd value. So for \(2, 3, 4, 4, 5\), the median (Q1) is \(4\). Wait, wait, let's re - calculate. The position of the median (Med) is \(\frac{n + 1}{2}=\frac{11+ 1}{2}=6\)th value. The lower half is the values from the 1st to the 5th (since the 6th is the median). So the lower half data: \(2,3,4,4,5\). The median of a set with 5 values (odd number of values) is the 3rd value. So the 3rd value in \(2,3,4,4,5\) is \(4\). So Q1 \( = 4\).

Step4: Find the median (Med)

The median is the middle value. For \(n = 11\), the median is the \(\frac{11 + 1}{2}=6\)th value. The data set in order: \(2,3,4,4,5,5,8,10,13,13,14\). The 6th value is \(5\). So Med \(= 5\).

Step5: Find the third quartile (Q3)

The third quartile is the median of the upper half of the data. The upper half of the data (excluding the median) is the values from the 7th to the 11th: \(8,10,13,13,14\). The median of this upper half is the 3rd value. For \(8,10,13,13,14\), the median (Q3) is \(13\). Wait, no. Wait, the upper half: the data after the median (6th value) is the 7th to 11th values. The median of these 5 values (\(8,10,13,13,14\)) is the 3rd value, which is \(13\)? Wait, no, let's list them: \(8,10,13,13,14\). The middle value (3rd) is \(13\)? Wait, no, 8 (1st), 10 (2nd), 13 (3rd), 13 (4th), 14 (5th). So the median is the 3rd value, which is \(13\)? Wait, no, that's not correct. Wait, the upper half: when \(n = 11\), the upper half is the values from index 7 to 11 (1 - based). The values are \(8,10,13,13,14\). The median of this sub - set is the middle value. Since there are 5 values, the middle one is the 3rd value, which is \(13\)? Wait, no, let's do it properly. The formula for Q1: for a data set with \(n\) observations, the position of Q1 is \(\frac{n + 1}{4}\) when \(n\) is such that \(\frac{n+1}{4}\) is an integer, otherwise we interpolate. For \(n = 11\), \(\frac{11 + 1}{4}=3\). So the 3rd value? Wait, no, the formula for quartiles:

Another method: The data is ordered: \(x_1 = 2,x_2 = 3,x_3 = 4,x_4 = 4,x_5 = 5,x_6 = 5,x_7 = 8,x_8 = 10,x_9 = 13,x_{10}=13,x_{11}=14\)

The position of Q1 is \(n\times0.25=11\times0.25 = 2.75\). So we take the 2nd value plus 0.75 times the difference between the 3rd and 2nd values. \(x_2 = 3\), \(x_3 = 4\). So Q1 \(=x_2+0.75\times(x_3 - x_2)=3 + 0.75\times(4 - 3)=3.75\)? Wait, this is a different result. There are different methods for calculating quartiles. The two common methods are the inclusive and exclusive methods.

In the inclusive method (used when we include the median in both halves for odd \(n\)):

The data set is \(2,3,4,4,5,5,8,10,13,13,14\). The median (Med) is the 6th value, which is \(5\). The lower half (including the median? No, in inclusive method for odd \(n\), we split the data into two halves with \(\frac{n + 1}{2}\) in each? Wait, no. Let's use the method where for \(n = 11\), the median is at position 6. The lower half is positions 1 - 5: \(2,3,4,4,5\) (5 values), the upper half is positions 7 - 11: \(8,10,13,13,14\) (5 values).

For the lower half \(2,3,4,4,5\), the median (Q1) is the 3rd value, which is \(4\).

For the upper half \(8,10,13,13,14\), the median (Q3) is the 3rd value, which is \(13\).

The maximum (Max) is the largest value, which is \(14\).

Let's re - check:

- Min: The smallest number is \(2\).
- Q1: The median of the first 5 numbers (\(2,3,4,4,5\)) is \(4\) (since the 3rd number in this sub - set is \(4\)).
- Med: The 6th number in the ordered data set is \(5\).
- Q3: The median of the last 5 numbers (\(8,10,13,13,14\)) is \(13\) (the 3rd number in this sub - set is \(13\)).
- Max: The largest number is \(14\).

Answer:

Min: \(2\), Q1: \(4\), Med: \(5\), Q3: \(13\), Max: \(14\)

To create the box - and - whisker plot:

• The left whisker goes from Min (\(2\)) to Q1 (\(4\)).
• The box goes from Q1 (\(4\)) to Q3 (\(13\)), with a line inside the box at the median (\(5\)).
• The right whisker goes from Q3 (\(13\)) to Max (\(14\)).