Question

question 1
find the first and third quartile for the quantitative data.
x
6.4
6.9
7.1
8.6
14.9
15.4
18.4
22.5
22.7
23.9
$q_1 = $
$q_3 = $
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question 2
data that lie beyond the fences are possible

Explanation:

Step1: Order the data

First, we confirm the data is already ordered: \(6.4, 6.9, 7.1, 8.6, 14.9, 15.4, 18.4, 22.5, 22.7, 23.9\). The number of data points \(n = 10\).

Step2: Find the median (to split into lower and upper halves)

The median for \(n = 10\) (even) is the average of the 5th and 6th values.
5th value: \(14.9\), 6th value: \(15.4\).
Median \(= \frac{14.9 + 15.4}{2} = 15.15\).

Step3: Find \(Q_1\) (first quartile, median of lower half)

Lower half data (first 5 values): \(6.4, 6.9, 7.1, 8.6, 14.9\).
Number of values in lower half \(n_1 = 5\) (odd), so \(Q_1\) is the 3rd value.
3rd value: \(7.1\).

Step4: Find \(Q_3\) (third quartile, median of upper half)

Upper half data (last 5 values): \(15.4, 18.4, 22.5, 22.7, 23.9\).
Number of values in upper half \(n_2 = 5\) (odd), so \(Q_3\) is the 3rd value.
3rd value: \(22.5\).

Answer:

\(Q_1 = 7.1\)
\(Q_3 = 22.5\)