Question

consider the following sample data, which represent weights of sea mussels grown on the california coast, in grams: { 2.8, 2.6, 2.3, 3.2, 3.4, 3.3, 3.0, 2.6, 2.8, 3.1, 2.9, 3.0, 4.4, 3.2, 1.9, 2.9, 3.0, 2.4, 3.1 }. enter the five - number summary below. do not round any answers on this problem.

minimum	$q_1$	$q_2$	$q_3$	maximum

Explanation:

Step1: Sort the data

First, we sort the given data in ascending order: \(1.9, 2.3, 2.4, 2.6, 2.6, 2.8, 2.8, 2.9, 2.9, 3.0, 3.0, 3.0, 3.1, 3.1, 3.2, 3.2, 3.3, 3.4, 4.4\)

Step2: Find the Minimum

The minimum value is the first value in the sorted data, so Minimum \(= 1.9\)

Step3: Find \(Q_2\) (Median)

The number of data points \(n = 19\) (odd). The median is the \(\frac{n + 1}{2}=\frac{19+ 1}{2}=10^{th}\) value. Looking at the sorted data, the \(10^{th}\) value is \(3.0\), so \(Q_2=3.0\)

Step4: Find \(Q_1\) (First Quartile)

The first quartile is the median of the lower half of the data. The lower half consists of the first \(\frac{n - 1}{2}=9\) values (since \(n = 19\), the lower half is from the \(1^{st}\) to \(9^{th}\) value: \(1.9, 2.3, 2.4, 2.6, 2.6, 2.8, 2.8, 2.9, 2.9\)). The number of values in the lower half is \(9\) (odd), so the median of the lower half is the \(\frac{9 + 1}{2}=5^{th}\) value. The \(5^{th}\) value in the lower half is \(2.6\), so \(Q_1 = 2.6\)

Step5: Find \(Q_3\) (Third Quartile)

The third quartile is the median of the upper half of the data. The upper half consists of the last \(\frac{n - 1}{2}=9\) values (from the \(11^{th}\) to \(19^{th}\) value: \(3.0, 3.0, 3.1, 3.1, 3.2, 3.2, 3.3, 3.4, 4.4\)). The number of values in the upper half is \(9\) (odd), so the median of the upper half is the \(\frac{9+ 1}{2}=5^{th}\) value. The \(5^{th}\) value in the upper half is \(3.2\), so \(Q_3=3.2\)

Step6: Find the Maximum

The maximum value is the last value in the sorted data, so Maximum \(= 4.4\)

Answer:

Minimum: \(1.9\), \(Q_1\): \(2.6\), \(Q_2\): \(3.0\), \(Q_3\): \(3.2\), Maximum: \(4.4\)