Question

jayden is a running back for his high school football team. in the first 3 games of the season he got 8 carries. the yards gained for each carry are given below: 11 2 4 17 5 0 3 6 use desmos commands to find the 5 - number summary. min \boxed{0} q1 \boxed{} median \boxed{4} q3 \boxed{} max \boxed{17}

Explanation:

Step1: Order the data

First, we order the data set: \(0, 2, 3, 4, 5, 6, 11, 17\)

Step2: Find Q1 (First Quartile)

The first quartile is the median of the lower half of the data. The lower half is \(0, 2, 3, 4\). The median of this set is \(\frac{2 + 3}{2}= 2.5\)

Step3: Find Q3 (Third Quartile)

The third quartile is the median of the upper half of the data. The upper half is \(5, 6, 11, 17\). The median of this set is \(\frac{6 + 11}{2}= 8.5\)

Answer:

min: \(0\), Q1: \(2.5\), median: \(4.5\) (Wait, earlier median calculation was wrong. Wait, the data set has 8 values, so the median is the average of the 4th and 5th values. The ordered data is \(0, 2, 3, 4, 5, 6, 11, 17\). The 4th value is \(4\), the 5th is \(5\), so median is \(\frac{4 + 5}{2}= 4.5\). Then Q1: lower half is \(0, 2, 3, 4\), median of lower half is \(\frac{2 + 3}{2}= 2.5\). Q3: upper half is \(5, 6, 11, 17\), median of upper half is \(\frac{6 + 11}{2}= 8.5\). max is \(17\). So correct 5 - number summary: min \(0\), Q1 \(2.5\), median \(4.5\), Q3 \(8.5\), max \(17\)

So filling the boxes:

min: \(0\)

Q1: \(2.5\)

median: \(4.5\)

Q3: \(8.5\)

max: \(17\)