Question

the stem-and-leaf plot below gives the test scores for mr. uzuns english classes. there were 18 students in fourth period and 17 in fifth period. use the plot to answer the questions. score on mr. uzuns english test fourth period fifth period 9 5 2 6 7 7 6 3 1 6 1 3 9 9 9 8 7 1 4 6 6 9 8 4 3 2 0 8 1 3 4 5 9 9 4 2 9 0 1 5 (a) which class had more test scores in the 60s? fourth period fifth period each had the same (b) what were the ranges of test scores for the two classes? fourth period fifth period (c) which class had the greater median test score? fourth period fifth period the medians were the same

Explanation:

Part (a)

Step1: Count 60s in Fourth period

In the stem - and - leaf plot, for the fourth period and stem 6, the leaves are 7, 7, 6, 3, 1. So the number of scores in the 60s is 5.

Step2: Count 60s in Fifth period

For the fifth period and stem 6, the leaves are 1, 3, 9. So the number of scores in the 60s is 3.
Since \(5>3\), the fourth period has more test scores in the 60s.

Step1: Find range for Fourth period

The range is calculated as \( \text{Maximum}-\text{Minimum} \). For the fourth period, the minimum score is 59 (stem 5, leaf 9) and the maximum score is 99 (stem 9, leaf 9). So the range is \(99 - 59=40\).

Step2: Find range for Fifth period

For the fifth period, the minimum score is 52 (stem 5, leaf 2) and the maximum score is 95 (stem 9, leaf 5). So the range is \(95 - 52 = 43\).

Step1: Find median for Fourth period

The fourth period has \(n = 18\) students. The median is the average of the \(\frac{n}{2}=9^{\text{th}}\) and \((\frac{n}{2}+ 1)=10^{\text{th}}\) values.
First, we list the scores in order. After counting the number of scores in each stem:

• Stem 5: 1 score
• Stem 6: 5 scores (total so far: \(1 + 5=6\))
• Stem 7: 3 scores (total so far: \(6+3 = 9\))
• Stem 8: 4 scores (the \(9^{\text{th}}\) score is the last score in stem 7, and the \(10^{\text{th}}\) score is the first score in stem 8. The score at stem 7, last leaf: 79; score at stem 8, first leaf: 80. The median is \(\frac{79 + 80}{2}=79.5\))

Step2: Find median for Fifth period

The fifth period has \(n = 17\) students. The median is the \(\frac{n + 1}{2}=9^{\text{th}}\) value.
Counting the number of scores in each stem:

• Stem 5: 2 scores
• Stem 6: 3 scores (total so far: \(2+3 = 5\))
• Stem 7: 5 scores (total so far: \(5 + 5=10\)) Wait, no, let's count correctly:
• Stem 5: 2 scores
• Stem 6: 3 scores (total: \(2+3 = 5\))
• Stem 7: 5 scores (total: \(5+5 = 10\)) - no, we need the \(9^{\text{th}}\) value. Let's list the cumulative counts:

Stem 5: 2 (positions 1 - 2)
Stem 6: 3 (positions 3 - 5)
Stem 7: 5 (positions 6 - 10). Wait, we need the \(9^{\text{th}}\) value. The \(6^{\text{th}}\) to \(10^{\text{th}}\) values are in stem 7. The \(9^{\text{th}}\) value in the ordered list:
After stem 5 (2 scores) and stem 6 (3 scores), we have 5 scores in stem 7. The \(9^{\text{th}}\) score is the \(9-(2 + 3)=4^{\text{th}}\) score in stem 7. The scores in stem 7 for fifth period: 71, 74, 76, 76, 79. The \(4^{\text{th}}\) score is 76.

Since \(79.5>76\), the fourth period has a greater median.

Answer:

A. Fourth period

Part (b)