Question

data analysis and probability using back - to - back stem - and - leaf displays to compare data sets
a stem - and - leaf display below gives the test scores for the students in prof. jones 10 a.m. and 11 a.m. management classes. there were 19 students in the 10 a.m. class and 16 students in the 11 a.m. class. use the display to answer the questions that follow.
test scores for prof. jones management classes
10 a.m. 11 a.m.
8 4 4 6 6
5 2 0 0 5 8
5 4 3 0 6 3 3 5 6 6 9
9 6 5 4 3 0 7 0 4 7 8 9 9
1 8 8
2 0 9
key: 44 - 4|4|6→46
(a) what were the ranges for the two classes?
10 a.m.
11 a.m.
(b) which class had the higher median score?
10 a.m. 11 a.m. the medians were the same
(c) which class had more scores in the 80s?
10 a.m. 11 a.m. each had the same

Explanation:

Step1: Find the range for 10 a.m. class

The range is calculated as \( \text{Max}-\text{Min} \).
For 10 a.m. class, \( \text{Max} = 92 \), \( \text{Min}=44 \).
So, \( 92 - 44=48 \).

Step2: Find the range for 11 a.m. class

For 11 a.m. class, \( \text{Max} = 88 \), \( \text{Min}=46 \).
So, \( 88 - 46 = 42 \).

Step3: Find the median for 10 a.m. class

There are \( n = 19 \) data points. The median is the \( \frac{n + 1}{2}=\frac{19+1}{2}=10^{th} \) value.
Ordering the data: \( 44,48,50,50,52,60,63,64,65,70,73,74,75,76,79,81,90,92 \). The \( 10^{th} \) value is \( 70 \).

Step4: Find the median for 11 a.m. class

There are \( n = 16 \) data points. The median is the average of \( \frac{n}{2}=8^{th} \) and \( \frac{n}{2}+1 = 9^{th} \) values.
Ordering the data: \( 46,46,58,63,63,65,66,69,70,74,76,78,79,79,88 \).
\( \text{Median}=\frac{69 + 70}{2}=69.5 \).
Since \( 70>69.5 \), 10 a.m. class has higher median.

Step5: Count scores in 80s

For 10 a.m. class: 1 score (\( 81 \)).
For 11 a.m. class: 1 score (\( 88 \)).

Answer:

(a) 10 a.m.: \( 48 \); 11 a.m.: \( 42 \)
(b) 10 a.m.
(c) Each had the same