Question 1187096: Given below is a back-to-back stem-and-leaf diagram in which common stems are placed at the center of the diagram while the leaves of the two data set are listed at the left- and right-hand side of the given stems. The following back-to-back stem-and-leaf diagram determines the time in seconds for the two groups of participants to solve a puzzle when exposed to different noise level. The first group of participant solves the puzzle in the classroom wherein the noise level is maintained at 60 decibel. For the other group, a decibel level of 45 was maintained while they solved the same problem.
Stem 3: Group1: 0 Group2: 4
Stem 4: Group1: 5 Group2: 3,3,6,7
Stem 5: Group1: 7,3,1 Group2: 1,4,4,5,6,8
Stem 6: Group1: 7,3,3,2 Group2: 3,4,5
Stem 7: Group1: 6,3,2,1 Group2: 3
Stem 8: Group1: 7,1 Group2: 0
i) How many participants in group 1 that needs greater than 45 seconds to solve the puzzle?
ii) How many participants in group 2 that requires 45 seconds to solve the puzzle?
iii) By examining the stem-and-leaf display, which of the two groups of participants seemed to have a greater mean in seconds required in solving the puzzle?
Answer by CPhill(1987)   (Show Source):
You can put this solution on YOUR website! Here's the breakdown of the stem-and-leaf diagram and the answers to your questions:
**Understanding the Diagram:**
The stem represents the tens digit of the time in seconds, and the leaves represent the units digit. For example, a stem of 5 and a leaf of 7 on the Group 1 side represents a time of 57 seconds.
**i) Group 1 participants taking more than 45 seconds:**
* Stem 5: 7, 3, 1 (3 participants)
* Stem 6: 7, 3, 3, 2 (4 participants)
* Stem 7: 6, 3, 2, 1 (4 participants)
* Stem 8: 7, 1 (2 participants)
Total: 3 + 4 + 4 + 2 = **13 participants**
**ii) Group 2 participants taking exactly 45 seconds:**
* Stem 4: 5 (1 participant)
So, **1 participant** took exactly 45 seconds.
**iii) Which group has a greater mean?**
Looking at the distribution of leaves, Group 2 has more values concentrated at the lower stems (and thus lower times). Group 1 has more data points at the higher stems (and thus higher times). This visual inspection suggests that **Group 1** likely has a greater mean time required to solve the puzzle. While we could calculate the exact means, the distribution makes it fairly clear.
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