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Eight students on a class treasure hunt found the following number of hidden chocolates:

| Ralph | Jim | Jack | Jennifer | Zach | Anna | Chris | Sam |
|-------|-----|------|----------|------|------|-------|-----|
| 5 | 2 | 6 | 1 | 2 | 0 | 3 | 7 |

Their teacher wanted to be sure that each student received the same number of chocolates after the event. Which measure(s) of central tendency can the teacher use to distribute the chocolates equally among all of the students?

A. the median
B. the mode
C. the mean
D. either the mean or the mode


Sagot :

To determine how to distribute the chocolates equally among all the students, let’s consider the available measures of central tendency: the median, the mode, and the mean.

### Given Data:
The number of chocolates found by each student is:
5, 2, 6, 1, 2, 0, 3, 7

### Calculating the Measures of Central Tendency:

1. Median:
- First, sort the data: 0, 1, 2, 2, 3, 5, 6, 7
- Since there are 8 numbers (an even number of data points), the median is the average of the 4th and 5th numbers in the sorted list.
- The 4th number is 2 and the 5th number is 3.
- Median = (2 + 3) / 2 = 2.5

2. Mode:
- The mode is the number that appears most frequently.
- In the sorted list: 0, 1, 2, 2, 3, 5, 6, 7, the number 2 appears most frequently.
- Mode = 2

3. Mean:
- The mean is the total number of chocolates divided by the number of students.
- Total chocolates = 5 + 2 + 6 + 1 + 2 + 0 + 3 + 7 = 26
- Number of students = 8
- Mean = 26 / 8 = 3.25

### Distributing Chocolates Equally:

Among the given measures of central tendency (median, mode, and mean):
- Median: By using the median (2.5), it would be impractical to distribute 2.5 chocolates equally to each student since we cannot have half a chocolate.
- Mode: Using the mode (2) would result in students getting only 2 chocolates each, which doesn't utilize all chocolates, leading to an uneven distribution for some.
- Mean: Using the mean (3.25), each student would theoretically receive 3.25 chocolates, which ensures an exact equal division mathematically.

Thus, the most appropriate measure to distribute the chocolates equally among all students is:
C) the mean.