Question

the weights, in pounds, of
packages on a delivery truck are
shown in the stem-and-leaf plot.
find the mean, the median, and
the mode of the data, if possible. if
any measure cannot be found or
does not represent the center of
the data, explain why

0 | 3 8
1 | 0 4 5 8
2 | 0 0 0 7 8 8 9
3 | 0 0 0 2 4 5 7 8 9
4 | 2 3 7 8 9
5 | 0
key: 3 | 0 = 30

a. the mean represents the center of the data set.
b. the mean does not represent the center because it is the greatest
data entry.
c. the mean does not represent the center because it is not a data
entry.
d. the mean does not represent the center because it is the least
data entry.
e. the data set does not have a mean.

find the median. select the correct choice below and, if necessary, fill in
the answer box to complete your choice.
a. the median is \square
(round to one decimal place as needed.)
b. the median cannot be calculated because the sample size is too
small.
c. the median cannot be calculated because the data are at the
nominal level of measurement.
d. the median cannot be calculated because there is an even
number of data entries

Explanation:

Part 1: Analyze the Mean (from the options)

To determine if the mean represents the center, we first recall: the mean is the sum of all data points divided by the number of data points. It is a measure of central tendency (represents the center) unless there are extreme outliers, but the options here don’t mention outliers. Let's check the options:

• Option B: Mean is not the greatest data entry (greatest data entry here is 50, mean will be a value around the center, not the max). Eliminate B.
• Option C: Mean can be a non - data entry (e.g., if data is 1,2,3, mean is 2 which is a data entry; if data is 1,2,4, mean is 7/3 ≈ 2.33, not a data entry, but it still represents the center). So this reasoning is wrong. Eliminate C.
• Option D: Mean is not the least data entry (least data entry here is 3, mean is around the center). Eliminate D.
• Option E: All numerical data sets have a mean (we can sum the values and divide by count). Eliminate E.
• Option A: The mean is a measure of central tendency and represents the center of the data set (when there are no extreme outliers skewing it severely, and here the data is about package weights with a stem - and - leaf plot that doesn't show extreme skew). So the correct option for the mean part is A.

Part 2: Find the Median

First, we need to list out all the data points from the stem - and - leaf plot:

• Stem 0: 3, 8 (2 values)
• Stem 1: 10, 14, 15, 18 (4 values)
• Stem 2: 20, 20, 20, 27, 28, 28, 29 (7 values)
• Stem 3: 30, 30, 30, 32, 34, 35, 37, 38, 39 (9 values)
• Stem 4: 42, 43, 47, 48, 49 (5 values)
• Stem 5: 50 (1 value)

Now, calculate the total number of data points: \(2 + 4+7 + 9+5 + 1=28\)

The median of a data set with \(n\) values (where \(n = 28\), an even number) is the average of the \(\frac{n}{2}\)-th and \((\frac{n}{2}+ 1)\)-th values when the data is ordered.

\(\frac{n}{2}=\frac{28}{2}=14\) and \(\frac{n}{2}+1 = 15\)

Now, we find the cumulative number of data points:

• After stem 0: 2 values
• After stem 1: \(2 + 4=6\) values
• After stem 2: \(6+7 = 13\) values
• After stem 3: \(13 + 9=22\) values

The 14th value: We have 13 values before stem 3. So the 14th value is the first value in stem 3? Wait, no. Wait, stem 3 has 9 values. The first value in stem 3 is the 14th value? Wait, cumulative up to stem 2 is 13. So the 14th value is the first value in stem 3 (30), the 15th value is the second value in stem 3 (30)? Wait, no, let's list the positions:

1: 3

2: 8

3: 10

4: 14

5: 15

6: 18

7: 20

8: 20

9: 20

10: 27

11: 28

12: 28

13: 29

14: 30

15: 30

16: 30

17: 32

18: 34

19: 35

20: 37

21: 38

22: 39

23: 42

24: 43

25: 47

26: 48

27: 49

28: 50

Wait, no, I made a mistake in counting. Let's recalculate the cumulative frequency:

• Stem 0: 2 values (positions 1 - 2)
• Stem 1: 4 values (positions 3 - 6)
• Stem 2: 7 values (positions 7 - 13)
• Stem 3: 9 values (positions 14 - 22)
• Stem 4: 5 values (positions 23 - 27)
• Stem 5: 1 value (position 28)

So the 14th value is the first value in stem 3 (30), and the 15th value is the second value in stem 3 (30)? Wait, no, stem 3 has values: 30, 30, 30, 32, 34, 35, 37, 38, 39. So the 14th value is 30 (position 14), 15th value is 30 (position 15)? Wait, no, cumulative up to stem 2 is 13. So position 14: first value of stem 3 (30), position 15: second value of stem 3 (30). Wait, but let's check again.

Wait, stem 0: 2 values (3,8)

Stem 1: 4 values (10,14,15,18) → total 6

Stem 2: 7 values (20,20,20,27,28,28,29) → total 13

Stem 3: 9 values (30,30,30,32,34,35,37,38,39) → positions 14 - 22 (14:30, 15:30, 16:30, 17:32…

The mean is a measure of central tendency. The other options have incorrect reasoning (mean is not max/min, all data sets have a mean, and mean can be non - data entry but still represent center).

Step1: Count total data points

Sum the number of values in each stem: \(2 + 4+7 + 9+5 + 1 = 28\).

Step2: Find positions of median values

For \(n = 28\) (even), median is average of 14th and 15th values.

Step3: Determine 14th and 15th values

Cumulative counts: stem 0 (2), stem 1 (6), stem 2 (13). 14th and 15th values are in stem 3 (30, 30).

Step4: Calculate median

\(\frac{30 + 30}{2}=30.0\)

Answer:

A. The mean represents the center of the data set.

For the median - related question: