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The table shows information about the heights of 80 plants.

\begin{tabular}{|c|c|}
\hline Height [tex]$(h \, \text{cm})$[/tex] & Frequency \\
\hline [tex]$10 \ \textless \ h \leq 20$[/tex] & 7 \\
\hline [tex]$20 \ \textless \ h \leq 30$[/tex] & 13 \\
\hline [tex]$30 \ \textless \ h \leq 40$[/tex] & 14 \\
\hline [tex]$40 \ \textless \ h \leq 50$[/tex] & 12 \\
\hline [tex]$50 \ \textless \ h \leq 60$[/tex] & 16 \\
\hline [tex]$60 \ \textless \ h \leq 70$[/tex] & 18 \\
\hline
\end{tabular}

1. Find the class interval that contains the median.

2. On the grid, draw a frequency polygon for the information in the table.



Answer :

GinnyAnswer
To find the class interval that contains the median and to draw a frequency polygon, let's follow these steps:

### Finding the Class Interval that Contains the Median:

1. List the Frequencies and Determine the Total Number of Plants:
- [tex]\(10 < h \leqslant 20\)[/tex]: [tex]\(7\)[/tex] plants
- [tex]\(20 < h \leqslant 30\)[/tex]: [tex]\(13\)[/tex] plants
- [tex]\(30 < h \leqslant 40\)[/tex]: [tex]\(14\)[/tex] plants
- [tex]\(40 < h \leqslant 50\)[/tex]: [tex]\(12\)[/tex] plants
- [tex]\(50 < h \leqslant 60\)[/tex]: [tex]\(16\)[/tex] plants
- [tex]\(60 < h \leqslant 70\)[/tex]: [tex]\(18\)[/tex] plants
- Total number of plants: [tex]\(7 + 13 + 14 + 12 + 16 + 18 = 80\)[/tex]

2. Calculate the Cumulative Frequency:
- For [tex]\(10 < h \leqslant 20\)[/tex]: [tex]\(7\)[/tex]
- For [tex]\(20 < h \leqslant 30\)[/tex]: [tex]\(7 + 13 = 20\)[/tex]
- For [tex]\(30 < h \leqslant 40\)[/tex]: [tex]\(20 + 14 = 34\)[/tex]
- For [tex]\(40 < h \leqslant 50\)[/tex]: [tex]\(34 + 12 = 46\)[/tex]
- For [tex]\(50 < h \leqslant 60\)[/tex]: [tex]\(46 + 16 = 62\)[/tex]
- For [tex]\(60 < h \leqslant 70\)[/tex]: [tex]\(62 + 18 = 80\)[/tex]

3. Find the Median Position:
- Median position is at [tex]\(\frac{total\ number\ of\ plants}{2} = \frac{80}{2} = 40\)[/tex]

4. Identify the Class Interval Containing the Median:
- The cumulative frequency just before 40 is 34, and the next cumulative frequency after 34 is 46.
- Therefore, [tex]\(\boldsymbol{40 < h \leqslant 50}\)[/tex] is the class interval that contains the median since [tex]\(34 < 40 \leqslant 46\)[/tex].

### Drawing the Frequency Polygon
To draw a frequency polygon, we first need to determine the midpoints for each class interval:

1. Calculate the Midpoints:
- For [tex]\(10 < h \leqslant 20\)[/tex]: midpoint = [tex]\(\frac{10 + 20}{2} = 15\)[/tex]
- For [tex]\(20 < h \leqslant 30\)[/tex]: midpoint = [tex]\(\frac{20 + 30}{2} = 25\)[/tex]
- For [tex]\(30 < h \leqslant 40\)[/tex]: midpoint = [tex]\(\frac{30 + 40}{2} = 35\)[/tex]
- For [tex]\(40 < h \leqslant 50\)[/tex]: midpoint = [tex]\(\frac{40 + 50}{2} = 45\)[/tex]
- For [tex]\(50 < h \leqslant 60\)[/tex]: midpoint = [tex]\(\frac{50 + 60}{2} = 55\)[/tex]
- For [tex]\(60 < h \leqslant 70\)[/tex]: midpoint = [tex]\(\frac{60 + 70}{2} = 65\)[/tex]

2. Plot the Points:
- [tex]\( (15, 7) \)[/tex]
- [tex]\( (25, 13) \)[/tex]
- [tex]\( (35, 14) \)[/tex]
- [tex]\( (45, 12) \)[/tex]
- [tex]\( (55, 16) \)[/tex]
- [tex]\( (65, 18) \)[/tex]

3. Draw the Frequency Polygon:
- Start by plotting these points on the grid.
- Join the points with straight lines in sequence.
- Optionally, you may extend lines to the baseline at the endpoints (these would be [tex]\((15, 0)\)[/tex] and [tex]\((65, 0)\)[/tex] to complete the polygon).

To summarize:

Class interval that contains the median: [tex]\( \boldsymbol{40 < h \leqslant 50} \)[/tex]

For the frequency polygon:
1. Calculate midpoints of each class.
2. Plot the midpoints against their frequencies.
3. Connect these points with straight lines.

Here is a simple depiction of how the frequency polygon should appear (points connected):

```
F
r 18 .
e / \
q 16 / \
u / .
e 14 . / \
n \ / \
c 12 \ / \
y . / .
7 . / \ /
\ / \ /
|_____/_____\_________/_____________>
15 25 35 45 55 65
```

Note: The grid and plotting would be done on graph paper or with a graphing tool for higher accuracy.

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