Michael recorded the maximum temperature every day in September. The table shows information about his results.

\begin{tabular}{|c|c|}
\hline Temperature [tex]$\left({ }^{\circ} C \right)$[/tex] & Frequency \\
\hline [tex]$14 \ \textless \ t \leqslant 18$[/tex] & 4 \\
\hline [tex]$18 \ \textless \ t \leqslant 20$[/tex] & 10 \\
\hline [tex]$20 \ \textless \ t \leqslant 22$[/tex] & 8 \\
\hline [tex]$22 \ \textless \ t \leqslant 24$[/tex] & 5 \\
\hline [tex]$24 \ \textless \ t \leqslant 28$[/tex] & 3 \\
\hline
\end{tabular}

1. Find the class interval that contains the median.

2. Calculate an estimate for the mean maximum temperature.

Sure! Let's carefully go through the steps to solve the problem.

### Step 1: Organize the Data
The data is already given in a table format:

| Temperature [tex]$ \left({ }^{\circ} C \right) $[/tex] | Frequency |
|----------------------------------------------|-----------|
| [tex]$14 < t \leq 18$[/tex] | 4 |
| [tex]$18 < t \leq 20$[/tex] | 10 |
| [tex]$20 < t \leq 22$[/tex] | 8 |
| [tex]$22 < t \leq 24$[/tex] | 5 |
| [tex]$24 < t \leq 28$[/tex] | 3 |

### Step 2: Calculate the Total Number of Observations
To find the total number of observations, we sum up all the frequencies:

[tex]\[ 4 + 10 + 8 + 5 + 3 = 30 \][/tex]

So, the total number of observations is 30.

### Step 3: Calculate the Cumulative Frequency
Next, we need to calculate the cumulative frequency for each class interval:

| Class Interval | Frequency | Cumulative Frequency |
|---------------------|-----------|----------------------|
| [tex]$14 < t \leq 18$[/tex] | 4 | 4 |
| [tex]$18 < t \leq 20$[/tex] | 10 | 4 + 10 = 14 |
| [tex]$20 < t \leq 22$[/tex] | 8 | 14 + 8 = 22 |
| [tex]$22 < t \leq 24$[/tex] | 5 | 22 + 5 = 27 |
| [tex]$24 < t \leq 28$[/tex] | 3 | 27 + 3 = 30 |

### Step 4: Find the Median Class Interval
The median position is calculated as:

[tex]\[ \frac{\text{Total Number of Observations}}{2} = \frac{30}{2} = 15 \][/tex]

Now, we look for the class interval in which the cumulative frequency just surpasses the median position of 15:

- Cumulative frequency of the first class interval (4) is less than 15.
- Cumulative frequency of the second class interval (14) is still less than 15.
- Cumulative frequency of the third class interval (22) is greater than 15.

So, the median class interval is [tex]$20 < t \leq 22$[/tex].

### Step 5: Calculate an Estimate for the Mean Maximum Temperature
To estimate the mean, we first need the midpoints of each class interval. These midpoints are calculated as follows:

- For [tex]$14 < t \leq 18$[/tex]: midpoint is [tex]$ \frac{14+18}{2} = 16.0$[/tex]
- For [tex]$18 < t \leq 20$[/tex]: midpoint is [tex]$ \frac{18+20}{2} = 19.0$[/tex]
- For [tex]$20 < t \leq 22$[/tex]: midpoint is [tex]$ \frac{20+22}{2} = 21.0$[/tex]
- For [tex]$22 < t \leq 24$[/tex]: midpoint is [tex]$ \frac{22+24}{2} = 23.0$[/tex]
- For [tex]$24 < t \leq 28$[/tex]: midpoint is [tex]$ \frac{24+28}{2} = 26.0$[/tex]

Next, we calculate the total sum of the products of frequencies and their corresponding midpoints:

[tex]\[ (4 \times 16.0) + (10 \times 19.0) + (8 \times 21.0) + (5 \times 23.0) + (3 \times 26.0) \][/tex]

[tex]\[ = 64.0 + 190.0 + 168.0 + 115.0 + 78.0 = 615.0 \][/tex]

Finally, the estimated mean maximum temperature is:

[tex]\[ \frac{\text{Total Sum of the Products}}{\text{Total Number of Observations}} = \frac{615.0}{30} = 20.5^\circ C \][/tex]

### Summary of Results
- The class interval that contains the median is [tex]$20 < t \leq 22$[/tex].
- The estimated mean maximum temperature is [tex]$20.5^\circ C$[/tex].