Find the mean of the data summarized in the given frequency distribution. Compare the computed mean to the actual mean of 51.7 miles per hour.

[tex]
\[
\begin{tabular}{l|ccccc}
\text{Speed (miles per hour)} & 42-45 & 46-49 & 50-53 & 54-57 & 58-61 \\
\hline
\text{Frequency} & 26 & 12 & 6 & 4 & 1 \\
\end{tabular}
\]
[/tex]

The mean of the frequency distribution is [tex]$\square$[/tex] miles per hour. (Round to the nearest tenth as needed.)

To find the mean (average) of the data summarized in the given frequency distribution, follow these steps:

1. Identify the midpoints of each interval:
The intervals and their midpoints are:
- [tex]$42-45$[/tex] -> midpoint = [tex]$ \frac{42 + 45}{2} = 43.5 $[/tex]
- [tex]$46-49$[/tex] -> midpoint = [tex]$ \frac{46 + 49}{2} = 47.5 $[/tex]
- [tex]$50-53$[/tex] -> midpoint = [tex]$ \frac{50 + 53}{2} = 51.5 $[/tex]
- [tex]$54-57$[/tex] -> midpoint = [tex]$ \frac{54 + 57}{2} = 55.5 $[/tex]
- [tex]$58-61$[/tex] -> midpoint = [tex]$ \frac{58 + 61}{2} = 59.5 $[/tex]

2. List the frequencies:
- Frequency for 43.5: 26
- Frequency for 47.5: 12
- Frequency for 51.5: 6
- Frequency for 55.5: 4
- Frequency for 59.5: 1

3. Multiply each midpoint by its corresponding frequency:
- [tex]$43.5 \times 26 = 1131$[/tex]
- [tex]$47.5 \times 12 = 570$[/tex]
- [tex]$51.5 \times 6 = 309$[/tex]
- [tex]$55.5 \times 4 = 222$[/tex]
- [tex]$59.5 \times 1 = 59.5$[/tex]

4. Sum these products:
[tex]\[
1131 + 570 + 309 + 222 + 59.5 = 2291.5
\][/tex]

5. Sum the frequencies to find the total number of observations:
[tex]\[
26 + 12 + 6 + 4 + 1 = 49
\][/tex]

6. Compute the mean of the frequency distribution:
[tex]\[
\text{Mean} = \frac{2291.5}{49} \approx 46.0
\][/tex]

Comparing this computed mean to the actual mean of 51.7 miles per hour, we note that the computed mean ([tex]$46.0 $[/tex] miles per hour) is lower than the actual mean.

Thus, the mean of the frequency distribution is approximately [tex]$ 46.0 $[/tex] miles per hour.