To find the mean of the data summarized in the frequency distribution, we use the formula for the mean of grouped dataTo find the mean of the data summarized in the frequency distribution, we use the formula for the mean of grouped

$\text{Mean} = \frac{\sum (f \times x)}{\sum f}$

where:

• $f$ is the frequency of each group.
• $x$ is the midpoint of each class interval.

Step 1: Find the Midpoints of Each Interval

1. For $42-45$, the midpoint is $\frac{42 + 45}{2} = 43.5$.
2. For $46-49$, the midpoint is $\frac{46 + 49}{2} = 47.5$.
3. For $50-53$, the midpoint is $\frac{50 + 53}{2} = 51.5$.
4. For $54-57$, the midpoint is $\frac{54 + 57}{2} = 55.5$.
5. For $58-61$, the midpoint is $\frac{58 + 61}{2} = 59.5$.

Step 2: Multiply Each Midpoint by Its Frequency

Next, we calculate the products of each midpoint and its corresponding frequency:

1. $43.5 \times 23 = 1000.5$
2. $47.5 \times 13 = 617.5$
3. $51.5 \times 6 = 309.0$
4. $55.5 \times 4 = 222.0$
5. $59.5 \times 2 = 119.0$

Step 3: Sum the Products and Frequencies

Sum of the products:

$1000.5 + 617.5 + 309.0 + 222.0 + 119.0 = 2268.0$

Sum of the frequencies:

$23 + 13 + 6 + 4 + 2 = 48$

Step 4: Compute the Mean

$\text{Mean} = \frac{2268.0}{48} \approx 47.3 \text{ miles per hour}$

Conclusion

The mean of the frequency distribution is approximately 47.3 miles per hour, which is lower than the actual mean of 51.5 miles per hour.

Would you like further details or explanations on any part? Here are some follow-up questions:

1. How do you calculate the midpoint of a class interval?
2. Why is the mean of a grouped frequency distribution different from the actual mean?
3. How would the mean change if the frequency of the highest class increased?
4. What does the comparison of the calculated and actual means suggest about the data distribution?
5. Can the mean of grouped data be a precise representation of the actual mean?

Tip: When comparing means from grouped data, remember that intervals can cause some loss of precision.