To calculate the mean of the frequency distribution, we use the formula for the mean of grouped dataTo calculate the mean of the frequency distribution, we use the formula for the mean of grouped

$\text{Mean} = \frac{\sum f_i x_i}{\sum f_i}$

Where:

• $f_i$ is the frequency of each group.
• $x_i$ is the midpoint of each group (class).

Step 1: Find the midpoints of the speed intervals.

To get the midpoint ($x_i$) of each class, use the formula:

$\text{Midpoint} = \frac{\text{Lower limit} + \text{Upper limit}}{2}$

For each interval:

• For 42−45: Midpoint = $\frac{42+45}{2} = 43.5$
• For 46−49: Midpoint = $\frac{46+49}{2} = 47.5$
• For 50−53: Midpoint = $\frac{50+53}{2} = 51.5$
• For 54−57: Midpoint = $\frac{54+57}{2} = 55.5$
• For 58−61: Midpoint = $\frac{58+61}{2} = 59.5$

Step 2: Multiply each midpoint by the frequency.

$f_i \times x_i =$

$(29 \times 43.5), (15 \times 47.5), (7 \times 51.5), (4 \times 55.5), (1 \times 59.5)$

Step 3: Compute the sum of $f_i \times x_i$ and the sum of the frequencies $\sum f_i$.

Step 4: Calculate the mean.

Let's compute this.The mean of the frequency distribution is approximately 46.7 miles per hour (rounded to the nearest tenth).

Comparison with the actual mean:

The computed mean of 46.7 miles per hour is lower than the actual mean of 51.65 miles per hour.

Would you like more details or have any questions?

Here are 5 related questions to expand on this topic:

1. How does grouping data affect the accuracy of the mean calculation?
2. What are the pros and cons of using class midpoints in frequency distributions?
3. Can you explain how the mean changes if one of the frequencies is altered?
4. How would you calculate the median and mode for this frequency distribution?
5. How could outliers affect the mean in a frequency distribution?

Tip: When comparing actual and calculated values, always check if rounding or grouping affects the final result.