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Let's start by solving part (a) of the problem, where we need to estimate the mean waiting time using the provided table. The steps for estimating the mean are as follows:

### Step 1: Find the midpoint for each time interval.
- For \(0 < x \leq 10\), the midpoint = \(\frac{0 + 10}{2} = 5\)
- For \(10 < x \leq 20\), the midpoint = \(\frac{10 + 20}{2} = 15\)
- For \(20 < x \leq 30\), the midpoint = \(\frac{20 + 30}{2} = 25\)
- For \(30 < x \leq 40\), the midpoint = \(\frac{30 + 40}{2} = 35\)
- For \(40 < x \leq 50\), the midpoint = \(\frac{40 + 50}{2} = 45\)
- For \(50 < x \leq 60\), the midpoint = \(\frac{50 + 60}{2} = 55\)

### Step 2: Multiply the midpoint by the corresponding frequency to get the total for each class interval.
- \(5 \times 1 = 5\)
- \(15 \times 9 = 135\)
- \(25 \times 11 = 275\)
- \(35 \times 20 = 700\)
- \(45 \times 27 = 1215\)
- \(55 \times 32 = 1760\)

### Step 3: Add up the totals and the frequencies.
- Total = \(5 + 135 + 275 + 700 + 1215 + 1760 = 4090\)
- Total frequency = \(1 + 9 + 11 + 20 + 27 + 32 = 100\)

### Step 4: Estimate the mean.
\[
\text{Mean} = \frac{\text{Total sum of midpoint × frequency}}{\text{Total frequency}} = \frac{4090}{100} = 40.9
\]

So, the estimated mean waiting time is **40.9 minutes** (to 1 decimal place).

---

### Part (b): Benjamin's New Estimated Mean
Benjamin rewrites the intervals into 3 intervals of 20 minutes each. The new intervals are:

- \(0 < x \leq 20\): Combine the first two rows. Frequency = \(1 + 9 = 10\)
- \(20 < x \leq 40\): Combine the next two rows. Frequency = \(11 + 20 = 31\)
- \(40 < x \leq 60\): Combine the last two rows. Frequency = \(27 + 32 = 59\)

### Step 1: Midpoints for the new intervals.
- For \(0 < x \leq 20\), the midpoint = \(\frac{0 + 20}{2} = 10\)
- For \(20 < x \leq 40\), the midpoint = \(\frac{20 + 40}{2} = 30\)
- For \(40 < x \leq 60\), the midpoint = \(\frac{40 + 60}{2} = 50\)

### Step 2: Multiply the midpoint by the corresponding frequency.
- \(10 \times 10 = 100\)
- \(30 \times 31 = 930\)
- \(50 \times 59 = 2950\)

### Step 3: Add up the totals and the frequencies.
- Total = \(100 + 930 + 2950 = 3980\)
- Total frequency = \(10 + 31 + 59 = 100\)

### Step 4: New estimated mean.
\[
\text{New mean} = \frac{3980}{100} = 39.8
\]

### Step 5: Find the difference between the original and new mean estimates.
\[
\text{Difference} = 40.9 - 39.8 = 1.1
\]

So, the difference between the original estimate and Benjamin's new estimate is **1.1 minutes** (to 1 decimal place).

---

Do you have any questions or would you like further details on any step?

Here are 5 related questions you might consider:
1. How does combining intervals affect the accuracy of the mean estimate?
2. What would happen to the mean if we used more intervals with narrower widths?
3. Can you use the weighted mean formula to check the calculations?
4. How would the median waiting time differ from the mean in this case?
5. Why is it important to round the final answers to 1 decimal place in such problems?

**Tip:** When combining intervals for a new estimation, the accuracy can decrease as details are lost from the original data.

Work out an estimate of the mean waiting time. Benjamin rewrites the table with 3 intervals, each 20 minutes wide, and calculates a new estimated mean. Work out the difference between the original estimate and Benjamin's new estimated mean.

Learn how to estimate the mean waiting time from grouped data, calculate new means after grouping intervals, and find the difference between original and new estimates. This statistical problem involves mean calculations, suitable for Grades 8-10.

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