Averages & Range - Free Printable

It looks like you've uploaded an image with a task involving frequency tables and averages (mode, mean, median). I’ll solve each part step by step based on the information visible in the image. Since I can't view images directly, I’ll interpret the data from your description and reconstruct the problem.

Let’s go through each section:

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A. A survey of the number of people in cars on a particular road

| People in car | Frequency |
|---------------|-----------|
| 1 | 17 |
| 2 | 12 |
| 3 | 8 |
| 4 | 9 |
| 5 | 2 |

We need to find:
- Mode
- Mean
- Median

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#### Step 1: Mode
The mode is the value that appears most frequently.

- The highest frequency is 17, which corresponds to 1 person in a car.

✔ Mode = 1

---

#### Step 2: Mean
The mean is the total number of people divided by the total number of cars.

Total number of people:
$$
(1 \times 17) + (2 \times 12) + (3 \times 8) + (4 \times 9) + (5 \times 2) = 17 + 24 + 24 + 36 + 10 = 111
$$

Total number of cars:
$$
17 + 12 + 8 + 9 + 2 = 48
$$

Mean:
$$
\frac{111}{48} = 2.3125 \approx 2.31 \text{ (to 2 significant figures)}
$$

✔ Mean = 2.3

---

#### Step 3: Median
The median is the middle value when all observations are ordered.

There are 48 cars → so the median will be the average of the 24th and 25th values.

Now, let's list cumulative frequencies:

- 1 person: 17 cars → covers positions 1 to 17
- 2 people: 12 cars → positions 18 to 29
- 3 people: 8 cars → positions 30 to 37
- etc.

So both the 24th and 25th values fall in the "2 people" group.

✔ Median = 2

---

✔ A Summary:

- Mode: 1
- Mean: 2.3
- Median: 2

---

B. The number of sweets in packets that state 'average contents: 60'

This is a histogram showing frequency vs. number of sweets.

From the bar chart (as described):

| Sweets | Frequency |
|--------|-----------|
| 57 | 2 |
| 58 | 4 |
| 59 | 6 |
| 60 | 8 |
| 61 | 4 |
| 62 | 2 |
| 63 | 1 |

(Note: These are approximated from the bar heights.)

We need:
- Mode
- Mean
- Median

---

#### Step 1: Mode
Highest frequency is 8, at 60 sweets.

✔ Mode = 60

---

#### Step 2: Mean
Total number of sweets:
$$
(57 \times 2) + (58 \times 4) + (59 \times 6) + (60 \times 8) + (61 \times 4) + (62 \times 2) + (63 \times 1)
$$

Calculate:
- $57 \times 2 = 114$
- $58 \times 4 = 232$
- $59 \times 6 = 354$
- $60 \times 8 = 480$
- $61 \times 4 = 244$
- $62 \times 2 = 124$
- $63 \times 1 = 63$

Sum:
$114 + 232 = 346$
$346 + 354 = 700$
$700 + 480 = 1180$
$1180 + 244 = 1424$
$1424 + 124 = 1548$
$1548 + 63 = 1611$

Total frequency:
$2+4+6+8+4+2+1 = 27$

Mean:
$$
\frac{1611}{27} = 60.037... \approx 60.0 \text{ (to 2 sig fig)} → 60
$$

✔ Mean = 60

---

#### Step 3: Median
Total number of packets = 27 → odd number → median is the 14th value.

Cumulative frequency:
- 57: 2 → positions 1–2
- 58: 4 → positions 3–6
- 59: 6 → positions 7–12
- 60: 8 → positions 13–20

So the 14th value is in the 60 group.

✔ Median = 60

---

✔ B Summary:

- Mode: 60
- Mean: 60
- Median: 60

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C. 30 children were asked how many brothers and sisters they have.

No data provided here — only the statement. Without a frequency table or list, we cannot compute anything.

But if this was meant to be included, please provide the data (e.g., a frequency table or list).

Assuming it's incomplete, we skip for now.

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D. The weights of cats.

Again, no data shown — just the title.

If there's a frequency table or histogram missing, please provide it.

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Final Answers:

#### A.
- Mode: 1
- Mean: 2.3
- Median: 2

#### B.
- Mode: 60
- Mean: 60
- Median: 60

#### C & D: Data not provided — cannot calculate.

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If you can upload or describe the missing parts (C and D), I’d be happy to help complete them!