Draw a stem-and-leaf display for this data Organise the data into class intervals. Calculate the actual mean for this data. Calculate an estimated mean for this data.

• Mean: The sum of all ratings is $1 \times 1 + 4 \times 2 + 12 \times 3 + 12 \times 4 + 7 \times 5 = 1 + 8 + 36 + 48 + 35 = 128$. The total number of ratings is 36. $$\text{Mean} = \frac{128}{36} \approx \boxed{3.56}$$
• Mode: The values that appear most frequently are 3 and 4, both appearing 12 times. $$\text{Mode} = \boxed{3 \text{ and } 4}$$
• Median: With 36 data points, the median is the average of the 18th and 19th values. After sorting, both the 18th and 19th values are 4. $$\text{Median} = \frac{4+4}{2} = \boxed{4}$$
• Range: The difference between the maximum and minimum values is $5 - 1 = 4$. $$\text{Range} = \boxed{4}$$

• Mean: $$\text{Mean} = \frac{\sum (\text{Score} \times \text{Frequency})}{\sum \text{Frequency}} = \frac{(1 \times 2820) + (2 \times 3130) + (3 \times 18080) + (4 \times 45922) + (5 \times 30048)}{100000}$$ $$\text{Mean} = \frac{2820 + 6260 + 54240 + 183688 + 150240}{100000} = \frac{397248}{100000} = \boxed{3.97248}$$
• Mode: The score with the highest frequency (45 922) is 4. $$\text{Mode} = \boxed{4}$$
• Median: The total frequency is 100 000. The median position is the 50 000th value. The cumulative frequency shows that the 50 000th value falls within the score of 4 (cumulative frequency up to score 3 is 24 030, up to score 4 is 69 952). $$\text{Median} = \boxed{4}$$
• Range: The difference between the maximum and minimum scores is $5 - 1 = 4$. $$\text{Range} = \boxed{4}$$

• Mean: $$\text{Mean} = \frac{\sum (\text{Score} \times \text{Frequency})}{\sum \text{Frequency}} = \frac{(1 \times 3910) + (2 \times 10050) + (3 \times 19050) + (4 \times 45890) + (5 \times 21100)}{100000}$$ $$\text{Mean} = \frac{3910 + 20100 + 57150 + 183560 + 105500}{100000} = \frac{370220}{100000} = \boxed{3.7022}$$
• Mode: The score with the highest frequency (45 890) is 4. $$\text{Mode} = \boxed{4}$$
• Median: The total frequency is 100 000. The median position is the 50 000th value. The cumulative frequency shows that the 50 000th value falls within the score of 4 (cumulative frequency up to score 3 is 33 010, up to score 4 is 78 900). $$\text{Median} = \boxed{4}$$
• Range: The difference between the maximum and minimum scores is $5 - 1 = 4$. $$\text{Range} = \boxed{4}$$

(1) Why is it not possible to calculate the actual mean for Station C?
It is not possible to calculate the actual mean because the exact values of the data points within each class interval are unknown. Only the frequency of values within a given range is provided.
(2) Calculate an estimated mean for Station C.
We use the midpoint of each class interval ($x$) and its frequency ($f$).
Total frequency $N = 124+326+402+678+2402+857+211 = 5000$.
Sum of $(f \times x)$ = $(124 \times 35) + (326 \times 45) + (402 \times 55) + (678 \times 65) + (2402 \times 75) + (857 \times 85) + (211 \times 95)$
$= 4340 + 14670 + 22110 + 44070 + 180150 + 72845 + 20045 = 358230$.

(3) In what interval is the mode for Station C?
The modal class is the class interval with the highest frequency. The highest frequency is 2402, which corresponds to the interval $70 \le x < 80$.

(4) In what interval is the median for Station C?
The total number of data points is $N = 5000$. The median position is $\frac{N}{2} = \frac{5000}{2} = 2500$.
Cumulative frequencies:
$30 \le x < 40$: 124
$40 \le x < 50$: $124 + 326 = 450$
$50 \le x < 60$: $450 + 402 = 852$
$60 \le x < 70$: $852 + 678 = 1530$
$70 \le x < 80$: $1530 + 2402 = 3932$
The 2500th value falls within the interval $70 \le x < 80$.

(1) Draw a stem-and-leaf display for this data.
First, sort the data:
44, 49, 50, 52, 54, 55, 56, 57, 59, 64, 64, 65, 66, 68, 68, 71, 71, 72, 74, 75, 77, 78, 80, 82, 84, 88, 88, 96, 97, 98.

Key: $4 | 4 = 44$ litres

(2) Organise the data into class intervals.
Using a class width of 10:

(3) Calculate the actual mean for this data.
Sum of all data points = $82+64+55+50+49+44+52+59+68+74+71+78+88+98+96+77+75+54+57+56+64+66+80+84+88+72+71+65+68+97 = 2009$.
Total number of data points $N = 30$.

(4) Calculate an estimated mean for this data.
Using the class intervals from (d)(2) and their midpoints:
Sum of $(f \times x)$ = $(2 \times 45) + (7 \times 55) + (6 \times 65) + (7 \times 75) + (5 \times 85) + (3 \times 95)$
$= 90 + 385 + 390 + 525 + 425 + 285 = 2100$.
Total frequency $N = 30$.