Ordered stem-and-leaf, averages, spread and grouped means

(a) $1$

1. The mode is the value with the greatest frequency.
2. The greatest frequency is 8, which belongs to 1 umbrella. The mode is 1.

• B1 Correct answer: $1$

(a) $12$ cm

1. Put the lengths in order: 7, 9, 11, 12, 15, 16, 20.
2. Put the lengths in order: 7, 9, 11, 12, 15, 16, 20.
3. The median of seven values is the fourth value, so the median length is 12 cm.

• P1 Put the lengths in order: 7, 9, 11, 12, 15, 16, 20.
• A1 Correct answer: $12$ cm

(a) 0.9

1. Total goals.
   $$0 \times 3 + 1 \times 5 + 2 \times 2 = 9$$
2. Divide by the number of matches.
   $$9 \div 10 = 0.9$$

• M1 Finding the total, 9 goals.
• A1 Correct answer: 0.9.

(a) $1.25$ books

1. $$0\times 4+1\times 9+2\times 5+3\times 2$$
2. $$25/20$$
3. For a frequency table, total the products of each value and its frequency.
4. The total number of books is 0 $\times$ 4 + 1 $\times$ 9 + 2 $\times$ 5 + 3 $\times$ 2 = 25.
5. Mean = total $\div$ number of students = 25 $\div$ 20 = 1.25 books.

• P1 Establishing $0\times 4+1\times 9+2\times 5+3\times 2$ or an equivalent valid method.
• P1 Establishing $25/20$ or an equivalent valid method.
• A1 Correct answer: $1.25$ books

(a) $26$

1. $$6\times 18$$
2. $$108-(12+17+20+19+14)$$
3. Total = mean $\times$ number of values, so the six scores total 18 $\times$ 6 = 108.
4. The five known scores total 12 + 17 + 20 + 19 + 14 = 82.
5. The sixth score is 108 $-$ 82 = 26.

• P1 Establishing $6\times 18$ or an equivalent valid method.
• P1 Establishing $108-(12+17+20+19+14)$ or an equivalent valid method.
• A1 Correct answer: $26$

(a) 21 minutes

1. Use the midpoints 5, 15, 25 and 35.
2. Multiply by the frequencies and add.
   $$4(5) + 9(15) + 12(25) + 5(35) = 20 + 135 + 300 + 175 = 630$$
3. Divide by the total frequency, 30.
   $$630 \div 30 = 21$$

• P1 Using the midpoints.
• P1 The sum of frequency times midpoint, 630.
• P1 Dividing the total by 30.
• A1 Correct answer: 21.

(b) The exact times are not known; each class is represented by its midpoint.

1. Grouping loses the individual values, so the midpoint stands in for every time in its class.

• C1 The actual times within each class are unknown (midpoints are used).

(a) $40$ kg

1. $$18\times 42+12\times 37$$
2. $$1200/30$$
3. Recover each group total using total = mean $\times$ number.
4. The combined mass is 18 $\times$ 42 + 12 $\times$ 37 = 756 + 444 = 1200 kg.
5. Divide by the combined number: 1200 $\div$ 30 = 40 kg. Averaging 42 and 37 directly would ignore the unequal group sizes.

• P1 Establishing $18\times 42+12\times 37$ or an equivalent valid method.
• P1 Establishing $1200/30$ or an equivalent valid method.
• A1 Correct answer: $40$ kg

(a) $27.5$

1. $$24n+32(n+4)=29(2n+4)$$
2. $$29-24/16$$
3. Group totals are 24n and 32(n + 4), and there are 2n + 4 readings.
4. Use the combined mean: 24n + 32(n + 4) = 29(2n + 4).
5. This gives 56n + 128 = 58n + 116, so n = 6. There are 16 readings in total.
6. Their recorded total is 29 $\times$ 16 = 464. Correcting 38 to 14 reduces it by 24 to 440.
7. The corrected mean is 440 $\div$ 16 = 27.5.

• P1 Establishing $24n+32(n+4)=29(2n+4)$ or an equivalent valid method.
• P1 Establishing $29-24/16$ or an equivalent valid method.
• A1 Correct answer: $27.5$