Box plots and distribution comparisons

(a) 9 minutes

1. Median: the 6th value, 11.
2. Lower quartile: the median of the lower five values (3, 5, 6, 8, 9), which is 6. Upper quartile: the median of 12, 14, 15, 18, 21, which is 15.
3. $$15 - 6 = 9$$

• M1 Both quartiles, 6 and 15.
• A1 Correct answer: 9.

(a) 7.5

1. $$\frac{7+8}{2}$$
2. The lower half is 2, 5, 7, 8, 10, 13.
3. For six values, the median is the mean of the third and fourth values.
4. The lower quartile is (7 + 8) $\div$ 2 = 7.5.

• P1 Establishing $\frac{7+8}{2}$ or an equivalent valid method.
• A1 Correct answer: 7.5

(a) $(10,6)$

1. Subtract the lower quartile from the upper quartile for each group.
   $$26-16$$
2. Therefore $(10,6)$.

• M1 Subtract the lower quartile from the upper quartile for each group.
• A1 Correct answer: $(10,6)$

(b) A is typically faster because its median time is lower. B is more consistent in its middle half because its IQR is smaller.

1. A is typically faster because its median time is lower. B is more consistent in its middle half because its IQR is smaller.

• C1 Correct conclusion with supporting reasoning: A is typically faster because its median time is lower. B is more consistent in its middle half because its IQR is smaller.

(a) $67$

1. A positive linear transformation preserves order; transform the median.
   $$3\times 23-2$$
2. Therefore $67$.

• M1 A positive linear transformation preserves order; transform the median.
• A1 Correct answer: $67$

(b) $30$

1. The common subtraction cancels from the difference; scale the original IQR.
   $$3\times (27-17)$$
2. Therefore $30$.

• M1 The common subtraction cancels from the difference; scale the original IQR.
• A1 Correct answer: $30$

(a) $\frac{19}{2}$

1. Average the middle pair of the lower half.
   $$\frac{4+7}{2}$$
2. Average the middle pair of the upper half, then subtract Q1.
   $$(14+16)/2-(11/2)$$
3. Therefore $\frac{19}{2}$.

• M1 Average the middle pair of the lower half.
• M1 Average the middle pair of the upper half, then subtract Q1.
• A1 Correct answer: $\frac{19}{2}$

(b) $\frac{21}{2}$

1. An even number of observations has two central values.
   $$\frac{9+12}{2}$$
2. Therefore $\frac{21}{2}$.

• M1 An even number of observations has two central values.
• A1 Correct answer: $\frac{21}{2}$

(a) $15$ minutes

1. Subtract the lower quartile from the upper quartile.
   $$35-20$$
2. Therefore $15$ minutes.

• M1 Subtract the lower quartile from the upper quartile.
• A1 Correct answer: $15$ minutes

(b) $10$

1. A quarter of the values lie above the upper quartile.
   $$40/4$$
2. Therefore $10$.

• M1 A quarter of the values lie above the upper quartile.
• A1 Correct answer: $10$

(c) Yes: 35 + 1.5 × 15 = 57.5, and 58 is greater than 57.5.

1. Find the outlier limit.
   $$35+1.5\times 15$$
2. Yes: 35 + 1.5 × 15 = 57.5, and 58 is greater than 57.5.

• M1 Find the outlier limit.
• C1 Correct conclusion with supporting reasoning: Yes: 35 + 1.5 × 15 = 57.5, and 58 is greater than 57.5.

(a) 16

1. Let the lower quartile be $L$ and the upper quartile $U$.
2. $$U - L = 12, \qquad U - 20 = 2(20 - L)$$
3. From the second, $U = 60 - 2L$. Substitute.
   $$60 - 2L - L = 12 \;\Rightarrow\; L = 16$$

• P1 Both equations in $L$ and $U$.
• P1 Solving to one value.
• A1 Correct answer: 16.

(b) 35

1. Range $=$ max $-$ min: $5 + 30 = 35$.

• B1 Correct answer: 35.

(a) 29

1. $$9\times 18$$
2. $$162-(4\times 6+2\times 22+36)$$
3. $$58/2$$
4. The total is 9 $\times$ 18 = 162. The smallest number is 36 $-$ 30 = 6.
5. Let the seventh number be x. The first four numbers are each at least 6; the fifth and sixth are each at least 22; the seventh and eighth are each at least x.
6. Including the largest number, 162 $\ge$ 4 $\times$ 6 + 2 $\times$ 22 + 2x + 36 = 104 + 2x. Hence x $\le$ 29.
7. The list 6, 6, 6, 6, 22, 22, 29, 29, 36 has total 162, median 22 and range 30. Therefore 29 is attainable and is the greatest possible value.

• P1 Establishing $9\times 18$ or an equivalent valid method.
• P1 Establishing $162-(4\times 6+2\times 22+36)$ or an equivalent valid method.
• P1 Establishing $58/2$ or an equivalent valid method.
• C1 Correct conclusion with the complete supporting argument: 29