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13 Here are the numbers of eggs laid by some hens on each of 11 days in June.
9 10 12 15 17 18 19 19 20 25 26
Work out the interquartile range of the numbers of eggs.
(2)
| Scheme | Marks |
|---|---|
| 20 (−) 12 | M1 |
| Correct answer scores full marks (unless from obvious incorrect working) Answer: 8 | A1 |
| (2) | |
| (2 marks) |
M1: for both values unambiguously identified
8 The mean of 7 numbers is 60
The mean of 3 of the numbers is 46
Work out the mean of the other 4 numbers.
(3)
| Scheme | Marks |
|---|---|
| 60 × 7 (= 420) or 46 × 3 (= 138) | M1 |
| “420” – “138” (= 282) | M1 |
| Correct answer scores full marks (unless from obvious incorrect working) Answer: 70.5 | A1 |
| (3) | |
| (3 marks) |
M1: may be embedded within an equation
M1: for a method to find the sum of the 4 numbers
Allow this mark if they do further incorrect work using 282
A1: allow \(\dfrac{141}{2}\) oe eg \(\dfrac{282}{4}\) or \(70\dfrac{1}{2}\)
1 Rio has six cards.
He writes a number on each card so that
the range of the numbers is 10
the median of the numbers is 7.5
the mode of the numbers is 6
Rio arranges the cards so that the numbers are in order of size.
Three of the numbers are hidden.
Complete the cards above to show the three numbers that are hidden.
(3)
| Scheme | Marks |
|---|---|
| 1st card = 4 | B1 |
| 2nd card = 6 | B1 |
| 4th card = 9 | B1 |
| (3) | |
| (3 marks) |
B1: or a list of 6 numbers with a mode of 6
SCB2 for 4, 6 and 9 in the incorrect order
8 Kim puts 4 bananas in a bag.
The mean weight of the 4 bananas in the bag is 145 grams.
Andy puts one more banana into the bag.
The mean weight of the 5 bananas in the bag is 142 grams.
Work out the weight of the banana that Andy puts into the bag.
(3)
| Scheme | Marks |
|---|---|
4 × 145 (= 580) or 5 × 142 (= 710) or \(\dfrac{145 + 145 + 145 + 145 + x}{5} = 142\) oe | M1 |
| 5 × 142 – 4 × 145 or “710” – “580” or \(145 + 145 + 145 + 145 + x = 5 \times 142\) | M1 |
| Correct answer scores full marks (unless from obvious incorrect working) Answer: 130 | A1 |
| (3) | |
| (3 marks) |
M1: for one correct product
or
for a correct equation for the weight of the last banana
M1: A fully correct method to find the weight of the 5th banana
or
a fully correct equation to find the missing weight with no denominator
1 The table shows some information about the hourly rates of pay of 60 workers.
| Hourly rate of pay (\(p\) dollars) | Frequency |
|---|---|
| \(10 \lt p \leqslant 15\) | 18 |
| \(15 \lt p \leqslant 20\) | 16 |
| \(20 \lt p \leqslant 25\) | 14 |
| \(25 \lt p \leqslant 30\) | 8 |
| \(30 \lt p \leqslant 35\) | 4 |
| Scheme | Marks |
|---|---|
| \(10 \lt p \leqslant 15\) | B1 |
| (1) |
B1: allow \(10 \leqslant p \leqslant 15\) or \(10 \lt p \lt 15\) or
\(10 \leqslant p \lt 15\) or 10 – 15
| Scheme | Marks |
|---|---|
| 12.5 × 18 + 17.5 × 16 + 22.5 × 14 + 27.5 × 8 + 32.5 × 4 (= 1170) or 225 + 280 + 315 + 220 + 130 (= 1170) [lower bound products are: 180, 240, 280, 200, 120] [upper bound products are: 270, 320, 350, 240, 140] Sum of lower bound products = 1020 Sum of upper bound products = 1320 Sum of products using 11, 16, 21, 26 and 31 = 1080 Sum of products using 12, 17, 22, 27 and 32 = 1140 Sum of products using 13, 18, 23, 28 and 33 = 1200 Sum of products using 14, 19, 24, 29 and 34 = 1260 | M2 |
| “1170” ÷ 60 | M1 |
| Correct answer scores full marks (unless from obvious incorrect working) Answer: 19.5 | A1 |
| (4) | |
| (5 marks) |
M2: for at least 4 correct products added (need not be evaluated) or
If not M2 then award:
M1 for consistent use of value within interval (including end points) for at least 4 products which must be added
or
correct midpoints used for at least 4 products and not added
M1: dep on at least M1
Allow division by their \(\Sigma f\) provided addition or total under column seen
A1: oe
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