CIE iGCSE Maths C8.2 Relative and expected frequencies Exam Style Practice Questions- Paper 3
The Patel family flies from their home town, H, to Kiruna, K, in Lapland.
(a) The scale drawing shows their journey. The scale is 1 centimetre represents 40 kilometres.
(i) Measure the bearing of K from H.
(ii) Work out the distance in kilometres from H to K.
(iii) The average speed of the plane is 450km/h. Find the average speed in m/s.
(b) The probability that the plane arrives on time is 0.15.
(i) Write down the probability that the plane does not arrive on time.
(ii) Every year there are 240 flights from H to K. Calculate the expected number of flights that arrive on time.
(c) The Patel family has six suitcases. The number of items in each suitcase is shown below.
15 16 16 18 19 21
(i) Find the range.
(ii) Write down the mode.
(iii) Work out the median.
(iv) Calculate the mean.
(v) Find the probability that a suitcase chosen at random has more than 18 items.
(d) Mr Patel buys a bag of sweets. The bag of sweets costs $3.25.
(i) Calculate the cost of the sweets in euros (€) when the exchange rate is €1 = $1.24.
(ii) The weight, w grams, of the bag of sweets is 250g correct to the nearest 10 g. Complete this statement about the value of w.
▶️ Answer/Explanation
(a)(i) Ans: 292
Using a protractor, measure the angle clockwise from North to the line HK. The bearing is 292°.
(a)(ii) Ans: 380 km
Measure the line HK (9.5 cm) and multiply by the scale factor (40 km/cm): \(9.5 \times 40 = 380\) km.
(a)(iii) Ans: 125 m/s
Convert km/h to m/s: \(450 \times \frac{1000}{3600} = 125\) m/s.
(b)(i) Ans: 0.85
Subtract the on-time probability from 1: \(1 – 0.15 = 0.85\).
(b)(ii) Ans: 36
Multiply the total flights by the on-time probability: \(240 \times 0.15 = 36\).
(c)(i) Ans: 6
Range = Maximum – Minimum = \(21 – 15 = 6\).
(c)(ii) Ans: 16
The mode is the most frequent number, which is 16 (appears twice).
(c)(iii) Ans: 17
Arrange in order: 15, 16, 16, 18, 19, 21. Median = average of 3rd and 4th terms = \(\frac{16 + 18}{2} = 17\).
(c)(iv) Ans: 17.5
Mean = \(\frac{15 + 16 + 16 + 18 + 19 + 21}{6} = \frac{105}{6} = 17.5\).
(c)(v) Ans: \(\frac{2}{6}\) oe
There are 2 suitcases (19, 21) with more than 18 items out of 6 total, so probability = \(\frac{2}{6}\).
(d)(i) Ans: 2.62
Convert dollars to euros: \(\frac{3.25}{1.24} \approx 2.62\) €.
(d)(ii) Ans: 245 ≤ w < 255
When rounded to the nearest 10g, 250g implies the actual weight is in the range [245g, 255g).
Olga owns a fruit and vegetable shop.
(a) An apple weighs 70 g correct to the nearest 5g.
Complete the statement about the mass, m grams, of this apple.
(b) The number of strawberries in each of 12 boxes is shown below.
23 21 21 20 21 20
22 22 21 20 20 20
(i) Find the range.
(ii) Write down the mode.
(iii) Find the median.
(iv) Calculate the mean.
(v) Find the probability that a box chosen at random has 22 or 23 strawberries in the box.
(c) The shop sells potatoes in bags A, B and C.
(d) The price of plums is $2.40 per kilogram.
Olga reduces this price by 35%.
Calculate the new price per kilogram.
▶️ Answer/Explanation
(a) Ans: 67.5 ≤ m < 72.5
For rounding to nearest 5g, the lower bound is 70-2.5=67.5g and upper bound is 70+2.5=72.5g.
(b)
(i) Ans: 3 → Range = Max(23) – Min(20) = 3
(ii) Ans: 20 → Mode is most frequent value (appears 5 times)
(iii) Ans: 21 → Median of 12 values is average of 6th and 7th when ordered
(iv) Ans: 20.92 → Mean = (23+21+…+20)/12 = 251/12 ≈ 20.92
(v) Ans: ¼ → Probability = (2 boxes of 22 + 1 of 23)/12 = 3/12 = ¼
(c) Ans: Bag B
Comparing value for money by calculating price per kg shows Bag B is cheapest.
(d) Ans: $1.56
New price = $2.40 – (35% of $2.40) = $2.40 × 0.65 = $1.56