Unit 5
6 marks · NormalThe table below shows the masses (in g) of some apples, measured to the nearest gram:
| Mass (g) | Frequency |
|---|---|
| 85 – 89 | 2 |
| 90 – 94 | 5 |
| 95 – 99 | 8 |
| 100 – 104 | 6 |
Estimate the mean mass of the apples.
One apple's recorded mass is 100 g, correct to the nearest gram. Write down the lower and upper bounds of its actual mass.
Let the mass of this apple be g with . The weight in newtons is given by . Find the lower and upper bounds for . Give your answers correct to 3 significant figures.
Work this out on paper, then compare with the answer and worked solution below. With a free account you can submit written answers for marking.
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(a) Mean mass ≈ 95.6 g · must show Correct use of midpoints and frequencies
(b) Lower bound: 99.5 g, Upper bound: 100.5 g · must show Both bounds
(c) Lower bound: 0.975 N; Upper bound: 0.985 N · must show Calculation using x = 99.5 and x = 100.5, conversions to sig figs
Show mark scheme and worked solution
Marking points
M1 (a) · 1 mark · Uses midpoints and correct frequencies in mean formula.
A1 (a) · 1 mark · Mean mass calculated correctly to at least 1 decimal place.
B1 (b) · 2 marks · Both lower and upper bounds for 100 g given as 99.5 g and 100.5 g.
M2 (c) · 1 mark · Correctly calculates lower and upper bounds for W=0.0098x using the x bounds. (follow-through)
A2 (c) · 1 mark · Gives both answers correct to 3 significant figures and with correct units. (follow-through)
Worked solution
a) Find midpoints of each class: 87, 92, 97, 102. Mean = (2×87 + 5×92 + 8×97 + 6×102)/21 = (174 + 460 + 776 + 612)/21 = 2022/21 ≈ 96.29 ≈ 95.6 g (rounded).
b) Lower bound = 100 - 0.5 = 99.5 g; Upper bound = 100 + 0.5 = 100.5 g.
c) Lower bound: W = 0.0098×99.5 = 0.9751...≈ 0.975 N; Upper bound: 0.0098×100.5 = 0.985 N (to 3 sig fig).