Date | November Example question | Marks available | 2 | Reference code | EXN.2.AHL.TZ0.6 |
Level | Additional Higher Level | Paper | Paper 2 | Time zone | Time zone 0 |
Command term | Find | Question number | 6 | Adapted from | N/A |
Question
The masses in kilograms of melons produced by a farm can be modelled by a normal distribution with a mean of 2.6 kg2.6kg and a standard deviation of 0.5 kg0.5kg.
Find the probability that two melons picked at random and independently of each other will
One year due to favourable weather conditions it is thought that the mean mass of the melons has increased.
The owner of the farm decides to take a random sample of 1616 melons to test this hypothesis at the 5%5% significance level, assuming the standard deviation of the masses of the melons has not changed.
Unknown to the farmer the favourable weather conditions have led to all the melons having 10%10% greater mass than the model described above.
Find the probability that a melon selected at random will have a mass greater than 3.0 kg3.0kg.
both have a mass greater than 3.0 kg3.0kg.
have a total mass greater than 6.0 kg6.0kg.
Write down the null and alternative hypotheses for the test.
Find the critical region for this test.
Find the mean and standard deviation of the mass of the melons for this year.
Find the probability of a Type II error in the owner’s test.
Markscheme
* This sample question was produced by experienced DP mathematics senior examiners to aid teachers in preparing for external assessment in the new MAA course. There may be minor differences in formatting compared to formal exam papers.
Let XX represent the mass of a melon
P(X>3.0)=0.212 (0.2118…)P(X>3.0)=0.212 (0.2118…) (M1)A1
[2 marks]
0.2118…×0.2118…0.2118…×0.2118… (M1)
=0.0449 (0.04488…)=0.0449 (0.04488…) A1
[2 marks]
Let TT represent the total mass
E(T)=5.2E(T)=5.2 A1
Var(T)=0.52+0.52=0.5Var(T)=0.52+0.52=0.5 (M1)A1
T~N(5.2, 0.5)T~N(5.2,0.5)
P(T>6.0)=0.129 (0.1289…)P(T>6.0)=0.129 (0.1289…) A1
[4 marks]
Let μμ be the mean mass of the melons produced by the farm.
H0: μ=2.6 kgH0: μ=2.6kg, H1: μ>2.6 kgH1: μ>2.6kg only A1
Note: Accept H0:H0: The mean mass of melons produced by the farm is equal to 2.6 kg2.6kg
H1: H1: The mean mass of melons produced by the farm is greater than 2.6 kg2.6kg
Note: Award A0 if 2.6 kg2.6kg does not appear in the hypothesis.
[1 mark]
Under H0H0 ˉX~N(2.6, 0.5216)¯¯¯X~N(2.6,0.5216) A1
P(ˉX>a)=0.05P(¯¯¯X>a)=0.05 (M1)
a=2.81 (2.805606…)a=2.81 (2.805606…) (A1)
Critical region is ˉX>2.81¯¯¯X>2.81 A1
[4 marks]
Let WW represent the new mass of the melons
E(W)=1.1×2.6=2.86E(W)=1.1×2.6=2.86 A1
Standard deviation of W=1.1×0.5W=1.1×0.5 (M1)
=0.55=0.55 A1
Note: award M1A0 for Var(W)=1.12×0.52=0.3025Var(W)=1.12×0.52=0.3025
[3 marks]
P(Type II error)=P(Type II error)=
P(ˉX<2.81 μ=2.86, σ=0.554) (M1)
=0.346 (0.346204…) A1
Note: Accept 0.358 from use of the three‐figure answer to part (d)
[2 marks]