The weights of apples on a tree can be modelled by a normal distribution with a mean of85 grams and a standard deviation of 7.5 grams.A sample of apples are taken from 2 trees, A and B, in different parts of the orchard.The data is shown in the table below.The owner of the orchard wants to know whether the mean weight of the apples from treeA(μA) is greater than the mean weight of the apples from tree B(μB) so sets up thefollowing test:H0: μA=μBandH1: μA>μB

Question

SLPaper 1

The weights of apples on a tree can be modelled by a normal distribution with a mean of $85$ grams and a standard deviation of $7.5$ grams.

A sample of apples are taken from $2$ trees, $A$ and $B$ , in different parts of the orchard.

The data is shown in the table below.

The owner of the orchard wants to know whether the mean weight of the apples from tree $A (μ_{A})$ is greater than the mean weight of the apples from tree $B (μ_{B})$ so sets up thefollowing test:

$H_{0} : μ_{A} = μ_{B}$ and $H_{1} : μ_{A} > μ_{B}$

Find the probability that an apple from the tree has a weight greater than $90$ grams.

[2]

Verified

Solution

* 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 the weight of an apple be $X$

$P (X > 90) = 0.252 (0.252492 \dots)$ (M1)A1

[2 marks]

Find the $p$ -value for the owner’s test.

[2]

Verified

Solution

$p$ -value $= 0.0189 (0.018947 \dots)$ (M1)A1

[2 marks]

The test is performed at the $5 %$ significance level.

State the conclusion of the test, giving a reason for your answer.

[2]

Verified

Solution

$0.0189 < 0.05$ R1

Sufficient evidence to reject the null hypothesis (that the weights of apples from thetwo trees are equal) A1

[2 marks]

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