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Date	May 2017	Marks available	3	Reference code	17M.2.AHL.TZ2.H_3
Level	Additional Higher Level	Paper	Paper 2	Time zone	Time zone 2
Command term	Calculate	Question number	H_3	Adapted from	N/A

Question

Packets of biscuits are produced by a machine. The weights $X$ $X$ , in grams, of packets of biscuits can be modelled by a normal distribution where $X \sim N (μ, σ^{2})$ $X \sim {\text{N}}(\mu ,{\text{ }}{\sigma ^2})$ . A packet of biscuits is considered to be underweight if it weighs less than 250 grams.

The manufacturer makes the decision that the probability that a packet is underweight should be 0.002. To do this $μ$ $\mu$ is increased and $σ$ $\sigma$ remains unchanged.

The manufacturer is happy with the decision that the probability that a packet is underweight should be 0.002, but is unhappy with the way in which this was achieved. The machine is now adjusted to reduce $σ$ $\sigma$ and return $μ$ $\mu$ to 253.

Given that $μ = 253$ $\mu = 253$ and $σ = 1.5$ $\sigma = 1.5$ find the probability that a randomly chosen packet of biscuits is underweight.

[2]

Calculate the new value of $μ$ $\mu$ giving your answer correct to two decimal places.

[3]

Calculate the new value of $σ$ $\sigma$ .

[2]

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$P (X < 250) = 0.0228$ ${\text{P}}(X < 250) = 0.0228$ (M1)A1

[2 marks]

$\frac{250 - μ}{1.5} = - 2.878 \dots$ $\frac{{250 - \mu }}{{1.5}} = - 2.878 \ldots$ (M1)(A1)

$\Rightarrow μ = 254.32$ $\Rightarrow \mu = 254.32$ A1

Notes: Only award A1 here if the correct 2dp answer is seen. Award M0 for use of ${1.5}^{2}$ ${1.5^2}$ .

[3 marks]

$\frac{250 - 253}{σ} = - 2.878 \dots$ $\frac{{250 - 253}}{\sigma } = - 2.878 \ldots$ (A1)

$\Rightarrow σ = 1.04$ $\Rightarrow \sigma = 1.04$ A1

[2 marks]

Examiners report

[N/A]

Syllabus sections

Topic 4—Statistics and probability » SL 4.9—Normal distribution and calculations

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