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Date May 2017 Marks available 2 Reference code 17M.2.hl.TZ2.3
Level HL only Paper 2 Time zone TZ2
Command term Find Question number 3 Adapted from N/A

Question

Packets of biscuits are produced by a machine. The weights \(X\), in grams, of packets of biscuits can be modelled by a normal distribution where \(X \sim {\text{N}}(\mu ,{\text{ }}{\sigma ^2})\). 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 \(\mu \) is increased and \(\sigma \) 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 \(\sigma \) and return \(\mu \) to 253.

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

[2]
a.

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

[3]
b.

Calculate the new value of \(\sigma \).

[2]
c.

Markscheme

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

[2 marks]

a.

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

\( \Rightarrow \mu = 254.32\)     A1

 

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

 

[3 marks]

b.

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

\( \Rightarrow \sigma = 1.04\)     A1

[2 marks]

c.

Examiners report

[N/A]
a.
[N/A]
b.
[N/A]
c.

Syllabus sections

Topic 5 - Core: Statistics and probability » 5.7
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