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Date May 2011 Marks available 2 Reference code 11M.2.sl.TZ2.6
Level SL only Paper 2 Time zone TZ2
Command term Find Question number 6 Adapted from N/A

Question

Let the random variable X be normally distributed with mean 25, as shown in the following diagram.


The shaded region between 25 and 27 represents \(30\% \) of the distribution.

Find \({\rm{P}}(X > 27)\) .

[2]
a.

Find the standard deviation of X .

[5]
b.

Markscheme

symmetry of normal curve     (M1)

e.g. \({\rm{P}}(X < 25) = 0.5\)

\({\rm{P}}(X > 27) = 0.2\)     A1     N2

[2 marks]

a.

METHOD 1

finding standardized value     (A1)

e.g. \(\frac{{27 - 25}}{\sigma }\)

evidence of complement     (M1)

e.g. \(1 - p\) , \({\rm{P}}(X < 27)\) , 0.8

finding z-score     (A1)

e.g. \(z = 0.84 \ldots \)

attempt to set up equation involving the standardized value     M1

e.g. \(0.84 = \frac{{27 - 25}}{\sigma }\) , \(0.84 = \frac{{X - \mu }}{\sigma }\)

\(\sigma = 2.38\)     A1     N3

METHOD 2

set up using normal CDF function and probability     (M1)

e.g. \({\rm{P}}(25 < X < 27) = 0.3\) , \({\rm{P}}(X < 27) = 0.8\)

correct equation     A2

e.g. \({\rm{P}}(25 < X < 27) = 0.3\) , \({\rm{P}}(X > 27) = 0.2\)

attempt to solve the equation using GDC     (M1)

e.g. solver, graph, trial and error (more than two trials must be shown)

\(\sigma = 2.38\)     A1     N3

[5 marks]

b.

Examiners report

This question proved challenging for many candidates. A surprising number did not use the symmetry of the normal curve to find the probability required in (a). While many students were able to set up a standardized equation in (b), far fewer were able to use the complement to find the correct z-score. Others used 0.8 as the z-score. A common confusion when approaching parts (a) and (b) was whether to use a probability or a z-score. Additionally, many candidates seemed unsure of appropriate notation on this problem which would have allowed them to better demonstrate their method.

a.

This question proved challenging for many candidates. A surprising number did not use the symmetry of the normal curve to find the probability required in (a). While many students were able to set up a standardized equation in (b), far fewer were able to use the complement to find the correct z-score. Others used 0.8 as the z-score. A common confusion when approaching parts (a) and (b) was whether to use a probability or a z-score. Additionally, many candidates seemed unsure of appropriate notation on this problem which would have allowed them to better demonstrate their method.

b.

Syllabus sections

Topic 5 - Statistics and probability » 5.9 » Normal distributions and curves.
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