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Date May 2013 Marks available 3 Reference code 13M.2.sl.TZ2.2
Level SL only Paper 2 Time zone TZ2
Command term Find Question number 2 Adapted from N/A

Question

The random variable \(X\) is normally distributed with mean \(20\) and standard deviation \(5\).

Find \({\rm{P}}(X \le 22.9)\) .

[3]
a.

Given that \({\rm{P}}(X < k) = 0.55\) , find the value of \(k\) .

[3]
b.

Markscheme

evidence of appropriate approach     (M1)

eg   \(z = \frac{{22.9 - 20}}{5}\)

\(z = 0.58\)     (A1)

\({\rm{P}}(X \le 22.9) = 0.719\)     A1     N3

[3 marks]

a.

\(z\)-score for \(0.55\) is \(0.12566…\)     (A1)

valid approach (must be with \(z\)-values)     (M1)

eg using inverse normal, \(0.1257 = \frac{{k - 20}}{5}\)

\(k = 20.6\)     A1     N3

[3 marks]

b.

Examiners report

The normal distribution was handled better than in previous years with many candidates successful in both parts and very few blank responses. Some candidates used tables and \(z\)-scores while others used the GDC directly; the GDC approach earned full marks more often than the \(z\)-score approach.

a.

The normal distribution was handled better than in previous years with many candidates successful in both parts and very few blank responses. Some candidates used tables and \(z\)-scores while others used the GDC directly; the GDC approach earned full marks more often than the \(z\)-score approach. A common error in part (b) was to set the expression for \(z\)-score equal to the probability. Many candidates had difficulty giving answers correct to three significant figures; this was particularly an issue if no working was shown.

b.

Syllabus sections

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