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Date May 2015 Marks available 3 Reference code 15M.1.sl.TZ1.1
Level SL only Paper 1 Time zone TZ1
Command term Find Question number 1 Adapted from N/A

Question

A discrete random variable \(X\) has the following probability distribution.

Find \(p\).

[3]
a.

Find \({\text{E}}(X)\).

[3]
b.

Markscheme

summing probabilities to 1     (M1)

eg,\(\;\;\;\sum { = 1,{\text{ }}3 + 4 + 2 + x = 10} \)

correct working     (A1)

\(\frac{3}{{10}} + \frac{4}{{10}} + \frac{2}{{10}} + p = 1,{\text{ }}p = 1 - \frac{9}{{10}}\)

\(p = \frac{1}{{10}}\)     A1     N3

[3 marks]

a.

correct substitution into formula for \({\text{E}}(X)\)     (A1)

eg\(\;\;\;0\left( {\frac{3}{{10}}} \right) +  \ldots  + 3(p)\)

correct working     (A1)

eg\(\;\;\;\frac{4}{{10}} + \frac{4}{{10}} + \frac{3}{{10}}\)

\({\text{E}}(X) = \frac{{11}}{{10}}\;\;\;(1.1)\)     A1     N2

[3 marks]

Total [6 marks]

b.

Examiners report

Most candidates were able to find \(p\), however expectation emerged as surprisingly more difficult. Quite often \({\text{E}}(X)/4\) was found or candidates wrote the formula with no further work.

a.

Most candidates were able to find \(p\), however expectation emerged as surprisingly more difficult. Quite often \({\text{E}}(X)/4\) was found or candidates wrote the formula with no further work.

b.

Syllabus sections

Topic 5 - Statistics and probability » 5.7 » Expected value (mean), \({\text{E}}(X)\) for discrete data.
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