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Date	May 2022	Marks available	2	Reference code	22M.2.SL.TZ1.4
Level	Standard Level	Paper	Paper 2	Time zone	Time zone 1
Command term	Hence and Find	Question number	4	Adapted from	N/A

Question

A discrete random variable, $X$ , has the following probability distribution:

Show that $2 k^{2} - k + 0.12 = 0$ .

[1]

Find the value of $k$ , giving a reason for your answer.

[3]

Hence, find $E (X)$ .

[2]

Markscheme

$0.41 + k - 0.28 + 0.46 + 0.29 - 2 k^{2} = 1$ OR $k - 2 k^{2} + 0.01 = 0.13$ (or equivalent) A1

$2 k^{2} - k + 0.12 = 0$ AG

[1 mark]

one of $0.2$ OR $0.3$ (M1)

$k = 0.3$ A1

reasoning to reject $k = 0.2$ eg $P (1) = k - 0.28 \geq 0$ therefore $k \neq 0.2$ R1

[3 marks]

attempting to use the expected value formula (M1)

$E (X) = 0 \times 0.41 + 1 \times (0.3 - 0.28) + 2 \times 0.46 + 3 \times (0.29 - 2 \times 0 . 3^{2})$

$= 1.27$ A1

Note: Award M1A0 if additional values are given.

[2 marks]

Examiners report

Part (a) was well done in this question, with most candidates recognising that the probabilities needed to sum to 1. Many candidates also approached part (b) appropriately. While many did so by graphing the quadratic on the GDC and identifying the zeros, most solved the equation analytically. Those that used the GDC, often assumed there was only one x-intercept and did not investigate the relevant area of the graph in more detail. While some who found the two required values of k recognised that k = 0.2 should be rejected by referring to the original probabilities, most had lost sight of the context of the question, and were unable to give a valid reason using P(X = 1) to reject this solution. Those that obtained one solution in part (b), were generally able to find the expected value successfully in part (c).

[N/A]

Syllabus sections

Topic 2—Functions » SL 2.7—Solutions of quadratic equations and inequalities, discriminant and nature of roots

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22M.1.AHL.TZ1.10b.iii:
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22M.1.SL.TZ2.6b:
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22M.1.AHL.TZ2.11c:
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22M.2.AHL.TZ1.8a:
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22M.2.AHL.TZ1.8b:
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16N.1.AHL.TZ0.H_5:
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21N.3.AHL.TZ0.2a.iii:
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