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Date May 2017 Marks available 2 Reference code 17M.2.sl.TZ1.9
Level SL only Paper 2 Time zone TZ1
Command term Find Question number 9 Adapted from N/A

Question

A random variable \(X\) is normally distributed with mean, \(\mu \). In the following diagram, the shaded region between 9 and \(\mu \) represents 30% of the distribution.

M17/5/MATME/SP2/ENG/TZ1/09

The standard deviation of \(X\) is 2.1.

The random variable \(Y\) is normally distributed with mean \(\lambda \) and standard deviation 3.5. The events \(X > 9\) and \(Y > 9\) are independent, and \(P\left( {(X > 9) \cap (Y > 9)} \right) = 0.4\).

Find \({\text{P}}(X < 9)\).

[2]
a.

Find the value of \(\mu \).

[3]
b.

Find \(\lambda \).

[5]
c.

Given that \(Y > 9\), find \({\text{P}}(Y < 13)\).

[5]
d.

Markscheme

valid approach     (M1)

eg\(\,\,\,\,\,\)\({\text{P}}(X < \mu ) = 0.5,{\text{ }}0.5 - 0.3\)

\({\text{P}}(X < 9) = 0.2\) (exact)     A1     N2

[2 marks]

a.

\(z =  - 0.841621\) (may be seen in equation)     (A1)

valid attempt to set up an equation with their \(z\)     (M1)

eg\(\,\,\,\,\,\)\( - 0.842 = \frac{{\mu  - X}}{\sigma },{\text{ }} - 0.842 = \frac{{X - \mu }}{\sigma },{\text{ }}z = \frac{{9 - \mu }}{{2.1}}\)

10.7674

\(\mu  = 10.8\)     A1     N3

[3 marks]

b.

\({\text{P}}(X > 9) = 0.8\) (seen anywhere)     (A1)

valid approach     (M1)

eg\(\,\,\,\,\,\)\({\text{P}}(A) \times {\text{P}}(B)\)

correct equation     (A1)

eg\(\,\,\,\,\,\)\(0.8 \times {\text{P}}(Y > 9) = 0.4\)

\({\text{P}}(Y > 9) = 0.5\)     A1

\(\lambda  = 9\)     A1     N3

[5 marks]

c.

finding \({\text{P}}(9 < Y < 13) = 0.373450\) (seen anywhere)     (A2)

recognizing conditional probability     (M1)

eg\(\,\,\,\,\,\)\({\text{P}}(A|B),{\text{ P}}(Y < 13|Y > 9)\)

correct working     (A1)

eg\(\,\,\,\,\,\)\(\frac{{{\text{0.373}}}}{{0.5}}\)

0.746901

0.747     A1     N3

[5 marks]

d.

Examiners report

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

Syllabus sections

Topic 5 - Statistics and probability » 5.6 » Conditional probability; the definition \(P\left( {\left. A \right|B} \right) = \frac{{P\left( {A\mathop \cap \nolimits B} \right)}}{{P\left( B \right)}}\) .
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