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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 XX is normally distributed with mean, μμ. In the following diagram, the shaded region between 9 and μμ represents 30% of the distribution.

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

The standard deviation of XX is 2.1.

The random variable YY is normally distributed with mean λλ and standard deviation 3.5. The events X>9X>9 and Y>9Y>9 are independent, and P((X>9)(Y>9))=0.4P((X>9)(Y>9))=0.4.

Find P(X<9)P(X<9).

[2]
a.

Find the value of μμ.

[3]
b.

Find λλ.

[5]
c.

Given that Y>9Y>9, find P(Y<13)P(Y<13).

[5]
d.

Markscheme

valid approach     (M1)

egP(X<μ)=0.5, 0.50.3P(X<μ)=0.5, 0.50.3

P(X<9)=0.2P(X<9)=0.2 (exact)     A1     N2

[2 marks]

a.

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

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

eg0.842=μXσ, 0.842=Xμσ, z=9μ2.10.842=μXσ, 0.842=Xμσ, z=9μ2.1

10.7674

μ=10.8μ=10.8     A1     N3

[3 marks]

b.

P(X>9)=0.8P(X>9)=0.8 (seen anywhere)     (A1)

valid approach     (M1)

egP(A)×P(B)P(A)×P(B)

correct equation     (A1)

eg0.8×P(Y>9)=0.40.8×P(Y>9)=0.4

P(Y>9)=0.5P(Y>9)=0.5     A1

λ=9λ=9     A1     N3

[5 marks]

c.

finding P(9<Y<13)=0.373450P(9<Y<13)=0.373450 (seen anywhere)     (A2)

recognizing conditional probability     (M1)

egP(A|B), P(Y<13|Y>9)P(A|B), P(Y<13|Y>9)

correct working     (A1)

eg0.3730.50.3730.5

0.746901

0.747     A1     N3

[5 marks]

d.

Examiners report

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

Syllabus sections

Topic 5 - Statistics and probability » 5.6 » Conditional probability; the definition P(A|B)=P(AB)P(B)P(A|B)=P(AB)P(B) .
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