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Date November 2019 Marks available 4 Reference code 19N.1.SL.TZ0.S_7
Level Standard Level Paper Paper 1 (without calculator) Time zone Time zone 0
Command term Find Question number S_7 Adapted from N/A

Question

Let X and Y be normally distributed with  X N ( 14 a 2 ) and  Y N ( 22 a 2 ) a > 0 .

Find b so that  P ( X > b ) = P ( Y < b ) .

[2]
a.

It is given that  P ( X > 20 ) = 0.112 .

Find P ( 16 < Y < 28 ) .

[4]
b.

Markscheme

METHOD 1

recognizing that b is midway between the means of 14 and 22 .          (M1)

eg    b = 14 + 22 2

b = 18         A1   N2

 

METHOD 2

valid attempt to compare distributions          (M1)

eg     b 14 a = b 22 a b 14 = 22 b

b = 18         A1   N2

 

[2 marks]

a.

valid attempt to compare distributions (seen anywhere)       (M1)

eg    Y is a horizontal translation of X of 8 units to the right,

P ( 16 < Y < 28 ) = P ( 8 < X < 20 ) , P ( Y > 22 + 6 ) = P ( X > 14 + 6 )

valid approach using symmetry       (M1)

eg   12P(X>20)12P(Y<16)2×P(14<x<20), P(X<8)=P(X>20)

correct working          (A1)

eg    1 2 ( 0.112 ) 2 × ( 0.5 0.112 ) 2 × 0.388 0.888 0.112

P ( 16 < Y < 28 ) = 0.776         A1   N3

[4 marks]

b.

Examiners report

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

Syllabus sections

Topic 4—Statistics and probability » SL 4.9—Normal distribution and calculations
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Topic 4—Statistics and probability

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