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Date May 2019 Marks available 4 Reference code 19M.1.SL.TZ1.S_9
Level Standard Level Paper Paper 1 Time zone Time zone 1
Command term Find Question number S_9 Adapted from N/A

Question

A random variable Z is normally distributed with mean 0 and standard deviation 1. It is known that P( z < −1.6) = a and P( z > 2.4) = b . This is shown in the following diagram.

A second random variable X is normally distributed with mean m and standard deviation s .

It is known that P( x < 1) = a .

Find P(−1.6 < z < 2.4). Write your answer in terms of a and b .

[2]
a.

Given that z > −1.6, find the probability that z < 2.4 . Write your answer in terms of a and b .

[4]
b.

Write down the standardized value for  x = 1 .

[1]
c.

It is also known that P( x > 2) = b .

Find  s .

[6]
d.

Markscheme

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

recognizing area under curve = 1        (M1)

eg    a + x + b = 1 ,   100 a b ,   1 a + b

P ( 1.6 < z < 2.4 ) = 1 a b ( = 1 ( a + b ) )                A1  N2

[2 marks]

a.

P ( z > 1.6 ) = 1 a  (seen anywhere)        (A1)

recognizing conditional probability        (M1)

eg    P ( A | B ) ,   P ( B | A )

correct working       (A1)

eg    P ( z < 2.4 z > 1.6 ) P ( z > 1.6 ) P ( 1.6 < z < 2.4 ) P ( z > 1.6 )  

P ( z < 2.4 | z > 1.6 ) = 1 a b 1 a       A1  N4

Note: Do not award the final A1 if correct answer is seen followed by incorrect simplification.

[4 marks]

b.

z = 1.6 (may be seen in part (d))     A1  N1

Note: Depending on the candidate’s interpretation of the question, they may give  1 m s as the answer to part (c). Such answers should be awarded the first (M1) in part (d), even when part (d) is left blank. If the candidate goes on to show z = 1.6 as part of their working in part (d), the A1 in part (c) may be awarded.

[1 mark]

c.

attempt to standardize x (do not accept x μ σ )       (M1)

eg    1 m s  (may be seen in part (c)),  m 2 s ,   x m σ

correct equation with each z -value      (A1)(A1)

eg    1.6 = 1 m s ,   2.4 = 2 m s ,   m + 2.4 s = 2

valid approach (to set up equation in one variable)       M1

eg    2.4 = 2 ( 1.6 s + 1 ) s ,    1 m 1.6 = 2 m 2.4

correct working        (A1)

eg    1.6 s + 1 = 2 2.4 s ,   4 s = 1 ,   m = 7 5

s = 1 4        A1 N2

[6 marks]

d.

Examiners report

[N/A]
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c.
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d.

Syllabus sections

Topic 4—Statistics and probability » SL 4.6—Combined, mutually exclusive, conditional, independence, prob diagrams
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Topic 4—Statistics and probability » SL 4.9—Normal distribution and calculations
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