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Date May Specimen paper Marks available 2 Reference code SPM.2.SL.TZ0.8
Level Standard Level Paper Paper 2 Time zone Time zone 0
Command term Find Question number 8 Adapted from N/A

Question

The length, X mm, of a certain species of seashell is normally distributed with mean 25 and variance, σ 2 .

The probability that X is less than 24.15 is 0.1446.

A random sample of 10 seashells is collected on a beach. Let Y represent the number of seashells with lengths greater than 26 mm.

Find P(24.15 < X < 25).

[2]
a.

Find σ , the standard deviation of X.

[3]
b.i.

Hence, find the probability that a seashell selected at random has a length greater than 26 mm.

[2]
b.ii.

Find E(Y).

[3]
c.

Find the probability that exactly three of these seashells have a length greater than 26 mm.

[2]
d.

A seashell selected at random has a length less than 26 mm.

Find the probability that its length is between 24.15 mm and 25 mm.

[3]
e.

Markscheme

attempt to use the symmetry of the normal curve        (M1)

eg   diagram, 0.5 − 0.1446

P(24.15 < X < 25) = 0.3554        A1

[2 marks]

a.

use of inverse normal to find z score      (M1)

z = −1.0598

correct substitution  24.15 25 σ = 1.0598        (A1)

σ = 0.802       A1   

[3 marks]

b.i.

P(X > 26) = 0.106       (M1)A1  

[2 marks]

b.ii.

recognizing binomial probability      (M1)

E(Y) = 10 × 0.10621       (A1)

= 1.06         A1  

[3 marks]

c.

P(Y = 3)        (M1)

= 0.0655        A1  

[2 marks]

d.

recognizing conditional probability        (M1)

correct substitution        A1

0.3554 1 0.10621

= 0.398       A1

[3 marks]

 

e.

Examiners report

[N/A]
a.
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b.i.
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b.ii.
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c.
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d.
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e.

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

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