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Date November 2014 Marks available 2 Reference code 14N.2.hl.TZ0.2
Level HL only Paper 2 Time zone TZ0
Command term Find Question number 2 Adapted from N/A

Question

The wingspans of a certain species of bird can be modelled by a normal distribution with mean 60.260.2 cm and standard deviation 2.4 cm.

According to this model, 99% of wingspans are greater than x cm.

Find the value of x.

[2]
a.

In a field experiment, a research team studies a large sample of these birds. The wingspans of each bird are measured correct to the nearest 0.1 cm.

Find the probability that a randomly selected bird has a wingspan measured as 60.2 cm.

[3]
b.

Markscheme

P(X>x)=0.99(=P(X<x)=0.01)     (M1)

x=54.6 (cm)     A1

[2 marks]

a.

P(60.15X60.25)     (M1)(A1)

=0.0166     A1

[3 marks]

Total [5 marks]

b.

Examiners report

Many candidates did not use the symmetry of the normal curve correctly. Many, for example, calculated the value of x for which P(X<x)=0.99 rather than P(X<x)=0.01.

a.

Most candidates did not recognize that the required probability interval was P(60.15X60.25). A large number of candidates simply stated that P(X=60.2)=0.166. Some candidates used P(60.1X60.3) while a number of candidates bizarrely used probability intervals not centred on 60.2, for example, P(60.15X60.24).

b.

Syllabus sections

Topic 5 - Core: Statistics and probability » 5.7 » Normal distribution.
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