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Date November 2011 Marks available 2 Reference code 11N.2.hl.TZ0.11
Level HL only Paper 2 Time zone TZ0
Command term Find Question number 11 Adapted from N/A

Question

Jan and Sia have been selected to represent their country at an international discus throwing competition. Assume that the distance thrown by each athlete is normally distributed. The mean distance thrown by Jan in the past year was 60.33 metres with a standard deviation of 1.95 metres.

In the past year, 80 % of Jan’s throws have been longer than x metres. Find x correct to two decimal places.

[2]
a.

In the past year, 80 % of Sia’s throws have been longer than 56.52 metres. If the mean distance of her throws was 59.39 metres, find the standard deviation of her throws.

[3]
b.

This year, Sia’s throws have a mean of 59.50 metres and a standard deviation of 3.00 metres. The mean and standard deviation of Jan’s throws have remained the same. In the competition, an athlete must have at least one throw of 65 metres or more in the first round to qualify for the final round. Each athlete is allowed three throws in the first round.

(i)     Determine whether Jan or Sia is more likely to qualify for the final on their first throw.

(ii)     Find the probability that both athletes qualify for the final.

[10]
c.

Markscheme

\(X \sim {\text{N(60.33, 1.9}}{{\text{5}}^2})\)

\({\text{P}}(X < x) = 0.2 \Rightarrow x = 58.69{\text{ m}}\)     (M1)A1

[2 marks]

a.

\(z = - 0.8416 \ldots \)     (A1)

\( - 0.8416 = \frac{{56.52 - 59.39}}{\sigma }\)     (M1)

\(\sigma \approx 3.41\)     A1

[3 marks]

b.

Jan \(X \sim {\text{N(60.33, 1.9}}{{\text{5}}^2})\); Sia \(X \sim {\text{N(59.50, 3.0}}{{\text{0}}^2})\)

(i)     Jan: \({\text{P}}(X > 65) \approx 0.00831\)     (M1)A1

Sia: \({\text{P}}(Y > 65) \approx 0.0334\)     A1

Sia is more likely to qualify     R1

Note: Only award R1 if (M1) has been awarded.

 

(ii)     Jan: \({\text{P}}(X \geqslant 1) = 1 - {\text{P}}(X = 0)\)     (M1)

\( = 1 - {(1 - 0.00831 \ldots )^3} \approx 0.0247\)     (M1)A1

Sia: \({\text{P}}(Y \geqslant 1) = 1 - {\text{P}}(Y = 0) = 1 - {(1 - 0.0334 \ldots )^3} \approx 0.0968\)     A1

Note: Accept 0.0240 and 0.0969.

 

hence, \({\text{P}}(X \geqslant 1{\text{ and }}y \geqslant 1) = 0.0247 \times 0.0968 = 0.00239\)     (M1)A1

[10 marks]

c.

Examiners report

Parts (a) and (b) were generally accessible to many candidates. In (c)(i) quite a few candidates missed the wording ‘first throw’, and consequently in (ii) used the incorrect probabilities.

a.

Parts (a) and (b) were generally accessible to many candidates. In (c)(i) quite a few candidates missed the wording ‘first throw’, and consequently in (ii) used the incorrect probabilities.

b.

Parts (a) and (b) were generally accessible to many candidates. In (c)(i) quite a few candidates missed the wording ‘first throw’, and consequently in (ii) used the incorrect probabilities.

c.

Syllabus sections

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