2012 A-level H2 Mathematics (9740) Paper 2 Question 10 Suggested Solutions

By KS Teng

All solutions here are SUGGESTED. Mr. Teng will hold no liability for any errors. Comments are entirely personal opinions.

(i)
Firstly, gold coins are scattered randomly in a randomly chosen region of 1m^2.
Secondly, the mean number of gold coins found in any randomly chosen region of 1m^2 is constant.
Lastly, the finding of a gold coin in a randomly chosen region of area 1m^2 is independent of the finding of another gold coins.

(ii)
Let X denote the number of gold coins in a region of 1m^2.
\mathrm{P}(X \ge 3)

= 1 - \mathrm{P}(X \le 2)

= 0.0474

(iii)
Let Y denote the number of gold coins found in a region of x~m^2.
Y ~\sim~ \mathrm{Po} (0.8x)

\mathrm{P} (Y = 1) = 0.2

Using the Graphing Calculator, x = 0.324

(iv)
Let W denote the number of gold coins found in a region of 100m^2
W ~\sim~ \mathrm{Po} (80)

Since \lambda = 80 > 10, W ~\sim~ \mathrm{N} (80, 80) approximately

\mathrm{P} (W \ge 90)
= \mathrm{P} (W \ge 89.5) by continuity correction
\approx 0.144

(v)
Let C and S denote the number of gold coins and pottery shards in in a region of 50m^2 respectively.

C ~\sim~ \mathrm{Po} (40)
Since \lambda = 40 > 10, C ~\sim~ \mathrm{N} (40, 40) approximately.
S ~\sim~ \mathrm{Po} (150)
Since \lambda = 150 > 10, S ~\sim~ \mathrm{N} (150, 150) approximately.

C + S ~\sim~ \mathrm{N} (190, 190) approximately

\mathrm{P} (C + S \ge 200)
\mathrm{P} (C + S \ge 199.5)
\approx 0.245

(vi)
S + 3C ~\sim~ \mathrm{N} (30, 510) approximately

\mathrm{P} (S + 3C \ge 0)
\mathrm{P} (S + 3C \ge -.05) by continuity correction
\approx 0.912

KS Comments:

For (iii), students can also introduce the poisson formula from MF15 and solve by hand. Please rm. to apply continuity correction for (iv). I took awhile to understand (vi) too. :/ Students should read carefully if their english is as bad as mine. For (vi), students should note that it is necessary that they approximate to normal before consider the difference of the distributions. Be reminded that we cannot take the difference of poisson distributions.

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