HCI Prelim H2 Mathematics P2 Solutions
2010 HCI H2 Mathematics Preliminary Examination Paper 2 Solution
Qn / Solutions1 / Surface area of the tin and lid
= =
Volume of the container
or (rejected)
when
V is maximum when .
When ,
(or x = 6.67, y = 18.3).
2(i) / ei0 = 32ei2k
where
(ii) / The highest power in the equation is four since the terms with w5 are canceled out. Hence the equation has only four roots.
=
=
Or use GC, = 10.
3 /
Using GC,
, ,
Thus
Hence sequence is an AP.
4i /
OR
ii / Suppose O, X, Y are collinear.
Then
Thus O, X, Y are not collinear.
iii /
= (4 – 4 + t2 – 2t2)
= 0
For all tℝ\{0}, 0 < < 1.
Thus XOY can be 90 when .
iv /
5(i) /
(ii) /
(iii) /
For the equation ,
When , .
When , .
Method 1: Using
Method 2: Using
= { 28 – [(–12 + 27) – (–4 + 1)]}
= 10 units2
(iv) / Volume required
= vol. of cylinder – (vol. generated by curve from y = –2 to y = 1)
= 196 – 92
= 327 unit3(3 s.f.)
6 / Use random sampling method to select a sample from each class. The number of seats from each class would be proportional to the size of each stratum.
First Class / Business Class / Economy Class
4 / 16 / 60
Any 1 of the answers below:
Some passengers have booked a flight ticket but did not turn up or changed flight so some of the seats in the sample may not have a passenger.
OR
The flight is not fully booked so the chosen seat could be empty.
OR The passenger may ignore the questionnaire.
It is not appropriate to use simple random sampling as passengers from different classes may have different opinions on the service. The number of passengers in the first class is very small, so the passengers from the first class may not be chosen at all using the simple random sampling method.
7(i) / No. of ways =
7(ii) / Case 1: The 2 blue tiles and 1 yellow tile are in the 4th row with the 4th tile being red or green.
No. of ways
= no. of ways with B, B, Y, G in 4th row + no. of ways with B, B, Y, R in 4th row
=
Case 2: The 2 blue tiles and 1 yellow tile are in the third row.
No. of ways =
Total no. of ways = 3240 + 1890 – – 3!
= 5130 – 108 – 216 = 4806
7(iii) / No. of ways such that less than 3 yellow tiles are in the fourth row
= 12600 =12600 420 =12180
7 last part / No. of ways
= × 7C4 = 2100
8(i) /
H0 :
H1 :
Test Stat:
p–value = 0.01265 < 0.05
Since the –value = 0.01265 < 0.05, we reject and conclude that there is sufficient evidence, at 5% level of significance, that the mean life span of the electronic component has increased.
8(ii) / H0: vs H1:
Under H0, ~ N(9000, ) = N(9000, 62.5).
Test Statistic = ~ N(0, 1).
Level of significance = 1%
P(Z2.326347877) = 0.01
At the 1% significance level, rejectif z ≥ 2.326347877.
z = ≥ 2.326347877
≥ 9018.391395 = 9020.
Assumptions: The standard deviation of the life span remains unchanged after the change in process.
9
First part /
= 0.580
OR
A = X1 +... + X20 – 2(X21 +... + X30) ~ N(0, 34560)
= 0.580
9(i) / Let Y be the r.v. denoting the mass of a randomly chosen apple from Mark's orchard.
Since the shaded area is the same, using the symmetric property of the normal curve,
9(ii) / Probability that Mark will get an apple graded as 'large' chosen at random = = 0.09121128
Let A be the r.v. denoting the number of apples graded as large out of 65 randomly chosen apples.
10(a) / (i) P(A M ) =
(ii) P(M ' C ')
A and M are not independent.
10(b) / (i) No. of immigrants in the sample
=
P(voter supports Party A given voter is an immigrant)
(ii) Number of immigrants supporting Party C ==6
P(exactly one immigrant voter supporting Party C or
exactly one female voter supporting Party A (or both))
Alternative method:
Required Probability =
= 0.434
11(i) / Let X be the r.v. denoting the number of call–ins in a week. Hence .
We are looking for the such that P(X ≤ 9) = 0.701
From graph, the value of (to 3 sig.fig).
The condition is that the rate of call–ins received by the centre is constant throughout a month / the call–in occurs randomly / The call–ins occur in a month are independent of one another
11(ii) / Let Y be the r.v. denoting the number of call–ins in a week.
Since the mean is bigger than 10, hence
approximately.
= 0.810
11(iii) / Let S be the r.v. denoting the number of successful cases out of the n people in a support group.
Since the number of groups concerned, which is 70, is large, therefore by applying CLT,
approximately.
EITHER
n /
27 / 0.589
28 / 0.812
Hence minimum value of n is 28.
OR
Least n = 28.
12(i) / Location F should be omitted as the road distance cannot be smaller than the straight line distance, indicating that it is an incorrect data entry.
From the scatter diagram, another location that should be omitted is location H, as it is an outlier based on the scatter diagram.
12(ii) / The suitable regression line is the regression x on y:
When y = 20.0,
km
12(iii) /
Since the graph of is concave downwards whereas the graph of is concave upwards, the graph of will be more suitable to describe the scatter diagram of s and y. Hence model II is more suitable.
12(iv) / The appropriate regression line of s on ln y is ,
i.e. (to 3 s.f.)
12(v) / Since r for s and ln y is 0.992 close to 1, the linear correlation is strong between s and ln y. Furthermore, 170 cents is within the data range of the sample. Therefore the estimation using the line in (iv) is reliable.
Since y is the independent variable, the line found in (iv) is also suitable for the estimation.
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