Quod Erat Demonstrandum


Marker’s report on SFXC F.6 Applied Mathematics (II) Final Exam 2007-2008

Filed under: Additional / Applied Mathematics,HKALE — johnmayhk @ 6:13 上午

Marker’s report on SFXC F.6 Applied Mathematics (II) Final Exam 2007-2008

Passing rate : 53.33% (with regular test score considered)
Max : 64
Min : 12
Mean : 40.3
SD : 14.7

Click HERE to download the updated version of the suggested solution.

Question 1

The solving of the first order D.E. was quite straight forward. Few students made careless mistake in evaluating the integrating factor by giving: \exp \int \frac{dy}{2y} = \exp (2\ln y).

Question 2(a)

All students could find out the value of k by considering the total area under a p.d.f. = 1. Most of students could determine the expected arrival time by applying E(X) = \int xf(x)dx directly.

Question 2(b)

Only one student could obtain full mark in this part. To determine the expected value of “waiting time", we need explicitly the function of “waiting time". That is

and it is required to determine E(g(t)).

Few students knew the function g(t), however, they thought that E(g(t)) = \int tg(t)dt instead of E(g(t)) = \int g(t)f(t)dt (where f(t) is the p.d.f. of the arrival time, T, of Mr.X). Students should be reminded that g(t) is NOT a p.d.f. and hence \int tg(t)dt does not make sense.

Question 3(a)

Students performed well in general. Few mistook that P(\spadesuit no. > \heartsuit no.) =
_5C_3(\frac{5}{10})^3(\frac{5}{10}) + _5C_4(\frac{5}{10})^4.
However, the ‘\frac{5}{10}‘ should not be a constant since there is no replacement during the drawing of cards.

Question 3(b)(i)

Some could obtain full mark while some obtained no mark in surd kind of counting problem. Some gave \frac{(_2C_2)^5}{_{10}C_2 \times _8C_2 \times _6C_2 \times _4C_2 \times _2C_2}.

Question 3(b)(ii)

Few could obtain full mark while many did not know how to do the counting.

Question 4(a)(i)

No problem in finding the confidence interval for mean.

Question 4(a)(ii)

Even if the population distribution is normal, P(a_0 < k < b_0) is greater than 95% since the s.d. of sample mean \frac{\sigma}{\sqrt{n}} is less than the population s.d. \sigma. Now the distribution of the population is unknown, it is clearly that we cannot say anything about whether P(a_0 < k < b_0) = 95% or not.

Question 4(b)

It was quite straight forward to perform the hypothesis testing for mean.

Question 4(c)(i)

Some established H_1 as p > 0.2 (instead of \ne 0.2). Some gave wrong counting of ‘over-weighted’ staff and many did not considered the continuity correction when calculating the sample proportion.

Few gave \sqrt{\frac{\frac{5}{16} \times \frac{11}{16}}{16}} (instead of \sqrt{\frac{\frac{1}{5} \times \frac{4}{5}}{16}}) as the s.d. of the sample proportion. Students should be reminded of using the proportion appearing in H_0 in the calculation of s.d. corncerned.

Question 4(c)(ii)

All students did not use the upper limit of \frac{p(1 - p)}{n} (i.e. \frac{0.5(1 - 0.5)}{n}) to estimate the possible value of n.

Many wrote 1.96 \times \frac{3.5}{\sqrt{n}} < 0.05. Students should pay attention to the fact that the calculation is about C.I. for proportion, not C.I. for mean.

Question 4(d)

Only few students understood the question and tried to find P(\overline{x} > C | true mean = 24.2) where C is the critical value of rejecting H_0 originally.

Further, students ignored the change in sample size from 16 to 32.

Question 5(a)

Many students knew that X - Y ~ N(22 , 25), however, few mistook the s.d. to be 25 (instead of \sqrt{25} = 5).

Some students performed continuity correction which is not necessary, i.e. they thought that P(D < 20) = P(D < 19.5).

Question 5(b)

To calculate the probability of getting both coupons, many gave 2P(D < 30)P(D < 20). Students should be reminded of considering all possible cases, at least, both D being less than 20 is also a case.

Question 5(c)(i)

To calculate the probability of getting both coupons, many just gave 3!P(D < 30)P(D < 20)P(D < 10). There are more cases that students should take into account.

Question 5(c)(ii)

It is just a typical type question of using normal approximation to estimate binomial probabilities. Few forgot about the continuity correction.

Question 5(c)(iii)

Some students would identify that the change in E(X) makes the left shift of the p.d.f. of X (note: the left shift should not be too more, otherwise, P(D < 10) may be decreasing.). However, their claim that all P(D < 10), P(D < 20) and P(D < 30) are increased is not correct. They should point out the ‘dominating’ factor P(D < 10) is increased and hence the probability of ‘lucky’ will also be increased, because, it is hard to obtain coupon C but now we have greater chance to get it.

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