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
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The solving of the first order D.E. was quite straight forward. Few students made careless mistake in evaluating the integrating factor by giving: .
All students could find out the value of by considering the total area under a p.d.f. = 1. Most of students could determine the expected arrival time by applying directly.
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 .
Few students knew the function , however, they thought that instead of (where is the p.d.f. of the arrival time, , of Mr.X). Students should be reminded that is NOT a p.d.f. and hence does not make sense.
Students performed well in general. Few mistook that
However, the ‘‘ should not be a constant since there is no replacement during the drawing of cards.
Some could obtain full mark while some obtained no mark in surd kind of counting problem. Some gave .
Few could obtain full mark while many did not know how to do the counting.
No problem in finding the confidence interval for mean.
Even if the population distribution is normal, is greater than 95% since the s.d. of sample mean is less than the population s.d. . Now the distribution of the population is unknown, it is clearly that we cannot say anything about whether = 95% or not.
It was quite straight forward to perform the hypothesis testing for mean.
Some established as (instead of ). Some gave wrong counting of ‘over-weighted’ staff and many did not considered the continuity correction when calculating the sample proportion.
Few gave (instead of ) as the s.d. of the sample proportion. Students should be reminded of using the proportion appearing in in the calculation of s.d. corncerned.
All students did not use the upper limit of (i.e. ) to estimate the possible value of .
Many wrote . Students should pay attention to the fact that the calculation is about C.I. for proportion, not C.I. for mean.
Only few students understood the question and tried to find true mean where is the critical value of rejecting originally.
Further, students ignored the change in sample size from 16 to 32.
Many students knew that ~ , however, few mistook the s.d. to be (instead of ).
Some students performed continuity correction which is not necessary, i.e. they thought that .
To calculate the probability of getting both coupons, many gave . Students should be reminded of considering all possible cases, at least, both being less than 20 is also a case.
To calculate the probability of getting both coupons, many just gave . There are more cases that students should take into account.
It is just a typical type question of using normal approximation to estimate binomial probabilities. Few forgot about the continuity correction.
Some students would identify that the change in makes the left shift of the p.d.f. of (note: the left shift should not be too more, otherwise, may be decreasing.). However, their claim that all and are increased is not correct. They should point out the ‘dominating’ factor is increased and hence the probability of ‘lucky’ will also be increased, because, it is hard to obtain coupon but now we have greater chance to get it.