Ria

Stats Definition and Theory Handout

Stats Definitions (Updated)

2017 Specimen Stats Comments

  • 5(i) The “2!” in your working rearrange the couple among themselves. But we have 6 different couples, all of whom can rearrange themselves so we should have $2! \times 2! \times \cdots \times 2! = (2!)^6$. This gives us the final answer of $6! \times 2^6 = 46080$.
    5(ii),(iii) We’d discuss them together during class.
  • 7(i) The two assumptions we usually use are (1) constant probability and (2) each event being independent. For most of the time having a fixed number of trials is usually clear in the question, so we do not need an assumption there.
    6(ii) Good
    6(iii) Well done on observing that 2 being the mode means that $P(X=2) > P(X=3)$. However, there is also the other side where $P(X=2) > P(X=1)$. Try this side out to get the final answer of $\frac{2}{7} < p < \frac{3}{7}$.
  • 8(ii) Rather than expanding for tough looking algebra you may want to consider using graphs to straightaway jump from $\frac{6(3y+1)}{(4+y)(3+y)(2+y)}=\frac{7}{20}$ to the answer $y=2$.
    8 Well done otherwise for the rest of the question.
  • 9: Well done except 9(iii) seems to be missing
  • 10 Well done. We’d discuss 10(v) together in class

2018 Stats Comments

  • 5(i) Good. The best answer will elaborate more on what “random” means: that the probability of a fan being chosen is the same for each fan and that the fans are independently selected.
    5(ii) Correct up until the line $\sigma / \sqrt{43} > 468.1267609$. Careless mistake in the next step (should have multiplied but went to divide instead). Also, the final answer asks for variance, so we need to square the final result to get $\sigma^2 > 9\,420\,000$.
  • 6,7: Update me after you’ve attempted them from our discussions/the videos I uploaded. We can discuss this during class too. Q7(i),(ii) calculations are good.
  • 8(i)-(ii): Good
    8(iii) Give it a try again based on what I sent over text. See if you can get the answer of $g(n)=22n^2+78n+36$.
  • 11(i): We’d discuss in class
    11(ii)-(v) Good

2019 Stats Comments

  • 6(i) Good.
    6(ii) (Theory: We want our sample to be random. That is, each member of the population must have equal probability of being selected. And each selection should be independent of each other).

    He could obtain a sample of 50 supporters to interview. The sample should be random (by selecting 11, 12, 13 and 14 supporters from the clubs in Division One, Two, Three and Four respectively). Taking a random sample ensures that prevents bias in our results.

    6(iii) Multiplying is the correct approach. The answer is $7.24 \times 10^{18}$.
  • 7(i),(ii): Good

    7(iii). Notice that there is a new random variable about the number of days in which there are “at least 7 faulty mugs”. So we have $Y \sim B(5,0.10187)$. and the answer is $P(Y \leq 2) = 0.991$.

    7(iv) When there is 1 faulty mug, probability should be $(1-0.08)(0.08)\times 2!$ (you missed out the 2! to account for the arrangement of faulty vs non-faulty mugs). Similarly we missed out the 2! for case 3 for 1 faulty saucer. Make the change and see if you can get the answer of 0.0689 (no need to expand. use the GC graph when equations get complicated).
  • 8(i)Good

    8(ii)(a) Since both cases are identical (neither is orange, there is no need for 2!. Or, we can use the identical formula to get 2!/2! which is 1). For cases like one orange, one non-orange we will need the 2!.

    8(ii)(b) You missed out the case of one yellow dog and a non-yellow non-dog. See if adding that will get us the answer of $\frac{21}{110}$.

    We can also combine the 3 cases you have into 1 case since the question doesn’t care about horse, rider, nor birds. So the 3 cases can be combined as “one non-yellow dog and a yellow non-dog” for $\frac{10}{56} \times \frac{7}{55} \times 2!$.

    8(iii) Send me what you have when you’re done/or if you want hints/a discussion
  • 9(i) Good.

    9(ii): Circled the mistake in whatsapp text.

    9(iii): Since we no longer know if the population is normally distributed, in order to carry out our test we need to ensure that CLT can be used so that $\overline{X}$ is normally distributed approximately. Hence we need to take a large sample size $(n \geq 30)$ vs the original test that only took a sample size of 8.
  • 11(i)-(iii) Good.

    11(iv) Send me what you have when you’re done/or if you want hints/a discussion

Recommended Schedule

  • Thursday: 2016 P1 Q9,10,11
  • Friday: 2016 P2 Q1,2,3
  • Saturday: 2017 P1 Full paper
  • Sunday: 2017 P2 Q1-4
  • Monday: Breather/buffer
  • Tuesday: 2017 P2 Q5-10

HW 23rd September

TYS

Stats Practice

Prelim Papers

CJC 2019 Prelims

JPJC 2019 Prelims

Integration, volumes, DE discussion questions

Google Docs Link
Integration, volume, DE selected questions – Google Docs

2019 CJC Prelims

CJC 2019 Prelims

Recommended questions before next class: P1 Q2,4,5,6,7,8,9. P2 Q1,2,3,4,5(i),(ii),(iii),(v).

HW questions on vectors/hypothesis testing

Vectors_11th_Aug

HW questions on complex numbers

Complex

HW questions on transformations and Maclaurin series

HW 25th-26th July

HW 9th-12th July

5th July

HW 29th June-5th July

MJC 2018 MYE
CJC 2018 MYE

HW 5th-12th June

  • Calculus module 4: Q5 (DE)
  • Complex module 5: Q3
  • Probability and p and c module 6a: Q4-5
  • PnC, Integration

HW 29th May-

  • Calculus module 4: Q5 (DE)
  • Complex module 5: Q1-3
  • Probability and p and c module 6a: everything

Plan for 20-27th May

  • Normal Q3aii, 5iii, 5iv
  • Integration Q1b,4i
  • Binomial + DRV

HW Plan

  • 10th-17th May: Module 2 Graphs Q2,3a(ii),3b,5,7. Module 6c Normal distribution Q1-5
  • 17th-24th May: Module 4 Calculus. Module 6b DRV, Binomial distribution
  • 24th-31st May: Module 5 Complex numbers. Module 6a PnC, Probability
  • 31st May – June NYJC Prelims

Plan for 10th May

HW 6th-10th May

Plan for 3rd May

  • Review any questions from last week/school/tutorial
  • Start on sampling
  • Discuss holiday revision schedule

HW for 26th April-3rd May:

Plan for 26th April:

  • Review any questions from last week/school/tutorial
  • Past prelim normal questions: S5_Normal

HW for 22nd-26th April:

HW for 20th-22nd April:

Plan for 18th April:

Plan for 12th April:

Zoom meeting link

Zoom Meeting Link

Meeting ID: 417 847 7601
Password: 117964

HW 5th-12th April:

Plan for 12th April:

  • Review any questions you have
  • Discuss binomial distribution
  • Linear combinations of DRV

Google Drive I will be storing files in for this period of time: Google Drive link

Plan for 5th April:

  • Review any questions you have
  • Work on and discuss Probability Worksheet: Q1-7, 12-15 together.
  • Summary notes on probability: Probability Summary Notes
  • Start on the new topic: DRV.
    • Basic concepts of DRV
    • Pdf (table/function)
    • $E(X)$, $Var(X)$ and related formulas.
    • Linear combinations of DRV: e.g. $2X+3$, $X-5Y$.
  • Past prelims worksheet on DRV: DRV Worksheet
  • Selected questions for DRV: Q1-7.

For the probability worksheet, the questions with asterisks (**) are from the next topic of DRV. Q2c is a tough question. I’ve explained the technique in tackling these sort of questions in two TYS examples in the following youtube videos. You can either look at those videos or ask me to explain to you directly.