S2 Edexcel Tuesday 17th January SOLUTIONS IN FIRST POST

OP

Definitions:

Conditions for a binomial distribution
1. A fixed number of trials
2. Each trial must be success or failure
3. The trials are independent
4. The probability of success is constant

Conditions for a Poisson distribution
1. Events occure singly in space or time
2. Events occure independently of each other
3. Events occur at a constant rate

Conditions to approximate the binomial with the Poisson
1. number of trials is 'large'
2. probability of success is 'small'
3. Mean is approximately equal to variance

Conditions to approximate the binomial with the normal
1. number of trials is 'large'
2. Probability of success is 'close' to 0.5

Conditions to approximate the Poisson with the normal
lambda is 'large'

A population is a collection of all individual people or items in a particular group (this could be infinite)

A census is a survey of all the mambers of a population

A sample survey is a survey of a part of the population. In this case the surveyed members are called sampling units and a numbered or named list of the members of the population is the sampling frame.

A characteristic of the population measured from a census is a population parameter normally these cannot be measured as the population is too large in this case we estimate them using a sample. A measurement from a sample that contains no unknown parameters is a statistic. The sampling distribution of a statistic is all the possible values of the statistic along with their probabilities.

(edited 12 years ago)

Reply 81

Gibbo81

OP

1

They can be pretty liberal in their approach to what is large, small and close to 0.5.

Reply 82

Rahul.S

14

Original post by r_t

If ever there is/was/will be an exam I am/will be/have been ready for, it's this S2 paper :smile:

Reply 83

YingYang

guys, i know when to use continuity corrections, but how do you decide whether to +0.5 or -0.5?

Reply 84

hankypanky3

In a test studying reaction times, white dots appear at random on a black rectangular screen.
The continuous random variable X represents the distance, in centimetres, of the dot from the
left-hand edge of the screen. The distribution of X is rectangular over the interval [0, 20].
(a) Find P(2 < X < 3.6). (2 marks)
(b) Find the mean and variance of X. (3 marks)
The continuous random variable Y represents the distance, in centimetres, of the dot from the
bottom edge of the screen. The distribution of Y is rectangular over the interval [0, 16].
Find the probability that a dot appears
(c) in a square of side 4 cm at the centre of the screen, (4 marks)
(d) within 2 cm of the edge of the screen. (4)

Can someone please help me with c and d..? thank you

Reply 85

confused dot com

12

Original post by hankypanky3

In a test studying reaction times, white dots appear at random on a black rectangular screen.
The continuous random variable X represents the distance, in centimetres, of the dot from the
left-hand edge of the screen. The distribution of X is rectangular over the interval [0, 20].
(a) Find P(2 < X < 3.6). (2 marks)
(b) Find the mean and variance of X. (3 marks)
The continuous random variable Y represents the distance, in centimetres, of the dot from the
bottom edge of the screen. The distribution of Y is rectangular over the interval [0, 16].
Find the probability that a dot appears
(c) in a square of side 4 cm at the centre of the screen, (4 marks)
(d) within 2 cm of the edge of the screen. (4)

Can someone please help me with c and d..? thank you

What were the answers? I think I got c, but d has baffled me. Was this on a past paper?

Reply 86

hankypanky3

Original post by confused dot com

What were the answers? I think I got c, but d has baffled me. Was this on a past paper?

Its solomon paper B question 3. I dont get c nor d..help please!

Reply 87

confused dot com

12

Original post by hankypanky3

Its solomon paper B question 3. I dont get c nor d..help please!

Here are the solutions. They explain it better than I would :smile:

Original post by hankypanky3

Its solomon paper B question 3. I dont get c nor d..help please!

It's quite straightforward if you just think about it logically. The best thing to do is to draw a big diagram of the rectangle, so you can actually see what's going on. Try that, and if you need further help, quote me.

Reply 89

JoshC.

2

Resitting this, after getting a c in June.

I'm fine with all the content now, I fear silly mistakes will cost me the 90+ UMS I'm aiming for. I just did the Jan11 paper, got 69/75, thought that was great, and turned out to be 80 UMS - bit worried by that.. :s-smilie:

The papers don't take me too long now though, about an hour, so just a case of going through the questions a bit slower and I'll be fine I hope.. :biggrin:

Reply 90

hankypanky3

Original post by safmaster

It's quite straightforward if you just think about it logically. The best thing to do is to draw a big diagram of the rectangle, so you can actually see what's going on. Try that, and if you need further help, quote me.

No luck

so i drew two different rectangles..one of 20 cm and the other being 16. But, why did they times it? :s-smilie:

Reply 91

confused dot com

12

Original post by hankypanky3

No luck

so i drew two different rectangles..one of 20 cm and the other being 16. But, why did they times it? :s-smilie:

Since it's a square, one side has to be 4cm AND the other has to be 4cm, therefore you have to multiply the two values together

Reply 92

hankypanky3

How are these two linked?? Gosh, its killing me already.Just when I thought I was ready for Tuesday :/

maths.png0.8KB

Reply 93

hankypanky3

Original post by confused dot com

Since it's a square, one side has to be 4cm AND the other has to be 4cm, therefore you have to multiply the two values together

How are these two linked?? Gosh, its killing me already.Just when I thought I was ready for Tuesday :/

Original post by hankypanky3

No luck

so i drew two different rectangles..one of 20 cm and the other being 16. But, why did they times it? :s-smilie:

It's only one rectangle with width 20cm and height 16cm. Read the question again: especially the part regarding the two distributions X and Y. You should be able to do it from here.

Reply 95

Edwin Okli

3

So hello S2vians. I'm a retaker and gosh, how could it be possible that anyone who need to retake this exam? I had a bad day where I wasn't in the mood for S2. It's been pretty easy to get back into S2 even with very little revision.

Anyway, good luck to you all. :smile:

Reply 96

Edwin Okli

3

Original post by YingYang

guys, i know when to use continuity corrections, but how do you decide whether to +0.5 or -0.5?

X~Bin(n, p) approximated as Y~N(np, np(1-p))
Then, P(X < x) = P(Y <= x-0.5) and P(X <= x) = P(Y <= x+0.5).

We learnt it as with less than or equal to you are including "x" (for example) so you should have 0.5 in the opposite direction to the sign (i.e. >= -0.5, <= +0.5). Therefore the opposite is true for less than and more than.

Reply 97

hankypanky3

Original post by safmaster

It's only one rectangle with width 20cm and height 16cm. Read the question again: especially the part regarding the two distributions X and Y. You should be able to do it from here.

still dont get it !

Reply 98

nakz_

If anyones done the last question in the May/June 11 Paper the one where it talks about finding the value of K when P(μ - kσ < X < μ + kσ) = 0.5

can someone explain to me what its trying to ask in thie question?

Reply 99

safmaster

7

Original post by hankypanky3

still dont get it !

To be honest it's quite hard to explain qualitatively; but I'll try to do so.

Basically, for part (c), the first thing to do is to draw a 4x4 square at the center of your previously drawn rectangle. You are trying to calculate the probability the dot appears inside this square, so, in terms of width (X), that is 4/20, and in terms of height (Y), that is 4/16. So probability is 4/20 x 4/16 = 1/20.

For part (d), now draw inside your rectangle another rectangle measuring 16 (width) by 12 (height). This is because you want the dot now to be within 2cm of the edge, which is the area between both rectangles. The probability the dot will appear in this area is most easily calculated by doing 1 - P(dot does not appear) = 1 - (16/20 x 12/16) = 1 - 3/5 = 2/5.

Hope that helps. Have a look at the mark scheme for a better understanding.

(edited 12 years ago)

S2 Edexcel Tuesday 17th January SOLUTIONS IN FIRST POST

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