Independent observations imply that :

P(X₁ = A and X₂ = B) = P(X₁ = A) × P(X₂ = B)


Then,


P(X₁ < X₂ ∣ X₁ ≤ 2)

= P(X₁ ≤ 2 and X₁ < X₂) ÷ P(X₁ ≤ 2)

= [P(X₁ = 2 and X₂ = 3)
+ P(X₁ = 2 and X₂ = 4)
+ P(X₁ = 1 and X₂ = 2)
+ P(X₁ = 1 and X₂ = 3)
+ P(X₁ = 1 and X₂ = 4)
+ P(X₁ = 0 and X₂ = 1)
+ P(X₁ = 0 and X₂ = 2)
+ P(X₁ = 0 and X₂ = 3)
+ P(X₁ = 0 and X₂ = 4) ]

÷ [P(X₁ = 0) + P(X₁ = 1) + P(X₁ = 2)]

= [ ¾ × 0
+ ¾ × 1/6
+ 0 × ¾
+ 0 × 0
+ 0 × 1/6
+ 1/12 × 0
+ 1/12 × ¾
+ 1/12 × 0
+ 1/12 × 1/6]

÷ [1/12 + 0 + ¾]


= [⅛ + 1/16 + 1/72] ÷ 5/6

= 29/144 × 6/5

= 29/24 × 1/5

= 29/120

Alternatively, do the complement method.


P(X₁ ≤ 2 and X₁ < X₂) ÷ P(X₁ ≤ 2)

= [P(X₁ ≤ 2) - P(X₁ ≤ 2 and X₁ ≥ X₂)] ÷ P(X₁ ≤ 2)

= [P(X₁ = 0 + P(X₁ = 1) + P(X₁ = 2)
- P(X₁ = 0) × P(X₂ = 0)
- P(X₁ = 1) × P(X₂ = 0)
- P(X₁ = 1) × P(X₂ = 1)
- P(X₁ = 2) × P(X₂ = 0)
- P(X₁ = 2) × P(X₂ = 1)
- P(X₁ = 2) × P(X₂ = 2)]

÷ [P(X₁ = 0 + P(X₁ = 1) + P(X₁ = 2)]


= [1/12 + 0 + ¾
- 1/12 × 1/12
- 0 × 1/12
- 0 × 0
- ¾ × 1/12
- ¾ × 0
- ¾ × ¾]

÷ [1/12 + 0 + ¾]


= [5/6 - 1/144 - 1/16 - 9/16] ÷ 5/6

= 29/144 × 6/5

= 29/24 × 1/5

= 29/120

Second part of (iii)

If the average exceeds 2, the total of 50 observations exceeds (2 × 50) = 100

What we want to do is to sum up all the variances Var(X) and expected values E(X) of all 50 independent observations first.

Let Y = X1 + X2 + X3 + ... + X50

Since they are all independent,


E(Y) = E(X1) + E(X2) + E(X3) + ... E(X50)

= 50 E(X)
= 50 × 13/6
= 108 ⅓

Var(Y) = Var(X1) + Var(X2) + Var(X3) +... + Var(X50)

= 50 Var(X)
= 50 × 35/36
= 48 11/18


Assume the total score of 50 observations is normally distributed.

So we can say :

Y ~ N(108⅓,√(48 11/18))


Then, find P(Y > 100) using your GC

normalcdf (100,200,108⅓√(48 11/18))

≈ 0.884001
= 0.884 (3s.f)

(Lower bound is 100, upper bound is 200 because the maximum total is 4 points per observation x 50 observations = 200)


Or,

find 1 - P(Y< 100)

1 - normalcdf (0,100,108⅓√(48 11/18))

≈ 1 - 0.115999
≈ 0.884001
= 0.884 (3s.f)

Alternately, use the Central Limit Theorem since number of observations > 30

Define X̄ to be the average score of all the observations.


μ = E(X) = 13/6 = 2 1/6

σ = √Var(X)/√n
= √35/36 / √ 50
= √7/360


Then,

X̄~N(2 1/6, √7/360)


Probability that the average observation exceeds 2

= P(X̄ > 2)

normalcdf(2,4,2 1/6, √7/360)

(Upper bound here is 4 since the max average score is 4)

≈ 0.884001
= 0.884 (3s.f)

if we 'assume' Y is normal, theoretically allowed. But should we do a continuity correction when finding the probability? For CLT, no should requirement.

under sample mean distribution, x-bar is the sample mean. it is not the mean of "all score" (all scores in the population?)

So far, continuity correction is done when the we do a normal approximation to the binomial distribution. (This has been taken out of syllabus 9758)

Using CLT, don't need as you mentioned.

On second look, fhe first case can use CLT directly already.

And anyway, I wrote 'average score of all the observations'

That means all those 50 observations i.e sample mean.

There was no mention of population mean.

'all the observations' ≠ 'every single possible 'observation' ≠ 'all observations' ≠ 'all possible scores observed'


There isn't actually a population here too anyway.

Also, I noticed you used 1E99 for your upper bound.

But the highest possible mean/average is only 4.

So not sure how the teacher/marker would grade that.

Did the qn suggest only 50 observations were made. Qn suggested that 50 observations were being taken as a sample. Beside, mis-concept again! Continuity correction applied whenever we approximate a discrete rv using normal.

Im referring to the 9758 (and also 9740) syllabi, where the students are taught normal approximation to the binomial distribution in the latter.

Did I say 'continuity correction is ONLY used for binomial distribution?'

I didn't.


So what misconception are you referring to?

' Did question suggest only 50 observations were made?'

No. And neither did I imply that... So what are you onto actually ? What is your concern?

Oh no... starting i just wanted to highlight that we should not define x-bar as 'average of all observations'. All observations is not confined to the 50 observations being taken as sample. At least you know that there are more than 50 observations. Otherwise, it should not be called a sample.

So your main concern is the word 'all' in the phrase 'of all the observations'?

http://jc-math.com/gce-a-level-h2-math/statistics/central-limit-theorem-eggs

The example in this website uses 'mean mass of 60 randomly chosen size 1 eggs'


Is this considered wrong to you then?


What is the appropriate phrasing in your opinion ?

60 randomly chosen eggs = "all'? I don't see it. In fact x-bar is s standard notation for sample mean. The qn only needs to define x. No need to define x-bar. If want to define x-bar, 'the mean values of x in the sample of 50'

"mean mass of 60 randomly chosen eggs"

Isn't that mean of all those 60 randomly chosen ones?

If not those 60, then which 60?

This 60? That 60? are you implying that the word 'any' should be used?

It is important to write out the values of mean and sigma^2÷n in the N distritbution.

As mentioned above ,

'all the observations' does not equal to 'all observations'

i.e it is not 'all possible observations'

That has already been written in my working, I merely found them individually first.

Marks will not be lost and there is no strict or specific requirement that states that you must first state this :

X̄~N (μ,σ²/n)


You should also be aware of the different styles of presentation for the above.

Some styles go with stating the variance in the brackets, others state the S.D.

We can't say either is wrong. There's no fixed convention here. What is important is that it is clarified.

Let it be. Wasting time!

Well that is your opinion... wasting time or not is subjective.

I can also say your workings are too short and you skip steps... risking loss of marks...

And are you actually a full-time tutor or teacher or lecturer or something?

I see you have answered plenty of questions, but your profile is not listed.