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Edexcel Further Statistics 1 - UNOFFICIAL MARK SCHEME [18 June 2019]

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but q1(a) said the probability of her winning A prize, so shouldnt it be P(X=3), cuz a prize = 1 prize.
Original post by freebirdk
The answers I got were:


1a) X~B(40, 0.02), P(X >= 3) = 0.0457
1b) Y~NB(3, 0.02), P(Y=40) = 0.00281
1c) E(Y) = 150

2a) X~Po(20/3), P(X>4) = 0.794
2b) P(exactly 3/4 5 min breaks had no calls) = 0.0219
2c) P(1 call in 5m break)*P(1 call in 15m break) = 0.0106

3) X~N(3, 2.6/80), P(X > 3.25) = 0.0828

4a) B(6, 0.55) - Although I suspect B(6, p) got you full marks
4b) v=3, X2 = 8.31 > 7.82, So reject H0 (not 100% on the numbers here, would appreciate confirmation)
4c) s=17.67, t=9.62 (or the other way around maybe)
4d) v=4, X2 = 8.749 < 9.488, Accept H0
4e) Wild not cultivated, because poisson

5a) λ=7.5, critical region: X<=2, X>=14
5b) P(Type I) = 0.0418
5c) X~Bin(8, 0.0418), P(X = 2) = 0.0414
5d) P(Type 2 | λ=6.3) = 0.945

6a) 1/ln2
6b) (2ln2 - 1)/(ln2)^2 OR 2/ln2 - 1/(ln2)^2
6c) P(X=3) = 1/24ln2 OR 0.0601

7a) P(B=4) = 8/81, P(B<=5) = 211/243
7b) E(B2) = 15
7c) E(B2) = 15, E(eR) = 19.297, so red is better than blue

I've compared these answers with several people so I'm pretty confident these are all correct
Original post by freebirdk
The answers I got were:


1a) X~B(40, 0.02), P(X >= 3) = 0.0457
1b) Y~NB(3, 0.02), P(Y=40) = 0.00281
1c) E(Y) = 150

2a) X~Po(20/3), P(X>4) = 0.794
2b) P(exactly 3/4 5 min breaks had no calls) = 0.0219
2c) P(1 call in 5m break)*P(1 call in 15m break) = 0.0106

3) X~N(3, 2.6/80), P(X > 3.25) = 0.0828

4a) B(6, 0.55) - Although I suspect B(6, p) got you full marks
4b) v=3, X2 = 8.31 > 7.82, So reject H0 (not 100% on the numbers here, would appreciate confirmation)
4c) s=17.67, t=9.62 (or the other way around maybe)
4d) v=4, X2 = 8.749 < 9.488, Accept H0
4e) Wild not cultivated, because poisson

5a) λ=7.5, critical region: X<=2, X>=14
5b) P(Type I) = 0.0418
5c) X~Bin(8, 0.0418), P(X = 2) = 0.0414
5d) P(Type 2 | λ=6.3) = 0.945

6a) 1/ln2
6b) (2ln2 - 1)/(ln2)^2 OR 2/ln2 - 1/(ln2)^2
6c) P(X=3) = 1/24ln2 OR 0.0601

7a) P(B=4) = 8/81, P(B<=5) = 211/243
7b) E(B2) = 15
7c) E(B2) = 15, E(eR) = 19.297, so red is better than blue

I've compared these answers with several people so I'm pretty confident these are all correct


didnt you find out the mean
Reply 62
no because if they won 5 tickets for example, they would get a prize
Original post by evenakimoto
but q1(a) said the probability of her winning A prize, so shouldnt it be P(X=3), cuz a prize = 1 prize.
but q1(a) said the probability of her winning A prize, so shouldnt it be P(X=3), cuz a prize = 1 prize.
and 5(c) i remember it was at least 2 out of 8 months, shouldnt it be P(X>=2) ?
Original post by freebirdk
The answers I got were:


1a) X~B(40, 0.02), P(X >= 3) = 0.0457
1b) Y~NB(3, 0.02), P(Y=40) = 0.00281
1c) E(Y) = 150

2a) X~Po(20/3), P(X>4) = 0.794
2b) P(exactly 3/4 5 min breaks had no calls) = 0.0219
2c) P(1 call in 5m break)*P(1 call in 15m break) = 0.0106

3) X~N(3, 2.6/80), P(X > 3.25) = 0.0828

4a) B(6, 0.55) - Although I suspect B(6, p) got you full marks
4b) v=3, X2 = 8.31 > 7.82, So reject H0 (not 100% on the numbers here, would appreciate confirmation)
4c) s=17.67, t=9.62 (or the other way around maybe)
4d) v=4, X2 = 8.749 < 9.488, Accept H0
4e) Wild not cultivated, because poisson

5a) λ=7.5, critical region: X<=2, X>=14
5b) P(Type I) = 0.0418
5c) X~Bin(8, 0.0418), P(X = 2) = 0.0414
5d) P(Type 2 | λ=6.3) = 0.945

6a) 1/ln2
6b) (2ln2 - 1)/(ln2)^2 OR 2/ln2 - 1/(ln2)^2
6c) P(X=3) = 1/24ln2 OR 0.0601

7a) P(B=4) = 8/81, P(B<=5) = 211/243
7b) E(B2) = 15
7c) E(B2) = 15, E(eR) = 19.297, so red is better than blue

I've compared these answers with several people so I'm pretty confident these are all correct
Have I lost many marks?
Original post by Statistician384
I even did the same thing for the critical region :frown:
Reply 65
If I got the critical region wrong, do you think I will get error carried forward marks for working out type1 error and type 2 error
Original post by freebirdk
The answers I got were:


1a) X~B(40, 0.02), P(X >= 3) = 0.0457
1b) Y~NB(3, 0.02), P(Y=40) = 0.00281
1c) E(Y) = 150

2a) X~Po(20/3), P(X>4) = 0.794
2b) P(exactly 3/4 5 min breaks had no calls) = 0.0219
2c) P(1 call in 5m break)*P(1 call in 15m break) = 0.0106

3) X~N(3, 2.6/80), P(X > 3.25) = 0.0828

4a) B(6, 0.55) - Although I suspect B(6, p) got you full marks
4b) v=3, X2 = 8.31 > 7.82, So reject H0 (not 100% on the numbers here, would appreciate confirmation)
4c) s=17.67, t=9.62 (or the other way around maybe)
4d) v=4, X2 = 8.749 < 9.488, Accept H0
4e) Wild not cultivated, because poisson

5a) λ=7.5, critical region: X<=2, X>=14
5b) P(Type I) = 0.0418
5c) X~Bin(8, 0.0418), P(X = 2) = 0.0414
5d) P(Type 2 | λ=6.3) = 0.945

6a) 1/ln2
6b) (2ln2 - 1)/(ln2)^2 OR 2/ln2 - 1/(ln2)^2
6c) P(X=3) = 1/24ln2 OR 0.0601

7a) P(B=4) = 8/81, P(B<=5) = 211/243
7b) E(B2) = 15
7c) E(B2) = 15, E(eR) = 19.297, so red is better than blue

I've compared these answers with several people so I'm pretty confident these are all correct

Why is the degrees of freedom for 4b 3? Isn’t 4 (5-1)?
And for 1a) shouldnt the probability be X=3 as it the probability of winning a prize??
Reply 67
definately, but you may lose accuracy marks.
Original post by asdasd2
If I got the critical region wrong, do you think I will get error carried forward marks for working out type1 error and type 2 error
I got the same thing too it should be X=3 because it is for winning one prize :/
Original post by evenakimoto
but q1(a) said the probability of her winning A prize, so shouldnt it be P(X=3), cuz a prize = 1 prize.
Reply 69
for 5c, yh he just wrote =2 by accident but it was greater than yh
Original post by evenakimoto
but q1(a) said the probability of her winning A prize, so shouldnt it be P(X=3), cuz a prize = 1 prize.
and 5(c) i remember it was at least 2 out of 8 months, shouldnt it be P(X>=2) ?
Original post by Hangsung
Why is the degrees of freedom for 4b 3? Isn’t 4 (5-1)?
And for 1a) shouldnt the probability be X=3 as it the probability of winning a prize??

Degrees of freedom is 3 because there are two constraints (one for estimating p, one for expected frequencies adding up to 80). Degree of freedom = 5 - 2 = 3
omg finally a paper i actually got similar answers to others for... maybe i'll get the A i need in FM after all. i did so bad on core pure 2 so im glad this one can pull me up
Did it say that the estimate was calculated? Because when the parameter is just guessed for the estimation it isn't a restraint.
Original post by freebirdk
6 columns to begin with
Combining columns 5 and 6 so that E>5 makes 5 columns.
-1 because the total is constrained to 80 makes 4 DOF.
-1 because p has been estimated from the data makes 3 DOF.

In the poisson one, it's 4 DOF because you don't need to combine columns 5 and 6.
Reply 73
I don't think so, because then it would be 6>X≥3 as four still means one prize.

I think if they wanted one prize and only one prize, it'd be more explicit, if you win two prizes, you have still won "a prize".
Original post by Hangsung
I got the same thing too it should be X=3 because it is for winning one prize :/
Reply 74
I calculated the parameter in part a just in case and got p=0.55. then the expected values we were given in part b matched p=0.55 exactly, so I reckon it was calculated.
Original post by Fathector01
Did it say that the estimate was calculated? Because when the parameter is just guessed for the estimation it isn't a restraint.
Reply 75
What mark will it be for an A* do you reckon?
Original post by Heidi-Heather
Degrees of freedom is 3 because there are two constraints (one for estimating p, one for expected frequencies adding up to 80). Degree of freedom = 5 - 2 = 3


But it didnt say the p probability was estimated. It just said that there was a constant value that the person believed.
It was only the second part with poisson that the mean was estimated where you had to take away 1.
Reply 77
Overall for further maths, probably 73-80%. But really hard to say
Original post by lafee
What mark will it be for an A* do you reckon?
(edited 4 years ago)
Huh, but if it didn't explicitly say calculated then how can you just assume that it was, is that a thing you're just supposed to do?

Original post by freebirdk
I calculated the parameter in part a just in case and got p=0.55. then the expected values we were given in part b matched p=0.55 exactly, so I reckon it was calculated.
Reply 79
Perhaps, I felt it wasn't clear in the exam so it could be either way really. Maybe they'll accept both.

I decided to go with it because that meant the first test was rejected and then the second was accepted, which makes more sense. Otherwise they'd both be accepted (contradictory)
Original post by Hangsung
But it didnt say the p probability was estimated. It just said that there was a constant value that the person believed.
It was only the second part with poisson that the mean was estimated where you had to take away 1.

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