# Edexcel A level Maths 2023 paper 3 UOMS

Edexcel A level Maths 2023 paper 3 UOMS
Feel free to correct this
Venn diagram
A only:0.13,A∩B:0.25,B only:0.05,B∩C: p,C only:0.3,Outside the circles:q
1(a)P(A)=0.38
(b)p=0.2, q=0.07
(c)P(A|B')=0.325

Binomial
B(40,1/7), test n=110,9
2(a)P(T=6)=0.173
P(T<3)=0.0615
(b)P(T<3 exactly twice)=0.158
(c)Reject H₀

Large data set
Sum of x=9, Sum of x²=4341
3(a)Consider tr, treat tr as zero as ≤0.05mm rounds to 0
(b)Mean=2.12, SD=4.37
(c)May to Oct only not representable
The great storm so actual mean is less

Normal distribution
N(175.4,6.8²), 52 sample
4(a)P(X>180)=0.2494
(b)Accept H₀
(c)p=0.0562

Discrete distrbution
20,50,80,100
a,b,c,d
P(S|{X=x})=k/x
P(S∩{X=x}) constant for all x
5(a)bk/50=(S∩{X=50})
ck/80=(S∩{X=80})
As P(S∩{X=20}) constant for all x, bk/50=ck/80
c=8/5b
(b)d=2b,a=2/5b,a+b+c+d=1,b=1/5
2/25,1/5,8/25,2/5
(c)k/100=0.3,k=30,P(S|{X=20})=30/20=1.5>1 so not reasonable

Histogram
N(14.9,3.2²)sth like that, y=kxe⁻ˣ for 0≤x≤4, unit of x is ten years
6(a)0.538
(b)Not symmetrical, median not the same etc
(c)Show that ∫₀ⁿxe⁻ˣdxdx=1-(n+1)e⁻ⁿ
(d)k(∫₀⁴ xe⁻ˣdx)=90 so k=99
(e)99/90∫₁³xe⁻ˣdx=0.5902
(f)Sample size=90 is too small to estimate

1D suvat
u=0, t=5, a=3.2
1(a)v=16
(b)s=40

1D forces
5kg, Pulling force=28, a=1.4
2(a)R=49
(b)Friction=21
(c)mu=0.43

Parametric
v=(t²-3t+7)i+(t²-3)j
3(a)|v₀|=7.62
(b)t=2
(c)a=(2t-3)i+4tj
(d)t=1.5

Vector suvat
u=-16i-3j, v=2.4i+j, t=5, at time T s=4i+cj
4(a)4.47
(b)T=10,c=10

Projectile
u=28, x=40, y=20
5(a)Horizontal:s=vt
40=28cosθT
T=10/(7cosθ)
(b)Vertical:s=ut+1/2at²
20=(28sinθ)10/(7cosθ)+1/2(-9.8)(10/(7cosθ))²
20=40tanθ-10(1+tanθ²)
tan²θ-4tanθ+3=0
(c)36
(d)Spin of the ball, accurate value of g etc

Ladder
tanθ=3/4
6(a)Ssinθ(2a)=Mgcosθ(a)
S=1/2Mgcotθ
(b)Vertical:R=mg
F=μR=μmg
Horizontal:F=S
μmg=1/2Mgcotθ=3/2Mg
μ=2/3
(c)Magnitude=√13/3Mg
(d)Farther away so moment increases so S is bigger
(edited 10 months ago)
Original post by Anonymo us
Edexcel A level Maths 2023 paper 3 UOMS
Feel free to correct this
Venn diagram
1(a)0.38
(b)0.2
(c)0.07
(d)0.325

Binomial
B(40,1/7), test n=110,9
2(a)0.173
P(T<3)0.0615
(b)P(T<3 exactlytwice)=0.158
(c)Accept H1

Large data set
Sum of x=9, Sum of x²=4341
3(a)Consider tr, treat tr as zero as ≤0.05mm rounds to 0
(b)2.12,4.37
(c)May to Oct only not representable
The great storm so actual value is less

Normal distribution
N(175.4,6.8²), 52 sample
4(a)0.2494
(b)Accept H0
(c)0.0562

Discrete distrbution
20,50,80,100
a,b,c,d
P(S|{X=20})=k/x
P(S∩{X=20}) constant for all x
5(a)Show thtat c=5/8b
(b)2/25,1/5,8/25,2/5
(c)k/100=0.3,k=30,P(S|{X=20})=30/20=1.5>1 so not reasonable

Histogram
N(14.9,3.2²)sth like that, y=kxe⁻ˣ for 0≤x≤4, unit of x is ten years
6(a)0.538
(b)Not symmetrical, median not the same etc
(c)Show that ∫₀ⁿxe⁻ˣdxdx=1-(n+1)e⁻ⁿ
(d)k(∫₀⁴ xe⁻ˣdx)=90 so k=99
(e)0.5902
(f)Sample size=90 is too small to estimate

1D suvat
u=0, t=5, a=3.2
1(a)v=16
(b)s=40

1D forces
5kg, Pulling force=28, a=1.4
2(a)R=49
(b)Friction=21
(c)mu=0.43

Parametric
v=(t²-3t+7)i+(t²-3)j
3(a)|v0|=7.62
(b)t=2
(c)a=(2t-3)i+4tj
(d)t=1.5

Vector suvat
u=-16i-3j, v=2.4,1, t=5, at time T s=4i+cj
4(a)-4i+2j
(b)4.47
(c)T=10,c=10

Projectile
u=28, x=40, y=20
5(a)T=10/(7cosθ)
(b)tan²θ-4tanθ+3=0
(c)36
(d)Spin of the ball, accurate value of g etc

Ladder, tanθ=3/4
6(a)S=1/2Mgcotθ
(b)2/3
(c)√13/3Mg
(d)Farther away so moment increase so S bigger

https://docs.google.com/document/d/12l15wB_3NiOoR2Xb0NrrqihMfVlCsNDnJd-QCWdhw74/edit - here is a link to the UOMS
I wasnt given permission lol
requested at least 3 times
(edited 10 months ago)
The person who deleted everything needs to get a life
can someone make an UOMS without other people editing it
Quick question, for the normal distribution question was the standard deviation for the hypothesis test root (6.8^2/52) ?
Original post by JustTheFox
Quick question, for the normal distribution question was the standard deviation for the hypothesis test root (6.8^2/52) ?

yes
Original post by Person72
can someone make an UOMS without other people editing it

https://docs.google.com/document/d/1TQGNzDID2rKo_qW3dWAuxMfcLXUhwpU53QmLmVbFZO4/edit
Original post by JustTheFox
Quick question, for the normal distribution question was the standard deviation for the hypothesis test root (6.8^2/52) ?

you had to square root it. What u stated is the variance
Original post by Person72
you had to square root it. What u stated is the variance

Yeah I did? I put root in my message
Original post by JustTheFox
Yeah I did? I put root in my message

my bad, misread it
Last question, for the normal distribution hypothesis question, I just found the critical regions by using the inverse normal option on the calculator and inserting 0.025 and 0.975 as it was two tailed. Is this correct?