EDEXCEL AS LEVEL FURTHER MATHEMATICS
PAPER 23 FURTHER STATISTICS 1 NEW 2026-2027
UPDATE WITH ALL COMPREHENSIVE REAL EXAM
QUESTIONS AND A DETAILED BREAKDOWN OF ALL
CORRECT VERIFIED ANSWERS EXAM PLUS
CERTIFIED RATIONALES | COMPLETE TEST
SOLUTION | PASSED & REWARDED WITH
DISTINCTION FOR ORIGINALLY GRADE A+ BRAND
NEW!!!
Section A: Discrete Probability Distributions
1. A discrete random variable XX has the
probability
distribution P(X=x)=kxP(X=x)=kx for x=1,2,3,4
x=1,2,3,4. Find the value of kk.
k=110k=101
The sum of all probabilities must equal 1.
So, k(1+2+3+4)=10k=1k(1+2+3+4)=10k=1,
hence k=110k=101.
, 2. For the distribution in Question 1,
find E(X)E(X).
E(X)=3E(X)=3
E(X)=∑xP(X=x)=∑x⋅x10=110(12+22+32+42)=301
0=3E(X)=∑xP(X=x)=∑x⋅10x=101
(12+22+32+42)=1030=3.
3. For the distribution in Question 1,
find Var(X)Var(X).
Var(X)=1Var(X)=1
E(X2)=∑x2P(X=x)=110(13+23+33+43)=10010=10
E(X2)=∑x2P(X=x)=101(13+23+33+43)=10100=10.
Then Var(X)=E(X2)−[E(X)]2=10−32=1Var(X)=E(
X2)−[E(X)]2=10−32=1.
4. A discrete random variable YY has the
probability generating
function GY(t)=16(1+t+t2+t3+t4+t5)GY(t)=61
(1+t+t2+t3+t4+t5). Find P(Y=3)P(Y=3).
P(Y=3)=16P(Y=3)=61
,The coefficient of t3t3 in the probability generating
function is the probability that Y=3Y=3. Here, the
coefficient is 1661.
5. For the distribution in Question 4,
find E(Y)E(Y).
E(Y)=2.5E(Y)=2.5
GY′(t)=16(1+2t+3t2+4t3+5t4)GY′(t)=61
(1+2t+3t2+4t3+5t4).
Then E(Y)=GY′(1)=16(1+2+3+4+5)=156=2.5E(Y)=
GY′(1)=61(1+2+3+4+5)=615=2.5.
6. A discrete random
variable ZZ has E(Z)=4E(Z)=4 and Var(Z)=3Var
(Z)=3. Find E(2Z+1)E(2Z+1).
E(2Z+1)=9E(2Z+1)=9
Using the linearity of
expectation: E(aZ+b)=aE(Z)+bE(aZ+b)=aE(Z)+b.
So, E(2Z+1)=2(4)+1=9E(2Z+1)=2(4)+1=9.
7. For the distribution in Question 6,
find Var(2Z+1)Var(2Z+1).
Var(2Z+1)=12Var(2Z+1)=12
, Using the
property Var(aZ+b)=a2Var(Z)Var(aZ+b)=a2Var(Z).
So, Var(2Z+1)=22×3=12Var(2Z+1)=22×3=12.
8. A fair six-sided die is rolled. Let XX be the
number on the uppermost face. Find E(X)E(X).
E(X)=3.5E(X)=3.5
For a fair die, each outcome has probability 1661.
So, E(X)=16(1+2+3+4+5+6)=216=3.5E(X)=61
(1+2+3+4+5+6)=621=3.5.
9. For the die roll in Question 8,
find Var(X)Var(X).
Var(X)=3512Var(X)=1235
E(X2)=16(12+22+32+42+52+62)=916E(X2)=61
(12+22+32+42+52+62)=691.
Then Var(X)=916−(3.5)2=916−494=3512Var(X)=6
91−(3.5)2=691−449=1235.
10. A discrete random variable WW has the
cumulative distribution
function F(w)=w225F(w)=25w2
PAPER 23 FURTHER STATISTICS 1 NEW 2026-2027
UPDATE WITH ALL COMPREHENSIVE REAL EXAM
QUESTIONS AND A DETAILED BREAKDOWN OF ALL
CORRECT VERIFIED ANSWERS EXAM PLUS
CERTIFIED RATIONALES | COMPLETE TEST
SOLUTION | PASSED & REWARDED WITH
DISTINCTION FOR ORIGINALLY GRADE A+ BRAND
NEW!!!
Section A: Discrete Probability Distributions
1. A discrete random variable XX has the
probability
distribution P(X=x)=kxP(X=x)=kx for x=1,2,3,4
x=1,2,3,4. Find the value of kk.
k=110k=101
The sum of all probabilities must equal 1.
So, k(1+2+3+4)=10k=1k(1+2+3+4)=10k=1,
hence k=110k=101.
, 2. For the distribution in Question 1,
find E(X)E(X).
E(X)=3E(X)=3
E(X)=∑xP(X=x)=∑x⋅x10=110(12+22+32+42)=301
0=3E(X)=∑xP(X=x)=∑x⋅10x=101
(12+22+32+42)=1030=3.
3. For the distribution in Question 1,
find Var(X)Var(X).
Var(X)=1Var(X)=1
E(X2)=∑x2P(X=x)=110(13+23+33+43)=10010=10
E(X2)=∑x2P(X=x)=101(13+23+33+43)=10100=10.
Then Var(X)=E(X2)−[E(X)]2=10−32=1Var(X)=E(
X2)−[E(X)]2=10−32=1.
4. A discrete random variable YY has the
probability generating
function GY(t)=16(1+t+t2+t3+t4+t5)GY(t)=61
(1+t+t2+t3+t4+t5). Find P(Y=3)P(Y=3).
P(Y=3)=16P(Y=3)=61
,The coefficient of t3t3 in the probability generating
function is the probability that Y=3Y=3. Here, the
coefficient is 1661.
5. For the distribution in Question 4,
find E(Y)E(Y).
E(Y)=2.5E(Y)=2.5
GY′(t)=16(1+2t+3t2+4t3+5t4)GY′(t)=61
(1+2t+3t2+4t3+5t4).
Then E(Y)=GY′(1)=16(1+2+3+4+5)=156=2.5E(Y)=
GY′(1)=61(1+2+3+4+5)=615=2.5.
6. A discrete random
variable ZZ has E(Z)=4E(Z)=4 and Var(Z)=3Var
(Z)=3. Find E(2Z+1)E(2Z+1).
E(2Z+1)=9E(2Z+1)=9
Using the linearity of
expectation: E(aZ+b)=aE(Z)+bE(aZ+b)=aE(Z)+b.
So, E(2Z+1)=2(4)+1=9E(2Z+1)=2(4)+1=9.
7. For the distribution in Question 6,
find Var(2Z+1)Var(2Z+1).
Var(2Z+1)=12Var(2Z+1)=12
, Using the
property Var(aZ+b)=a2Var(Z)Var(aZ+b)=a2Var(Z).
So, Var(2Z+1)=22×3=12Var(2Z+1)=22×3=12.
8. A fair six-sided die is rolled. Let XX be the
number on the uppermost face. Find E(X)E(X).
E(X)=3.5E(X)=3.5
For a fair die, each outcome has probability 1661.
So, E(X)=16(1+2+3+4+5+6)=216=3.5E(X)=61
(1+2+3+4+5+6)=621=3.5.
9. For the die roll in Question 8,
find Var(X)Var(X).
Var(X)=3512Var(X)=1235
E(X2)=16(12+22+32+42+52+62)=916E(X2)=61
(12+22+32+42+52+62)=691.
Then Var(X)=916−(3.5)2=916−494=3512Var(X)=6
91−(3.5)2=691−449=1235.
10. A discrete random variable WW has the
cumulative distribution
function F(w)=w225F(w)=25w2
