S3: Chapter 1 – Combining
Variables
Dr J Frost ([email protected])
www.drfrostmaths.com
Last modified: 30th August 2015
Starter
Let 𝑋 represent the throw of a fair 6-sided die.
Let 𝑌 represent the throw of another fair die.
𝑋 + 𝑌 represents a new variable where
the values of two dice have been added.
𝟑𝟓
= 𝟐. 𝟗𝟏𝟔𝟔
𝑬 𝑿 = 𝟑.?𝟓 𝑽𝒂𝒓 𝑿 =
? …
𝟏𝟐
𝑋 − 𝑌 represents a new variable where the
values of the dice have been subtracted.
+ 1
2
3
4
5
6
-
1
2
3
4
5
6
1
2
3
4
5
6
7
1
0
-1
-2
-3
-4
-5
2
3
4
5
6
7
8
2
1
0
-1
-2
-3
-4
3
4
5
6
7
8
9
3
2
1
0
-1
-2
-3
4
5
6
7
8
9
10
4
3
2
1
0
-1
-2
5
6
7
8
9
10
11
5
4
3
2
1
0
-1
6
7
8
9
10
11
12
6
5
4
3
2
1
0
𝑬 𝑿+𝒀 =𝟕
?
𝟕𝟎
𝑽𝒂𝒓 𝑿 + 𝒀 =
= 𝟓. 𝟖𝟑𝟑𝟑
? …
𝟏𝟐
(You’re welcome to use the
STATS mode on your calculator)
𝑬 𝑿+𝒀 =𝟎
?
𝟕𝟎
𝑽𝒂𝒓 𝑿 + 𝒀 =
= 𝟓. 𝟖𝟑𝟑𝟑
? …
𝟏𝟐
Combining Variables
𝐸 𝑋±𝑌 =𝐸 𝑋 ±𝐸 𝑌
𝑉𝑎𝑟 𝑋 ± 𝑌 = 𝑉𝑎𝑟 𝑋 + 𝑉𝑎𝑟(𝑌)
𝑋 and 𝑌 must be independent.
Quickfire Questions:
𝑋~𝑁 100,80
𝑌~𝑁 110,70
𝑬 𝑿 + 𝒀 = 𝟐𝟏𝟎
?
𝑽𝒂𝒓 𝑿 − 𝒀 = 𝟏𝟓𝟎
?
𝑬 𝑿 − 𝒀 = −𝟏𝟎?
𝑽𝒂𝒓 𝑿 + 𝒀 = 𝟏𝟓𝟎
?
𝑋~𝑁 5,0.1
𝑌~𝑁 10,0.3
𝑽𝒂𝒓 𝑿 − 𝒀 = 𝟎. 𝟒?
𝑬 𝑿 + 𝒀 = 𝟏𝟓 ?
𝑽𝒂𝒓 𝑿 + 𝒀 = 𝟎. 𝟒?
More Generally:
𝐸 𝑎𝑋 ± 𝑏𝑌 = 𝑎𝐸 𝑋 ±?𝑏𝐸 𝑌
𝑉𝑎𝑟 𝑎𝑋 ± 𝑏𝑌 = 𝑎2 𝑉𝑎𝑟 𝑋 +
? 𝑏2 𝑉𝑎𝑟(𝑌)
>> Proof
Examples
𝑋1 , 𝑋2 , 𝑋3 are independent normal variables such that
𝑋1 ~𝑁 8,22 , 𝑋2 ~𝑁 13,22 , 𝑋3 ~𝑁 18,32 and 𝑌 = 3𝑋1 − 𝑋2 + 𝑋3 , find the
distribution of 𝑌
𝑌~𝑁 3 × 8 − 13 + 18, 32 ?× 4 + 4 + 9
𝑌~𝑁(29,49)
Quickfire Questions:
𝑋1 ~𝑁 1,1
𝑋2 ~𝑁 1,1
𝑌 = 𝑋1 + 2𝑋2
𝒀~𝑵?𝟑, 𝟓
𝑋1 ~𝑁 2,3
𝑋2 ~𝑁 4,5
𝑌 = 2𝑋1 − 𝑋2
𝒀~𝑵 ?𝟎, 𝟏𝟕
𝑋1 ~𝑁 5,32
𝑋2 ~𝑁 6,22
𝑌 = 2𝑋1 − 3𝑋2
𝒀~𝑵 −𝟖,
? 𝟕𝟐
4𝑋 vs 4 𝑋’s
Suppose 𝑿~𝑵(𝟏, 𝟎. 𝟏) represents
the thickness (in mm) of a randomly
chosen A4 piece of paper.
What does 4𝑋 represent?
What does 𝑋1 + 𝑋2 + 𝑋3 + 𝑋4 represent?
(where each 𝑋𝑖 has the same distribution as 𝑋)
Each piece of paper is folded twice so
that its thickness is 4 times as much.
The total thickness when 4 separate
pieces of paper are taken.
?
?
𝑉𝑎𝑟 𝑋1 + ⋯ + 𝑋4
= 𝑉𝑎𝑟 𝑋1 + ⋯ + 𝑉𝑎𝑟 𝑋4
= 4𝑉𝑎𝑟(𝑋)?
𝑉𝑎𝑟 4𝑋 = 16𝑉𝑎𝑟(𝑋)
?
So 𝟒𝑿 does not mean the same as 𝑿𝟏 + ⋯ + 𝑿𝟒 . Why do you think the variance in the
first case is so much larger?
If a paper’s thickness is say above the mean, then multiplying the thickness by 4 also multiplies the difference from the
mean by 4, resulting in a much larger spread. However, in the second case, a thicker piece of paper is ‘averaged out’ by
another potentially thinner piece of paper, so although the variance increases (as we have 4 sheets), the increase isn’t
quite as much.
?
Test Your Understanding
𝑋~𝑁 4,52 𝑌~𝑁 7,32
𝑍 = 3𝑋 − 𝑌
𝑉𝑎𝑟
𝑉𝑎𝑟
𝑉𝑎𝑟
𝑉𝑎𝑟
𝑉𝑎𝑟
𝑋1 + 𝑋2 + 𝑋3
3𝑋
𝑌1 + ⋯ + 𝑌𝑛
𝑛𝑌
𝑍
𝑉𝑎𝑟 𝑍1 + 𝑍2
𝑉𝑎𝑟 2𝑍
= 𝟑𝑽𝒂𝒓 𝑿 = 𝟕𝟓 ?
= 𝟗𝑽𝒂𝒓 𝑿 = 𝟐𝟐𝟓 ?
= 𝒏𝑽𝒂𝒓 𝒀 = 𝟗𝒏 ?
= 𝒏𝟐 𝑽𝒂𝒓 𝒀 = 𝟗𝒏𝟐 ?
= 𝑽𝒂𝒓 𝟑𝑿 − 𝒀 = 𝟗𝑽𝒂𝒓 𝑿 + 𝑽𝒂𝒓 𝒀
?
= 𝟐𝟐𝟓 + 𝟗 = 𝟐𝟑𝟒
= 𝑽𝒂𝒓 𝒁 + 𝑽𝒂𝒓 𝒁 ? = 𝟒𝟔𝟖
= 𝟒𝑽𝒂𝒓 𝒁 = 𝟗𝟑𝟔 ?
Beefier Example
Bottles of mineral water are delivered to shops in crates containing 12 bottles each.
The weights of bottles are normally distributed with mean weight 2kg and standard
deviation 0.05kg. The weights of empty crates are normally distributed with mean
2.5kg and standard deviation 0.3kg.
a) Assuming that all random variables are independent, find the probability that
the full crate will weigh between 26kg and 27kg.
b) Two bottles are selected at random from a crate. Find the probability that they
differ in weight by more than 0.1kg.
c) Find the maximum weight 𝑀, that a full crate could have on its label so that
there is only a 1% chance that it will weigh more than 𝑀.
Weight of full crate: 𝑊 = 𝐵1 + ⋯ + 𝐵12 + 𝐶
𝐵𝑖 ~𝑁 2,0.052 𝐶~𝑁 2.5, 0.32
𝐸 𝑊 = 12𝐸 𝐵 + 𝐸 𝐶 = 26.5
𝑉𝑎𝑟 𝑊 = 12𝑉𝑎𝑟 𝐵 + 𝑉𝑎𝑟 𝐶 = 0.12
𝑊~𝑁(26.5, 0.12)
? Find distribution for weight of full crate ?
?
a) 𝑃 26 < 𝑊 < 27 = ⋯ = 0.850 (using tables)
b) Find the distribution which represents the difference of the weights.
𝐸 𝑋1 − 𝑋2 = 0 𝑉𝑎𝑟 𝑋1 − 𝑋2 = 0.005
𝑃 𝑋1 − 𝑋2 > 0 = 2𝑃 𝑋1 − 𝑋2 > 0.01
= 0.0159
c) 𝑃 𝑊 > 𝑀 = 0.01
→ 𝑀 = 27.3
?
?
Test Your Understanding
S3 Edexcel June 2009 Q8
?
?
?
Exercise 1A
Q1, 3, 5, 7-10
And also…
S3 June 2007 Q7
?
?
?
Proofey-Woofs
Not examined
<< Back
Prove that 𝐸 𝑋 + 𝑌 = 𝐸 𝑋 + 𝐸(𝑌)
𝐸 𝑋 + 𝑌 = 𝛴𝑥 𝛴𝑦 𝑥 + 𝑦 𝑝 𝑥 𝑎𝑛𝑑 𝑦
Note assumption of
= 𝛴𝑥 𝛴𝑦 𝑥 + 𝑦 𝑝 𝑥 𝑝 𝑦
independence.
= 𝛴𝑥 𝛴𝑦 𝑥 𝑝 𝑥 𝑝 𝑦 + 𝑦 𝑝 𝑥 𝑝 𝑦
= 𝛴𝑥 𝛴𝑦 𝑥 𝑝 𝑥 𝑝 𝑦 + 𝛴𝑥 𝛴𝑦?𝑦 𝑝 𝑥 𝑝 𝑦
= 𝛴𝑥 𝑥 𝑝 𝑥 𝛴𝑦 𝑝 𝑦 + 𝛴𝑦 𝑦 𝑝 𝑦 𝛴𝑥 𝑝 𝑥
= 𝛴𝑥 𝑥 𝑝 𝑥 1 + 𝛴𝑦 𝑦 𝑝 𝑦 1
=𝐸 𝑋 +𝐸 𝑌
Prove that 𝑉𝑎𝑟 𝑋 − 𝑌 = 𝑉𝑎𝑟 𝑋 + 𝑉𝑎𝑟(𝑌)
𝑉𝑎𝑟 𝑋 − 𝑌 = 𝛴𝑥 𝛴𝑦 𝑥 − 𝑦
2
𝑝 𝑥 𝑝 𝑦 − 𝐸 𝑋 −𝐸 𝑌
2
2
= 𝛴𝑥 𝛴𝑦 𝑥 2 𝑝 𝑥 𝑝 𝑦 − 𝛴𝑥 𝛴𝑦 2𝑥𝑦 𝑝 𝑥 𝑝 𝑦 + 𝛴𝑥 𝛴𝑦 𝑦 2 𝑝 𝑥 𝑝 𝑦 − 𝐸 𝑋 − 𝐸 𝑌
= 𝛴𝑥 𝑥 2 𝑝 𝑥 𝛴𝑦 𝑝 𝑦 − 2𝛴𝑥 𝑥 𝑝 𝑥 𝛴𝑦 𝑝 𝑦 + 𝛴𝑦 𝑦 2 𝑝 𝑦 𝛴𝑥 𝑝 𝑥 − 𝐸 𝑋 2 + 2𝐸 𝑋 𝐸 𝑌 − 𝐸 𝑌
= 𝐸 𝑋 2 − 2𝐸 𝑋 𝐸 𝑌 + 𝐸 𝑌 2 − 𝐸 𝑋 2 + 2𝐸 𝑋 𝐸 𝑌 − 𝐸 𝑌 2
= 𝐸 𝑋 2 − 𝐸 𝑋 2 + 𝐸 𝑌2 − 𝐸 𝑌 2
= 𝑉𝑎𝑟 𝑋 + 𝑉𝑎𝑟 𝑌
?
2