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1
Discrete and Continuous Random
•
A variable X whose value depends on the outcome of a random process is called a random
variable.
ex: X is the outcome of a coin toss
ex: X is the 1st number drawn in the next lottery draw
ex: X is the age of an individual chosen at random from Zagreb population
Discrete Random Variables
•
•
A discrete variable is a variable which can only take a countable number of values.
Thus, a discrete random variable X has possible values x1 , x2 , x3, .....
The variable is said to be random if the sum of the probabilities is one.
Example:
A coin is tossed three times
X represents the number of heads that we throw
x
0
1
8
P(X = x)
•
1
3
8
2
3
8
3
1
8
The probability density function (p.d.f.) of X is a function which allocates probabilities. Put
simply, it is a function which tells you the probability of certain events occurring. The usual
notation that is used is P(X = x) = something. The random variable X is the event that we are
considering.
1. 0 ≤ 𝑃𝑃(𝑋𝑋 = 𝑥𝑥) ≤ 1
2. � 𝑃𝑃(𝑋𝑋 = 𝑥𝑥) = 1
𝑥𝑥
3. 𝑃𝑃(𝑋𝑋 = 𝑥𝑥𝑛𝑛 ) = 1 − � 𝑃𝑃(𝑋𝑋 = 𝑥𝑥𝑘𝑘 )
𝑘𝑘≠𝑛𝑛
𝑃𝑃 (𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 𝑥𝑥𝑛𝑛 𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜) = 1 − 𝑃𝑃(𝑎𝑎𝑎𝑎𝑎𝑎 𝑜𝑜𝑜𝑜ℎ𝑒𝑒𝑒𝑒 𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒𝑒 𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜)
Example:
A die is thrown repeatedly until a 6 is obtained. Find the probability density function for the number
times we throw the die.
Let X be the random variable representing the number of times we throw the die.
P(X = 1) = 1/6
(probability that we get 6 on our first throw)
P(X = 2) = (5/6) × (1/6) (probability that we get 6 on second throw; if we throw the die twice
before getting a 6, first throw is something that isn't a 6 (probability
is 5/6) and second throw is a 6
etc
5 𝑥𝑥−1 1
𝑃𝑃(𝑋𝑋 = 𝑥𝑥) = � �
6
6
2
Continuous Random Variables
•
•
A continuous random variable is a random variable that can assume any value in an interval.
ex: X is the length of time until the next time you are sick.
ex: X is the weight of someone chosen at random from the Croatian population.
Definition: If X is a continuous random variable, then a function f (x) with domain (a,b) is called a
probability density function for X if and only if
1.
𝑓𝑓(𝑥𝑥) ≥ 0
𝑏𝑏
2. � 𝑓𝑓(𝑥𝑥) = 1
𝑓𝑓𝑓𝑓𝑓𝑓 𝑎𝑎𝑎𝑎𝑎𝑎 𝑥𝑥 𝜖𝜖 (𝑎𝑎, 𝑏𝑏)
𝑎𝑎
N.B.
3. 𝑓𝑓𝑓𝑓𝑓𝑓 𝑎𝑎𝑎𝑎𝑎𝑎 𝑎𝑎 ≤ 𝑐𝑐 < 𝑑𝑑 ≤ 𝑏𝑏,
𝑑𝑑
𝑃𝑃(𝑐𝑐 < 𝑋𝑋 < 𝑑𝑑) = � 𝑓𝑓(𝑥𝑥)𝑑𝑑𝑑𝑑
𝑐𝑐
1. 𝑎𝑎 𝑎𝑎𝑎𝑎𝑎𝑎 𝑏𝑏 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 𝑏𝑏𝑏𝑏 − ∞ 𝑎𝑎𝑎𝑎𝑎𝑎 ∞. 𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀 𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜𝑜 𝑖𝑖𝑖𝑖 𝐼𝐼𝐼𝐼.
2. 𝐹𝐹𝐹𝐹𝐹𝐹 𝑎𝑎 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐 𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟𝑟 𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣, 𝑡𝑡ℎ𝑒𝑒 𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝𝑝 𝑜𝑜𝑜𝑜 𝑎𝑎𝑎𝑎𝑎𝑎 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣𝑣 𝑖𝑖𝑖𝑖 𝑧𝑧𝑧𝑧𝑧𝑧𝑧𝑧
𝑃𝑃(𝑋𝑋 = 𝑐𝑐) = 0 ⇒ 𝑃𝑃(𝑐𝑐 ≤ 𝑋𝑋 ≤ 𝑑𝑑) = 𝑃𝑃(𝑐𝑐 < 𝑋𝑋 < 𝑑𝑑) = 𝑃𝑃(𝑐𝑐 ≤ 𝑋𝑋 < 𝑑𝑑) 𝑒𝑒𝑒𝑒𝑒𝑒.
•
Expected value (or mean) is a weighted average of the possible values that X can take, each
value being weighted according to the probability of that event occurring. The expected value of X
is usually written as E(X) or μ.
Discrete Random Variables
𝑬𝑬(𝑿𝑿) = 𝝁𝝁 = � 𝒙𝒙 𝑷𝑷(𝑿𝑿 = 𝒙𝒙)
Continuous Random Variables
∞
𝑬𝑬(𝑿𝑿) = 𝝁𝝁 = � 𝒙𝒙 𝒇𝒇(𝒙𝒙)𝒅𝒅𝒅𝒅
sum of: [(each of the possible outcomes)
× (the probability of the outcome occurring)].
−∞
Example
What is the expected value when we roll a fair die?
There are six possible outcomes: 1, 2, 3, 4, 5, 6. Each of these has a probability of 1/6 of
occurring.
Let X represent the outcome of the experiment.
1
𝑃𝑃(𝑋𝑋 = 1) = 𝑃𝑃(𝑋𝑋 = 2) = ⋯ = 𝑃𝑃(𝑋𝑋 = 6) =
6
𝐸𝐸(𝑋𝑋) = 1 × 𝑃𝑃(𝑋𝑋 = 1) + 2 × 𝑃𝑃(𝑋𝑋 = 2) + 3 × 𝑃𝑃(𝑋𝑋 = 3)
+ 4 × 𝑃𝑃(𝑋𝑋 = 4) + 5 × 𝑃𝑃(𝑋𝑋 = 5) + 6 × 𝑃𝑃(𝑋𝑋 = 6)
𝐸𝐸(𝑋𝑋) =
1 2 3 4 5 6 7
+ + + + + = = 3.5
6 6 6 6 6 6 2
3
•
Properties of expected value E(X)
1. 𝐼𝐼𝐼𝐼 𝑎𝑎 𝑎𝑎𝑎𝑎𝑎𝑎 𝑏𝑏 𝑎𝑎𝑎𝑎𝑎𝑎 𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐𝑐, 𝑡𝑡ℎ𝑒𝑒𝑒𝑒 𝐸𝐸(𝑎𝑎𝑎𝑎 + 𝑏𝑏) = 𝑎𝑎𝑎𝑎(𝑋𝑋) + 𝑏𝑏
2. 𝐸𝐸(𝑋𝑋 + 𝑌𝑌 ) = 𝐸𝐸(𝑋𝑋) + 𝐸𝐸(𝑌𝑌 )
Expected Value of a Function of X
𝐸𝐸[ 𝑔𝑔(𝑋𝑋)], 𝑤𝑤ℎ𝑒𝑒𝑒𝑒𝑒𝑒 𝑔𝑔(𝑋𝑋) 𝑖𝑖𝑖𝑖 𝑎𝑎 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 𝑜𝑜𝑜𝑜 𝑋𝑋 𝑖𝑖𝑖𝑖:
Discrete Random Variables
Continuous Random Variables
𝐸𝐸[ 𝑔𝑔(𝑋𝑋)] = � 𝑔𝑔(𝑥𝑥)𝑃𝑃(𝑋𝑋 = 𝑥𝑥)
𝐸𝐸[ 𝑔𝑔(𝑋𝑋)] = ∫ 𝑔𝑔(𝑥𝑥)𝑓𝑓(𝑥𝑥)𝑑𝑑𝑑𝑑
Example
For the experiment with the die, calculate E(X2)
f(x) = x2
f(1)=1, f(2)=4, f(3)=9, f(4)=16, f(5)=25, f(6)=36
𝑃𝑃(𝑋𝑋 = 1) = 𝑃𝑃(𝑋𝑋 = 2) = ⋯ = 𝑃𝑃(𝑋𝑋 = 6) =
1
6
So E(X2) = 1/6 + 4/6 + 9/6 + 16/6 + 25/6 + 36/6 = 91/6 = 15.167
•
Mode is the most likely value of X.
Discrete Random Variables
The mode is the value of x with largest 𝑃𝑃(𝑋𝑋 = 𝑥𝑥) which can be different from the expected value
Continuous Random Variables
The mode is the value of x where f(x) is maximum (which may not be unique).
•
Median is the value with half the probabilities below and half above the median value.
Discrete Random Variables
The median is the middle value when there are an odd number of measurements listed, and it is the
average of the two middle values when there are an even number of measurements listed.
It is the value of X for which P(X < x) is greater than or equal to 0.5 and P(X > x) is greater than or equal
to 0.5.
Continuous Random Variables
The median of a random variable X is a number m such that
𝑚𝑚
� 𝑓𝑓(𝑥𝑥) =
𝑎𝑎
1
2
The median m is the number for which the probability is exactly ½ that the random variable will have
a value greater than m, and ½ that it will have a value less than m.
4
• Variance
The variance of a random variable tells us something about the spread of the possible values of the
variable. Variance, Var( X ), is defined as the average of the squared differences of X from the mean:
•
𝟐𝟐
𝑽𝑽𝑽𝑽𝑽𝑽(𝑿𝑿) = 𝝈𝝈𝟐𝟐 = 𝑬𝑬 �𝑿𝑿 – 𝝁𝝁�
= 𝐸𝐸(𝑋𝑋 2 )– 2𝜇𝜇 𝐸𝐸(𝑋𝑋)
��� + 𝜇𝜇2 = 𝑬𝑬�𝑿𝑿𝟐𝟐 � − 𝝁𝝁𝟐𝟐
𝜇𝜇
Discrete Random Variables
𝑽𝑽𝑽𝑽𝑽𝑽(𝑿𝑿) = 𝛴𝛴 (𝑥𝑥 – 𝜇𝜇)2 𝑃𝑃(𝑋𝑋 = 𝑥𝑥) = 𝛴𝛴𝑥𝑥 2 𝑃𝑃(𝑋𝑋 = 𝑥𝑥) − 2𝜇𝜇 𝛴𝛴 𝑥𝑥 𝑃𝑃(𝑋𝑋 = 𝑥𝑥) − 𝜇𝜇2 𝛴𝛴 𝑃𝑃(𝑋𝑋 = 𝑥𝑥)
= 𝛴𝛴𝑥𝑥 2 𝑃𝑃(𝑋𝑋 = 𝑥𝑥) − 2𝜇𝜇 ∙ 𝜇𝜇 + 𝜇𝜇2 = 𝛴𝛴𝑥𝑥 2 𝑃𝑃(𝑋𝑋 = 𝑥𝑥) − 𝜇𝜇2
Continuous Random Variables
𝟐𝟐
𝒃𝒃
𝟐𝟐
𝑏𝑏
𝑏𝑏
2
2
𝑏𝑏
• 𝑽𝑽𝑽𝑽𝑽𝑽(𝑿𝑿) = 𝝈𝝈 = �(𝒙𝒙 − 𝝁𝝁) 𝒇𝒇(𝒙𝒙)𝒅𝒅𝒅𝒅 = � 𝑥𝑥 𝑓𝑓(𝑥𝑥)𝑑𝑑𝑑𝑑 − 2𝜇𝜇 � 𝑥𝑥𝑥𝑥(𝑥𝑥)𝑑𝑑𝑑𝑑 + 𝜇𝜇 � 𝑓𝑓(𝑥𝑥)𝑑𝑑𝑑𝑑 =
•
𝒂𝒂
𝑎𝑎
𝑎𝑎
𝑏𝑏
𝒃𝒃
𝑎𝑎
= ∫𝑎𝑎 𝑥𝑥 2 𝑓𝑓(𝑥𝑥)𝑑𝑑𝑑𝑑 − 2𝜇𝜇 ∙ 𝜇𝜇 + 𝜇𝜇2 = ∫𝒂𝒂 𝒙𝒙𝟐𝟐 𝒇𝒇(𝒙𝒙)𝒅𝒅𝒅𝒅 − 𝝁𝝁𝟐𝟐
Properties of variance
Note that the variance does not behave in the same way as expectation when we multiply and add
constants to random variables. In fact:
•
𝑉𝑉𝑉𝑉𝑉𝑉[𝑎𝑎𝑎𝑎 + 𝑏𝑏] = 𝑎𝑎2 𝑉𝑉𝑉𝑉𝑉𝑉(𝑋𝑋)
𝑉𝑉 𝑎𝑎𝑎𝑎(𝑎𝑎𝑎𝑎 + 𝑏𝑏) = 𝐸𝐸[ (𝑎𝑎𝑎𝑎 + 𝑏𝑏)2 ] − (𝐸𝐸 [𝑎𝑎𝑎𝑎 + 𝑏𝑏])2
= 𝐸𝐸[ 𝑎𝑎 2 𝑋𝑋 2 + 2𝑎𝑎𝑎𝑎𝑎𝑎 + 𝑏𝑏 2 ] − (𝑎𝑎𝑎𝑎(𝑋𝑋) + 𝑏𝑏) 2
= 𝑎𝑎2 𝐸𝐸[𝑋𝑋 2 ] + 2𝑎𝑎𝑎𝑎𝑎𝑎(𝑋𝑋) + 𝑏𝑏 2 − 𝑎𝑎2 𝐸𝐸 2 (X) − 2𝑎𝑎𝑎𝑎𝑎𝑎(𝑋𝑋) − 𝑏𝑏 2
= 𝑎𝑎 2 𝐸𝐸[𝑋𝑋 2 ] − 𝑎𝑎2 𝐸𝐸 2 (X)
•
•
= 𝑎𝑎2 𝑉𝑉𝑉𝑉𝑉𝑉(𝑋𝑋)
𝐼𝐼𝐼𝐼 𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔𝑔 𝑉𝑉𝑉𝑉𝑉𝑉[𝑋𝑋 + 𝑌𝑌] ≠ 𝑉𝑉𝑉𝑉𝑉𝑉(𝑋𝑋) + 𝑉𝑉𝑉𝑉𝑉𝑉(𝑌𝑌)
(only for independent − not IB)
Standard deviation of X is the square root of Var(X).
𝜎𝜎 = �𝑉𝑉𝑉𝑉𝑉𝑉(𝑋𝑋)
5
Binomial Distribution - Discrete
There are three criteria that must be met in order for a random probability distribution to be a
binomial distribution.




The experiment consists of n repeated trials.
Each trial can result in just two possible outcomes. We call one of these outcomes a success and
the other, a failure.
The probability of success, denoted by p, is the same on every trial.
The trials are independent; that is, the outcome on one trial does not affect the outcome on
other trials – the probability of success is a constant in each trial
If a random variable X has a binomial distribution, we write
• 𝑿𝑿 ~ 𝑩𝑩(𝒏𝒏, 𝒑𝒑)
(~ means ‘has distribution…’).
and the probability density function is:
𝑛𝑛
• 𝑃𝑃(𝑋𝑋 = 𝑥𝑥) = � � 𝑝𝑝 𝑥𝑥 (1 − 𝑝𝑝)𝑛𝑛−𝑥𝑥
𝑥𝑥
• n is number of trials
• p is the probability of a success
• (1 – p) is the probability of a failure.
𝑥𝑥 = 0, 1, … , 𝑛𝑛
If X is a random variable is binomial with parameters n and p, then mean and variance are: •
𝐸𝐸(𝑋𝑋) = 𝜇𝜇 = 𝑛𝑛𝑛𝑛
• 𝑉𝑉𝑉𝑉𝑉𝑉(𝑋𝑋) = 𝜎𝜎 2 = 𝑛𝑛𝑛𝑛(1 − 𝑝𝑝)
Example
Suppose a die is tossed 5 times. What is the probability of getting exactly 2 fours?
Solution: This is a binomial experiment in which the number of trials is equal to 5, the number
of successes is equal to 2, and the probability of success on a single trial is 1/6 or about 0.167.
Therefore, the binomial probability is:
n = 5,
p = 0.167
x=2
5
𝑃𝑃(𝑋𝑋 = 2) = � � 0.1672 (0.833)3 = 0.161
2
6
Poisson Distribution
A discrete random variable X with a probability distribution function (p.d.f.) of the form:
• 𝑃𝑃(𝑋𝑋 = 𝑥𝑥) =
𝑚𝑚 𝑥𝑥 𝑒𝑒 −𝑚𝑚
,
𝑥𝑥!
𝑥𝑥 = 0, 1, 2, …
is said to be a Poisson random variable with parameter m. We write • 𝑋𝑋 ~ 𝑃𝑃𝑃𝑃(𝑚𝑚)
Mean and Variance
𝐼𝐼𝐼𝐼 𝑋𝑋 ~ 𝑃𝑃𝑃𝑃(𝑚𝑚), 𝑡𝑡ℎ𝑒𝑒𝑒𝑒
• 𝐸𝐸(𝑋𝑋) = 𝑚𝑚
• 𝑉𝑉𝑉𝑉𝑉𝑉(𝑋𝑋) = 𝑚𝑚
Random Events
The Poisson distribution is useful because many random events follow it.
If a random event has a mean number of occurrences m in a given time period, then the number of
occurrences within that time period will follow a Poisson distribution.
For example, the occurrence of earthquakes could be considered to be a random event. If there are 5
major earthquakes each year, then the number of earthquakes in any given year will have a Poisson
distribution with parameter 5.
There are three criteria that must be met in order for a random probability distribution to be a
Poisson distribution.



The average number of occurrences (m) is constant for every interval.
The probability of more than one occurrence in a given interval is very small.
The number of occurrences in disjoint intervals are independent of each other.
Sums of Poissons
Suppose X and Y are independent Poisson random variables with parameters ℓ and m respectively.
Then X + Y has a Poisson distribution with parameter ℓ + m.
In other words: If 𝑿𝑿 ~ 𝑷𝑷𝑷𝑷(𝓵𝓵), and 𝒀𝒀 ~ 𝑷𝑷𝑷𝑷(𝒎𝒎), ), then 𝑿𝑿 + 𝒀𝒀~ 𝑷𝑷𝑷𝑷(𝓵𝓵 + 𝒎𝒎)
Example
There are 50 misprints in a book which has 250 pages. Find the probability that page 100 has no
misprints.
The average number of misprints on a page is 50/250 = 0.2. If we let X be the random variable
denoting the number of misprints on a page, X will follow a Poisson distribution with parameter 0.2
m = 0.2
𝑃𝑃(𝑋𝑋 = 0) =
0.20 𝑒𝑒 −0.2
= 0.819 (3𝑠𝑠𝑠𝑠)
0!
7
Normal distribution. Standardization of normal variables.
Data can be "distributed" (spread out) in different ways. It can be spread out more on the left or more on
the right, but there are many cases where the data tends to be around a central value with no bias left
or right, and it gets close to a "Bell Curve". That is called Normal Distribution. Many things closely
follow a Normal Distribution:
•heights of people
•size of things produced by machines
•errors in measurements
•blood pressure
•marks on a test (not at North Hills)
▪ Most results are close to the mean,
while few are much left or right of the mean.
A continuous random variable X follows a normal distribution if it has the following probability
density function :
• 𝑓𝑓(𝑥𝑥) =
1
𝜎𝜎√2𝜋𝜋
𝑒𝑒
1 𝑥𝑥−𝜇𝜇 2
− �
�
2 𝜎𝜎
− ∞ < 𝑥𝑥 < ∞
The grand majority of continuous distributions are normal distributions, where the probability density
decreases according to how far the value is from the mean. This is particularly true for variables in
nature.
The parameters of the distribution are μ and 𝝈𝝈𝟐𝟐 , where μ is the mean (expectation) of the distribution
and 𝜎𝜎 2 is the variance.
𝑿𝑿 ~ 𝑵𝑵(𝝁𝝁, 𝝈𝝈𝟐𝟐 ) means that the random variable X has a normal distribution with parameters μ and 𝜎𝜎2 .
Properties
1.
2.
𝑑𝑑
1
𝑥𝑥 − 𝜇𝜇 −1�𝑥𝑥−𝜇𝜇�2
𝑓𝑓(𝑥𝑥) = −
�
� 𝑒𝑒 2 𝜎𝜎
=0
𝑑𝑑𝑑𝑑
𝜎𝜎
𝜎𝜎 2 √2𝜋𝜋
𝑥𝑥 − 𝜇𝜇 2 −1�𝑥𝑥−𝜇𝜇�2
𝑑𝑑
1
𝑓𝑓(𝑥𝑥) = −
� � 𝑒𝑒 2 𝜎𝜎
=0
�1 − �
𝜎𝜎
𝑑𝑑𝑑𝑑
𝜎𝜎 3 √2𝜋𝜋
⇒ 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 :
𝒙𝒙 = 𝝁𝝁
⇒ 𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖: 𝒙𝒙 = 𝝁𝝁 ± 𝝈𝝈
The curve is symmetrical about the line x = μ
▪
▪
lim 𝑓𝑓(𝑥𝑥) = 0
𝑥𝑥→±∞
∞
� 𝑓𝑓(𝑥𝑥) = 1
−∞
▪ 𝜇𝜇 = 𝑚𝑚𝑚𝑚𝑚𝑚{𝑓𝑓(𝑥𝑥)}
For a normal curve, standard deviation σ is uniquely determined as the horizontal distance from the
vertical line of symmetry 𝑥𝑥 = 𝜇𝜇 to the point of inflection.
In a normal distribution, 68.26% of values lie within one standard deviation of the mean, 95.4% of
values lie within two standard deviations of the mean and 99.74% of values lie within three standard
deviations of the mean.
8
Example: 95% of students at school are between 1.1m and 1.7m tall.
Assuming this data is normally distributed can you calculate the mean and standard deviation?
The mean is halfway between 1.1m and 1.7m:
Mean = (1.1m + 1.7m) / 2 = 1.4m
95% is 2 standard deviations either side of the mean
(a total of 4 standard deviations) so:
1 standard deviation = (1.7m-1.1m) / 4 = 0.6m / 4 = 0.15m
Standard score or z – score is the number of standard deviations from the mean.
Example: In that same school one of your friends is 1.85m tall
How far is 1.85m from the mean?
1.85 - 1.4 = 0.45m
How many standard deviations is that?
0.45m / 0.15m = 3 standard deviations
Your friend's height has a "z-score" of 3.0
The Standard Normal Distribution
If 𝑍𝑍 ~ 𝑁𝑁(0, 1), then Z is said to follow a standard normal distribution.
P(Z < z) is known as the cumulative distribution function of the random variable Z. For the standard
normal distribution, this is usually denoted by F(z). Normally, you would work out the c.d.f. by doing
some integration. However, it is impossible to do this for the normal distribution and so results have to
be looked up in statistical tables.
Standardising: Now, the mean and variance of the normal distribution can be any value and so clearly
there can't be a statistical table for each one. Instead, we convert to the standard normal distributionwe can also use statistical tables for the standard normal distribution to find the c.d.f. of any normal
distribution.
Cumulative Distribution Functions is defined as 𝑃𝑃(𝑋𝑋 ≤ 𝑥𝑥) and that what we need
The standard normal distribution, or Z-distribution, is the application of the transformation
𝑍𝑍 =
𝑋𝑋 − 𝜇𝜇
𝜎𝜎
to a normal X-distribution, such that the mean is at x = 0 and there is one standard deviation per unit
on the x-axis. Where the probability density function for normal distribution has two parameters and ,
the Z-distribution has none. This makes it useful when comparing results from two or more different
normal distributions, since comparing Z-values allows one to take into account the standard deviation
and mean when comparing results.
9
Example: Professor Einstein is marking a test.
Here are the students results (out of 60 points):
20, 15, 26, 32, 18, 28, 35, 14, 26, 22, 17
Most students didn't even get 30 out of 60, and most will fail.
The test must have been really hard, so the Prof decides to Standardize all the scores and only fail people 1
standard deviation below the mean.
The Mean is 23, and the Standard Deviation is 6.6, and these are the Standard Scores:
-0.45, -1.21, 0.45, 1.36, -0.76, 0.76, 1.82, -1.36, 0.45, -0.15, -0.91
Only 2 students will fail (the ones who scored 15 and 14 on the test)
Finding probabilities with a GDC involves using normalcdf(a,b,μ ,σ ) for lower limit a and upper limit b
(under “DISTR”). It is important to note that (𝑍𝑍 ≤ 𝑎𝑎) = 𝑃𝑃(𝑍𝑍 < 𝑎𝑎) .
QUANTILES OR k-VALUES
Consider a population of crabs where the length of a shell, X mm, is normally distributed with mean 70
mm and standard deviation 10 mm.
A biologist wants to protect the population by allowing only the largest 5% of crabs to be harvested.
He therefore asks the question:
“95% of the crabs have lengths less than what?” We need to find k such that 𝑃𝑃(𝑋𝑋 ≤ 𝑘𝑘) = 0.95 .
The number k is known as a quantile, and in this case the 95% quantile.
This is the inverse of finding probabilities, and we use the inverse normal function.
EXAMPLE: If Z is the standard normal distribution, find k such that P(Z > k) = 0.73 .
Interpret your result.
𝑃𝑃(𝑍𝑍 ≤ 𝑘𝑘) = 0.27
∴ 𝑘𝑘 ≈ −0.613
This means 73% of the Z-distribution values are
more than – 0.613 .
10
EXAMPLE: A university professor determines that no more than 80% of this year’s History candidates
should pass the final examination. The examination results were approximately normally distributed
with mean 62 and standard deviation 13. Find the lowest score necessary to pass the examination.
Let X denote the final examination result, 𝑠𝑠𝑠𝑠 𝑋𝑋 ~ 𝑁𝑁(62, 132 ).
We need to find k such that
𝑃𝑃(𝑋𝑋 ≥ 𝑘𝑘) = 08
So, the minimum pass mark is 52.
→
𝑃𝑃(𝑋𝑋 < 𝑘𝑘) = 0.2
∴ 𝑘𝑘 ≈ 51.059
THE IMPORTANCE OF QUANTILES
For some questions we must convert to z-scores in order to find the answer.
We always need to convert to z-scores if we are trying to find an unknown mean or standard
deviation.
EXAMPLE: An adult scallop population is known to be normally distributed with a standard deviation
of 5.9 g. If 15% of scallops weigh less than 58.2 g, find the mean weight of the population.
Let the mean weight of the population be µ g.
If X g denotes the weight of an adult scallop, then 𝑋𝑋 ~ 𝑁𝑁(𝜇𝜇, 5.92 ).
As we do not know µ we cannot use the invNorm directly, but we can convert to z-scores and use
the properties of N(0, 12).
𝑃𝑃(𝑋𝑋 ≤ 58.2) = 0.15
∴ 𝑃𝑃 �𝑍𝑍 ≤
58.2 − 𝜇𝜇
� = 0.15
5.9
Using invNorm for N(0, 12)
58.2 − 𝜇𝜇
≈ −1.0364
5.9
𝜇𝜇 ≈ 64.3
So, the mean weight is 64:3 g.
11
12
Binomial Distribution Problems
(1) A company owns 400 laptops. Each laptop has an 8% probability of not working. You
randomly select 20 laptops for your salespeople.
(a) What is the likelihood that 5 will be broken?
(b) What is the likelihood that they
will all work?
(c) What is the likelihood that they will all be broken?
(2) A study indicates that 4% of American teenagers have tattoos. You randomly sample
30 teenagers. What is the likelihood that exactly 3 will have a tattoo?
(3) An XYZ cell phone is made from 55 components. Each component has a .002
probability of being defective. What is the probability that an XYZ cell phone will not work
perfectly?
(4) The ABC Company manufactures toy robots. About 1 toy robot per 100 does not
work. You purchase 35 ABC toy robots. What is the probability that exactly 4 do not work?
(5) The LMB Company manufactures tires. They claim that only .007 of LMB tires are
defective. What is the probability of finding 2 defective tires in a random sample of 50 LMB
tires?
(6) An HDTV is made from 100 components. Each component has a .005 probability of
being defective. What is the probability that an HDTV will not work perfectly?
Poisson Distribution Problems
1. A life insurance salesman sells on the average 3 life insurance policies per week. Use Poisson's law
to calculate the probability that in a given week he will sell
a. Some policies
b. 2 or more policies but less than 5 policies.
c. Assuming that there are 5 working days per week, what is the probability that in a given day he
will sell one policy?
2. If electricity power failures occur according to a Poisson distribution with an average of 3 failures
every twenty weeks, calculate the probability that there will not be more than one failure during a
particular week.
3. A small life insurance company has determined that on the average it receives 6 death
claims per day. Find the probability that the company receives at least seven death claims
on a randomly selected day.
4. The number of traffic accidents that occurs on a particular stretch of road during a
month follows a Poisson distribution with a mean of 9.4. Find the probability that less than
two accidents will occur on this stretch of road during a randomly selected month.
Normal Distribution Problems
1. X is a normally normally distributed variable with mean μ = 30 and standard deviation σ
= 4. Find
a) P(x < 40)
13
b) P(x > 21)
c) P(30 < x < 35)
2. A radar unit is used to measure speeds of cars on a motorway. The speeds are normally
distributed with a mean of 90 km/hr and a standard deviation of 10 km/hr. What is the
probability that a car picked at random is travelling at more than 100 km/hr?
3. For a certain type of computers, the length of time bewteen charges of the battery is
normally distributed with a mean of 50 hours and a standard deviation of 15 hours. John
owns one of these computers and wants to know the probability that the length of time will
be between 50 and 70 hours.
4. Entry to a certain University is determined by a national test. The scores on this test are
normally distributed with a mean of 500 and a standard deviation of 100. Tom wants to be
admitted to this university and he knows that he must score better than at least 70% of the
students who took the test. Tom takes the test and scores 585. Will he be admitted to this
university?
5. The length of similar components produced by a company are approximated by a normal
distribution model with a mean of 5 cm and a standard deviation of 0.02 cm. If a
component is chosen at random
a) what is the probability that the length of this component is between 4.98 and 5.02 cm?
b) what is the probability that the length of this component is between 4.96 and 5.04 cm?
6. The length of life of an instrument produced by a machine has a normal ditribution with
a mean of 12 months and standard deviation of 2 months. Find the probability that an
instrument produced by this machine will last
a) less than 7 months.
b) between 7 and 12 months.
7. The time taken to assemble a car in a certain plant is a random variable having a normal
distribution of 20 hours and a standard deviation of 2 hours. What is the probability that a
car can be assembled at this plant in a period of time
a) less than 19.5 hours?
b) between 20 and 22 hours?
8. A large group of students took a test in Physics and the final grades have a mean of 70
and a standard deviation of 10. If we can approximate the distribution of these grades by a
normal distribution, what percent of the students
a) scored higher than 80?
b) should pass the test (grades≥60)?
c) should fail the test (grades<60)?
9. The annual salaries of employees in a large company are approximately normally
distributed with a mean of $50,000 and a standard deviation of $20,000.
a) What percent of people earn less than $40,000?
b) What percent of people earn between $45,000 and $65,000?
c) What percent of people earn more than $70,000?
14
Binomial Distribution SOLUTIONS
(1) (a)
20C
5
5 (.08)
(.92)15 = .0145 (b) 20C0 (.08)0(.92)20 = .1887
(c) 20C20 (.08)20(.92)0 = .0000000000000000000001
(2) 30C3 (.04)3 (.96)27 = .0863
(3) Probability that it will work (0 defective components) 55C0 (.002)0 (.998)55 = .896
Probability that it will not work perfectly is 1 - .896 = .104 or 10.4%
(4) 35C4 (.01)4 (.99)31 = .00038
(5) 50C2 (.007)2 (.993)48 = .0428
(6) Probability that it will work (0 defective components) 100C0 (.005)0 (.995)100 = .606
Probability that it will not work perfectly is 1 - .606 = .394 or 39.40%
Poisson Distribution SOLUTIONS
1. m = 3
(a) "Some policies" means "1 or more policies":
𝑃𝑃(𝑋𝑋 = 𝑥𝑥0 ) =
30 𝑒𝑒 −3
= 0.04.9787
0!
P(X > 0) = 1 − P(x0)
𝑃𝑃(𝑋𝑋 ≥ 1) = 0.95021
(b) The probability of selling 2 or more, but less than 5 policies is:
𝑃𝑃(2 ≤ 𝑋𝑋 < 5) =
32 𝑒𝑒 −3 33 𝑒𝑒 −3 34 𝑒𝑒 −3
+
+
= 0.61611
2!
3!
4!
(c) Average number of policies sold per day:
So on a given day, 𝑃𝑃(𝑥𝑥) =
0.61 𝑒𝑒 −0.6
1!
3
5
= 0.6
= 0.32929
2. The average number of failures per week is: m =3/20 = 0.15
"Not more than one failure" means we need to include the probabilities for "0 failures" plus "1 failure".
𝑃𝑃(𝑋𝑋 = 𝑥𝑥0 ) + 𝑃𝑃(𝑋𝑋 = 𝑥𝑥1 ) =
3.
4.
0.150 𝑒𝑒 −0.15 0.151 𝑒𝑒 −0.15 0.152 𝑒𝑒 −0.15
+
+
= 0.98981
0!
1!
2!
P(x ≥ 7) = 1 - P(x ≤ 6) = 0.393697
P(x < 2) = P(x = 0) + P(x = 1) = 0.000860
15
Normal Distribution SOLUTIONS
1.
a) For x = 40, the z-value z = (40 - 30) / 4 = 2.5
Hence P(x < 40) = P(z < 2.5) = [area to the left of 2.5] = 0.9938
b) For x = 21, z = (21 - 30) / 4 = -2.25
Hence P(x > 21) = P(z > -2.25) = [total area] - [area to the left of -2.25]
= 1 - 0.0122 = 0.9878
c) For x = 30 , z = (30 - 30) / 4 = 0 and for x = 35, z = (35 - 30) / 4 = 1.25
Hence P(30 < x < 35) = P(0 < z < 1.25) = [area to the left of z = 1.25] - [area to the left of 0]
= 0.8944 - 0.5 = 0.3944
2.
Let x be the random variable that represents the speed of cars. x has μ = 90 and σ = 10. We
have to find the probability that x is higher than 100 or P(x > 100)
For x = 100 , z = (100 - 90) / 10 = 1
P(x > 90) = P(z >, 1) = [total area] - [area to the left of z = 1]
= 1 - 0.8413 = 0.1587
The probability that a car selected at a random has a speed greater than 100 km/hr is equal to 0.1587
3.
Let x be the random variable that represents the length of time. It has a mean of 50 and a
standard deviation of 15. We have to find the probability that x is between 50 and 70 or P( 50< x < 70)
For x = 50 , z = (50 - 50) / 15 = 0
For x = 70 , z = (70 - 50) / 15 = 1.33 (rounded to 2 decimal places)
P( 50< x < 70) = P( 0< z < 1.33) = [area to the left of z = 1.33] - [area to the left of z = 0]
= 0.9082 - 0.5 = 0.4082
The probability that John's computer has a length of time between 50 and 70 hours is equal to 0.4082.
4.
Let x be the random variable that represents the scores. x is normally ditsributed with a mean
of 500 and a standard deviation of 100. The total area under the normal curve represents the total
number of students who took the test. If we multiply the values of the areas under the curve by 100,
we obtain percentages.
For x = 585 , z = (585 - 500) / 100 = 0.85
The proportion P of students who scored below 585 is given by
16
P = [area to the left of z = 0.85] = 0.8023 = 80.23%
Tom scored better than 80.23% of the students who took the test and he will be admitted to this
University.
5.
a) P(4.98 < x < 5.02) = P(-1 < z < 1)
= 0.6826
b) P(4.96 < x < 5.04) = P(-2 < z < 2)
= 0.9544
6.
a) P(x < 7) = P(z < -2.5)
= 0.0062
b) P(7 < x < 12) = P(-2.5 < z < 0)
= 0.4938
7.
a) P(x < 19.5) = P(z < -0.25)
= 0.4013
b) P(20 < x < 22) = P(0 < z < 1)
= 0.3413
8.
a) For x = 80, z = 1
Area to the right (higher than) z = 1 is equal to 0.1586 = 15.87% scored more that 80.
b) For x = 60, z = -1
Area to the right of z = -1 is equal to 0.8413 = 84.13% should pass the test.
c)100% - 84.13% = 15.87% should fail the test.
a) For x = 40000, z = -0.5
Area to the left (less than) of z = -0.5 is equal to 0.3085 = 30.85% earn less than $40,000.
b) For x = 45000 , z = -0.25 and for x = 65000, z = 0.75
Area between z = -0.25 and z = 0.75 is equal to 0.3720 = 37.20 earn between $45,000 and
$65,000.
c)For x = 70000, z = 1
Area to the right (higher) of z = 1 is equal to 0.1586 = 15.86% earn more than $70,000.