PROBABILITY DISTRIBUTIONS
Business Statistics
Probability distribution functions (discrete)
Characteristics of a discrete distribution
Example: uniform (discrete) distribution
Example: Bernoulli distribution
Example: binomial distribution
Probability density functions (continuous)
Characteristics of a continuous distribution
Example: uniform (continuous) distribution
Example: normal (or Gaussian) distribution
Example: standard normal distribution
Back to the normal distribution
Approximations to distributions
Old exam question
Further study
CONTENTS
Today we want to speed up. We will skip some slides or postpone a few. Prepare
well, we want to start the statistical topics as soon as
possible.
▪ A sample space is called discrete when its elements can be
counted
▪ We will code the elements of a discrete sample space 𝑆 as
1,2,3, … , 𝑛 or 0,1,2, … , 𝑛 − 1▪ Examples
▪ die 𝑥 ∈ 1,2,3,4,5,6 , so 𝑆 = 1,2,3,4,5,6▪ coin 𝑥 ∈ 0,1▪ number of broken TV sets 𝑥 ∈ 0,1,2,…
PROBABILITY DISTRIBUTION FUNCTIONS (DISCRETE)
Distribution function
𝑃 𝑥 = 𝑃 𝑋 = 𝑥
▪ the probability that the (discrete) random variable 𝑋assumes the value 𝑥
▪ alternative notation: 𝑃𝑋 𝑥
PROBABILITY DISTRIBUTION FUNCTIONS (DISCRETE)
Note our convention:
capital letters (𝑋) for random variables
lowercase letters (𝑥) for values
Example
▪ die: 𝑃 𝑥 =
1
6if 𝑥 = 1
1
6if 𝑥 = 2
1
6if 𝑥 = 3
1
6if 𝑥 = 4
1
6if 𝑥 = 5
1
6if 𝑥 = 6
0 otherwise
PROBABILITY DISTRIBUTION FUNCTIONS (DISCRETE)
Example: flipping a coin 3 times
▪ sample space 𝑆 = 𝐻𝐻𝐻,𝐻𝐻𝑇,𝐻𝑇𝐻, 𝑇𝐻𝐻,…▪ define the random variable 𝑋 = number of heads
▪ distribution function 𝑃 𝑥 =
1
8if 𝑥 = 0
3
8if 𝑥 = 1
3
8if 𝑥 = 2
1
8if 𝑥 = 3
0 otherwise
▪ or: 𝑃𝑋 0 =1
8, 𝑃𝑋 1 =
3
8, 𝑃𝑋 2 =
3
8, 𝑃𝑋 3 =
1
8
PROBABILITY DISTRIBUTION FUNCTIONS (DISCRETE)
▪ 𝑃 𝑥 is a (discrete) probability distribution function (pdf or
PDF)
▪ 𝑃 𝑥 = 𝑃 𝑋 = 𝑥 expresses the probability that 𝑋 = 𝑥▪ A random variable 𝑋 that is distributed with pdf 𝑃 is written
as
𝑋~𝑃
▪ Some properties of the pdf:▪ 0 ≤ 𝑃 𝑥 ≤ 1
▪ a probability is always between 0 and 1▪ σ𝑥∈𝑆𝑃 𝑥 = 1
▪ the probabilities of all elementary outcomes add up to 1
PROBABILITY DISTRIBUTION FUNCTIONS (DISCRETE)
▪ A pdf may have one or more parameters to denote a
collection of different but “similar”pdfs
▪ Example: a regular die with 𝑚 faces
▪ 𝑃 𝑋 = 𝑥;𝑚 = 𝑃𝑋 𝑥;𝑚 = 𝑃 𝑥;𝑚 =1
𝑚(for 𝑥 = 1,… ,𝑚)
▪ 𝑋~𝑃 𝑚
PROBABILITY DISTRIBUTION FUNCTIONS (DISCRETE)
𝑚 = 4 𝑚 = 6 𝑚 = 8 𝑚 = 12 𝑚 = 20
In addition to the (discrete) probability distribution function
(pdf)
▪ 𝑃 𝑋 = 𝑥 = 𝑃𝑋 𝑥 = 𝑃 𝑥we define the (discrete) cumulative distribution function (cdf or
CDF)
𝐹 𝑥 = 𝐹𝑋 𝑥 = 𝑃 𝑋 ≤ 𝑥
and therefore
𝐹 𝑥 =
𝑘=−∞
𝑥
𝑃 𝑋 = 𝑘 =
𝑘=−∞
𝑥
𝑃 𝑘
PROBABILITY DISTRIBUTION FUNCTIONS (DISCRETE)
Depending on how we
count, you may also start
at 𝑘 = 0 or 𝑘 = 1
Example
▪ die: 𝑃 𝑋 = 2 =1
6, but 𝑃 𝑋 ≤ 2 = 𝑃 𝑋 = 1 +
𝑃 𝑋 = 2 =1
3
▪ Some properties of the cdf:▪ 𝐹 −∞ = 0 and 𝐹 ∞ = 1▪ monotonously increasing
PROBABILITY DISTRIBUTION FUNCTIONS (DISCRETE)
▪ cdf
PROBABILITY DISTRIBUTION FUNCTIONS (DISCRETE)
Expected value of 𝑋
𝐸 𝑋 =
𝑖=1
𝑁
𝑥𝑖𝑃 𝑋 = 𝑥𝑖 =
𝑖=1
𝑁
𝑥𝑖𝑃 𝑥𝑖
▪ Example▪ die with 𝑃 1 = 𝑃 2 = ⋯ = 𝑃 6 =
1
6▪ 𝐸 𝑋 = 1 ×
1
6+ 2 ×
1
6+ 3 ×
1
6+ 4 ×
1
6+ 5 ×
1
6+ 6 ×
1
6=
7
2= 3
1
2▪ Interpretation: mean (average)
▪ alternative notation: 𝜇 or 𝜇𝑋▪ so 𝐸 𝑋 = 𝜇𝑋
▪ Note difference between 𝜇 and the sample mean ҧ𝑥▪ e.g., rolling a specific die 𝑛 = 100 times may return a mean ҧ𝑥 = 3.72 or
3.43▪ while 𝜇 = 7/2, always (property of die, property of “population”)
CHARACTERISTICS OF A DISCRETE DISTRIBUTION
Variance
var 𝑋 =
𝑖=1
𝑁
𝑥𝑖 − 𝐸 𝑋2𝑃 𝑥𝑖
▪ Interpretation: dispersion▪ alternative notation: 𝜎2 or 𝜎𝑋
2 or 𝑉 𝑋
▪ so var 𝑋 = 𝜎𝑋2
▪ Note difference between 𝜎2 and the sample variance 𝑠2
▪ e.g., rolling a specific die 100 times may return a variance 𝑠2 = 2.86 or 3.04
▪ while 𝜎2 =35
12, always (property of die, property of “population”)
▪ And of course: standard deviation 𝜎𝑋 = var 𝑋
CHARACTERISTICS OF A DISCRETE DISTRIBUTION
Transformation rules of random variable 𝑋 and 𝑌▪ For means:
▪ 𝐸 𝑘 + 𝑋 = 𝑘 + 𝐸 𝑋▪ 𝐸 𝑎𝑋 = 𝑎𝐸 𝑋▪ 𝐸 𝑋 + 𝑌 = 𝐸 𝑋 + 𝐸 𝑌
▪ For variances:▪ var 𝑘 + 𝑋 = var 𝑋▪ var 𝑎𝑋 = 𝑎2var 𝑋▪ if 𝑋 and 𝑌 independent (so if cov 𝑋, `𝑌 ):
▪ var 𝑋 + 𝑌 = var 𝑋 + var 𝑌▪ if 𝑋 and 𝑌 dependent:
▪ var 𝑋 + 𝑌 = var 𝑋 + 2cov 𝑋, 𝑌 + var 𝑌
CHARACTERISTICS OF A DISCRETE DISTRIBUTION
▪ Generalization of fair die:▪ equal probability of integer outcomes from 𝑎 through 𝑏
▪ conditions: 𝑎, 𝑏 ∈ ℤ, 𝑎 < 𝑏▪ zero probability elsewhere
▪ uniform discrete distribution
▪ pdf: 𝑃 𝑥; 𝑎, 𝑏 = ൝1
𝑏−𝑎+1𝑥 ∈ ℤ and 𝑥 ∈ 𝑎, 𝑏
0 otherwise▪ Examples:
▪ coin: 𝑎 = 0, 𝑏 = 1▪ die: 𝑎 = 1, 𝑏 = 6
▪ Random variable:▪ 𝑋~𝑈 𝑎, 𝑏
EXAMPLE: UNIFORM DISTRIBUTION
EXAMPLE: UNIFORM DISTRIBUTION
No need to memorize or even
discuss this sheet. Most
information is either on the
formula sheet or unimportant.
▪ Example: choose a random number from 1 through 100with equal probability and denote it by 𝑋▪ random variable: 𝑋~𝑈 1,100
▪ pdf: 𝑃 𝑥 = 𝑃 𝑋 = 𝑥 =1
100(𝑥 ∈ 1,2,… , 100 )
▪ cdf: 𝐹 𝑥 = 𝑃 𝑋 ≤ 𝑥 =𝑥
100(𝑥 ∈ 1,2,… , 100 )
▪ expected value: 𝐸 𝑋 = 501
2
▪ variance: var 𝑋 =9999
12≈ 833.25
▪ Sample (𝑛 = 1000): ▪ values (e.g.): 45, 96, 33, 7, 44, 96, 20, …▪ mean: ҧ𝑥 = 50.92 (e.g.)
▪ variance: 𝑠𝑥2 = 823.25 (e.g.)
EXAMPLE: UNIFORM DISTRIBUTION
Given are two dice, with outcomes 𝑋 and 𝑌.
a. Find 𝐸 𝑋 + 𝑌b. Find var 𝑋 + 𝑌
EXERCISE 1
▪ Bernoulli experiment▪ random experiment with 2 discrete outcomes (coin type)
▪ head, true, “success”, female: 𝑋 = 1▪ tail, false, “fail”, male: 𝑋 = 0▪ Bernoulli distribution
▪ Examples:▪ winning a price in a lottery (buying one ticket)
▪ your luggage arrives in time at a destination
▪ Probability of success is parameter 𝜋 (with 0 ≤ 𝜋 ≤ 1)▪ 𝑃 1 = 𝑃 𝑋 = 1 = 𝜋▪ 𝑃 0 = 𝑃 𝑋 = 0 = 1 − 𝜋
▪ Random variable▪ 𝑋~𝐵𝑒𝑟𝑛𝑜𝑢𝑙𝑙𝑖 𝜋 or 𝑋~𝑎𝑙𝑡 𝜋
EXAMPLE: BERNOULLI DISTRIBUTION
▪ Expected value▪ 𝐸 𝑋 = 𝜋 (obviously!)
▪ Variance▪ var 𝑋 = 𝜋 1 − 𝜋▪ variance zero when 𝜋 = 0 or 𝜋 = 1 (obviously!)
▪ variance maximal when 𝜋 = 1 − 𝜋 =1
2(obviously!)
▪ pdf: 𝑝 𝑥; 𝜋 = ቐ𝜋 if 𝑥 = 1
1 − 𝜋 if 𝑥 = 00 otherwise
▪ cdf: (not so interesting)
EXAMPLE: BERNOULLI DISTRIBUTION
▪ Repeating a Bernoulli experiment 𝑛 times▪ 𝑋 is total number of “successes”
▪ 𝑃 𝑋 = 𝑥 is probality of 𝑥 “successes” in sample
▪ 𝑋 = 𝑋1 + 𝑋2 +⋯+ 𝑋𝑛▪ where 𝑋𝑖 is the outcome of Bernoulli experiment number 𝑖 =1,2,… , 𝑛
▪ 𝑋 has a binomial distribution
EXAMPLE: BINOMIAL DISTRIBUTION
▪ Example▪ flip a coin 10 times:𝑋 is number of “heads up”
▪ roll 100 dice: 𝑋 is number of “sixes”
▪ produce 1000 TV sets: 𝑋 is number of broken sets
▪ What is important?▪ the number of repitions (𝑛)
▪ the probability of success (𝜋) per item
▪ the constancy of 𝜋▪ the independence of the “experiments”
EXAMPLE: BINOMIAL DISTRIBUTION
▪ Expected value▪ 𝐸 𝑋 = 𝑛𝜋 (obviously!)
▪ Variance▪ var 𝑋 = 𝑛𝜋 1 − 𝜋▪ minimum (0) when 𝜋 = 0 or 𝜋 = 1 (obviously!)
▪ maximum for given 𝑛 when 𝜋 = 1 − 𝜋 =1
2(obviously!)
▪ pdf:
▪ 𝑝 𝑥; 𝑛, 𝜋 =𝑛!
𝑥! 𝑛−𝑥 !𝜋𝑥 1 − 𝜋 𝑛−𝑥 (𝑥 ∈ 0,1,2,… , 𝑛 )
▪ cdf:▪ 𝐹 𝑥; 𝑛, 𝜋 = σ𝑘=0
𝑥 𝑝 𝑥; 𝑛, 𝜋
▪ Random variable:▪ 𝑋~𝑏𝑖𝑛 𝑛, 𝜋 or 𝑋~𝑏𝑖𝑛𝑜𝑚 𝑛, 𝜋
EXAMPLE: BINOMIAL DISTRIBUTION
Recall the factorial function:
5! = 5 × 4 × 3 × 2 × 1
▪ Example:▪ roll 10 dice: what is the distribution of 𝑋 = number of “sixes”?
▪ What is the probability model?▪ you repeat an experiment 10 times (𝑛 = 10)
▪ with a probability 𝜋 =1
6of success and a probability 1 − 𝜋 =
5
6of failure per
experiment
▪ What is the probability distribution?
▪ 𝑋~𝑏𝑖𝑛 10,1
6
▪ where the random variable 𝑋 represents the total number of sixes
▪ so 𝑋 is not the outcome of a roll of the die!
▪ 𝐸 𝑋 = 10 ×1
6= 1
2
3
▪ so we expect on average 12
3sixes in 10 rolls
▪ var 𝑋 = 10 ×1
6×
5
6=
25
18
EXAMPLE: BINOMIAL DISTRIBUTION
EXAMPLE: BINOMIAL DISTRIBUTION
No need to memorize or even discuss this
sheet. Most information is either on the
formula sheet or unimportant.
▪ Calculating pdf and cdf values
▪ Example: binomial distrbution with 𝑛 = 8, 𝜋 = 0.5▪ what is 𝑃 3 = 𝑃 𝑋 = 3 (pdf)?
▪ what is 𝐹 3 = 𝑃 𝑋 ≤ 3 (cdf)?
▪ Different methods:▪ using a graphical calculator (not at the exam)
▪ using the formula (see next slides)
▪ using a table (see next slides)
▪ using Excel (see the computer tutorials)
▪ using online calculators (figure out for yourself)
EXAMPLE: BINOMIAL DISTRIBUTION
▪ pdf using the formula
▪ 𝑃 3; 8,0.5 =8!
3! 8−3 !0.53 1 − 0.5 8−3 = 0.2188
▪ or
▪ 𝑃 3; 8,0.5 = 830.53 1 − 0.5 8−3 = 0.2188
▪ using the binomial coefficient 𝑛𝑘
= 𝑛𝐶𝑘 =𝑛!
𝑘! 𝑛−𝑘 !
EXAMPLE: BINOMIAL DISTRIBUTION
At the exam, you can just use the tables.
Much easier!
▪ pdf using the table in Appendix A▪ 𝑃 3; 8,0.50 = 0.2188
EXAMPLE: BINOMIAL DISTRIBUTION
▪ At the exam: non-cumulative table only
▪ Problem: how to do the cdf?
▪ Use the definition:
𝐹 𝑥 = 𝑃 𝑋 ≤ 𝑥 =
𝑘=0
𝑥
𝑃 𝑋 = 𝑘
▪ 𝑃 𝑋 ≤ 3 = 𝑃 𝑋 = 0 + 𝑃 𝑋 = 1 + 𝑃 𝑋 = 2 +𝑃 𝑋 = 3
▪ use table, four times
EXAMPLE: BINOMIAL DISTRIBUTION
▪ Example▪ 𝐹 3; 8,0.50 = 0.0039 + 0.0313 + 0.1094 + 0.2188
EXAMPLE: BINOMIAL DISTRIBUTION
Note that this table gives a
pdf, not a cdf
▪ Note that cdf is 𝐹 𝑥 = 𝑃 𝑋 ≤ 𝑥▪ How to find 𝑃 𝑋 < 𝑥 ?
▪ use 𝑃 𝑋 ≤ 𝑥 = 𝑃 𝑋 ≤ 𝑥 − 1▪ How to find 𝑃 𝑋 > 𝑥 ?
▪ use 𝑃 X > x = 1 − 𝑃 𝑋 ≤ 𝑥▪ How to find 𝑃 𝑥1 < 𝑋 < 𝑥2 ?
▪ use 𝑃 𝑥1 < 𝑋 < 𝑥2 = 𝑃 𝑋 < 𝑥2 − 𝑃 𝑋 ≤ 𝑥1▪ Etc.
EXAMPLE: BINOMIAL DISTRIBUTION
▪ Use such rules to efficiently use the (pdf) table (𝑛 = 8)▪ 𝑃 𝑋 ≤ 7 = 𝑃 0 + 𝑃 1 +⋯+ 𝑃 7
▪ Much easier:▪ 𝑃 𝑋 ≤ 7 = 1 − 𝑃 8
EXAMPLE: BINOMIAL DISTRIBUTION
Example:
▪ Context:▪ on average, 20% of the emergency room patients at Greenwood
General Hospital lack health insurance
▪ In a random sample of 4 patients, what is the probability
that at least 2 will be uninsured?
EXAMPLE: BINOMIAL DISTRIBUTION
▪ Binomial model (patient is uninsured or not, 𝜋uninsured =0.20)▪ 𝑋 is number of uninsured patients in sample
▪ 𝑃 𝑋 ≥ 2 = 𝑃 𝑋 = 2 + 𝑃 𝑋 = 3 + 𝑃 𝑋 = 4 =0.1536 + 0.0256 + 0.0016 = 0.1808
EXAMPLE: BINOMIAL DISTRIBUTION
Note that this table gives a
pdf, not a cdf
Discrete distributions▪ probability distribution function (pdf): 𝑃 𝑥 = 𝑃 𝑋 = 𝑥▪ probability of obtaining the value 𝑥
Continuous distributions▪ the probability of obtaining the value 𝑥 is 0▪ define probability density function (pdf): 𝑓 𝑥
▪ 𝑃 𝑎 ≤ 𝑋 ≤ 𝑏 = 𝑎𝑏𝑓 𝑥 𝑑𝑥
▪ probability of obtaining a value between 𝑎 and 𝑏
PROBABILITY DENSITY FUNCTION (CONTINUOUS)
Compare with the
probability distribution
function (pdf) 𝑃 𝑋 = 𝑥for the discrete case
The red curve is the pdf, 𝑓 𝑥The integral is the grey area
under the pdf
So pdf refers to two distinct but related things:▪ probability distribution function 𝑃 𝑥 (discrete case)
▪ probability density function 𝑓 𝑥 (continuous case)
Note also that the dimensions are different▪ 𝑃 is a dimensionless probability
▪ example:
▪ if 𝑋 is in kg, the discrete pdf 𝑃 𝑋 is dimensionless
▪ while the continuous pdf 𝑓 𝑥 is in 1/kg
PROBABILITY DENSITY FUNCTION (CONTINUOUS)
Because 𝑓 𝑥 𝑑𝑥 should be
dimensionless, and 𝑑𝑥 is in in kg
In addition to the probability density function ...▪ 𝑃 𝑥 = 𝑃𝑋 𝑥
... we define the cumulative distribution function (cdf or CDF)
𝐹 𝑥 = 𝑃 𝑋 ≤ 𝑥 = න
−∞
𝑥
𝑓 𝑦 𝑑𝑦
Some properties of the cdf:▪ 𝐹 −∞ = 0 and 𝐹 ∞ = 1▪ monotonously increasing
PROBABILITY DENSITY FUNCTION (CONTINUOUS)
Compare with
𝐹 𝑥 = 𝑃 𝑋 ≤ 𝑥 =
𝑘=−∞
𝑥
𝑃 𝑋 = 𝑘
for the discrete case
𝑥
𝐹 𝑥
▪ cdf
PROBABILITY DENSITY FUNCTION (CONTINUOUS)
𝑃 70 ≤ 𝑋 ≤ 75
= න
70
75
𝑓 𝑥 𝑑𝑥
𝑃 70 ≤ 𝑋 ≤ 75= 𝐹 75 − 𝐹 70
▪ Expected value
𝐸 𝑋 = න
−∞
∞
𝑥𝑓 𝑥 𝑑𝑥
▪ Example: let 𝑓 𝑥 = 1 for 𝑥 ∈ 0,1
▪ 𝐸 𝑋 = 01𝑥𝑑𝑥 = ቃ
1
2𝑥2
0
1=
1
2
▪ Interpretation: mean (average)▪ alternative notation for 𝐸 𝑋 : 𝜇 or 𝜇𝑋
CHARACTERISTICS OF A CONTINUOUS DISTRIBUTION
Compare with
𝐸 𝑋 =
𝑖=1
𝑛
𝑥𝑖𝑃 𝑥
for the discrete case
▪ Variance
var 𝑋 = න
−∞
∞
𝑥 − 𝐸 𝑋2𝑓 𝑥 𝑑𝑥
▪ Interpretation: dispersion▪ alternative notation for var 𝑋 : 𝜎2 or 𝜎𝑋
2 or V(𝑋)
CHARACTERISTICS OF A CONTINUOUS DISTRIBUTION
Compare with
var 𝑋 =
𝑖=1
𝑛
𝑥𝑖 − 𝐸 𝑋2𝑃 𝑥𝑖
for the discrete case
▪ Analogy with uniform discrete distribution▪ equal density for all outcomes between 𝑎 and 𝑏
▪ condition: 𝑎 < 𝑏▪ zero probability elsewhere
▪ uniform continuous distribution
▪ pdf: 𝑓 𝑥; 𝑎, 𝑏 = ൝1
𝑏−𝑎𝑥 ∈ 𝑎, 𝑏
0 otherwise
▪ or easier: 𝑓 𝑥; 𝑎, 𝑏 =1
𝑏−𝑎(𝑥 ∈ 𝑎, 𝑏 )
▪ Examples:▪ “standard” uniform deviate: 𝑎 = 0, 𝑏 = 1
EXAMPLE: UNIFORM (CONTINUOUS) DISTRIBUTION
Example: let 𝑋 be exam grade of randomly selected student▪ assume uniform distribution: 𝑋~𝑈 1,10▪ what is 𝑃 𝑋 ≥ 6.5 ?
Solution▪ use 𝑃 𝑋 ≥ 6.5 = 1 − 𝑃 𝑋 < 6.5 = 1 − 𝑃 𝑋 ≤ 6.5
▪ cdf: 𝑃 𝑋 ≤ 𝑥 = 𝐹 𝑥 = ∞−𝑥𝑓 𝑦 𝑑𝑦
▪ uniform continuous with 𝑎 = 1 and 𝑏 = 10
▪ pdf: 𝑓 𝑥 =1
9(𝑥 ∈ 1,10 )
▪ cdf: 𝑃 𝑋 ≤ 𝑥 = 1𝑥 1
9𝑑𝑦 =
1
9𝑥 − 1
▪ answer: 𝑃 𝑋 ≥ 6.5 = 1 −1
96.5 − 1
▪ or: area of black rectangle
EXAMPLE: UNIFORM (CONTINUOUS) DISTRIBUTION
For a continuous distribution
𝑃 𝑋 < 𝑥 = 𝑃 𝑋 ≤ 𝑥because 𝑃 𝑋 = 𝑥 = 0
1 6.5 10
1
9
𝑃 𝑋 ≥ 6.5 is the black area
▪ Expected value
▪ 𝐸 𝑋 =𝑎+𝑏
2
▪ Variance
▪ var 𝑋 =𝑏−𝑎 2
12𝑎)𝑏𝑥 −
𝑎+𝑏
2
2×
1
𝑏−𝑎𝑑𝑥 =
𝑏−𝑎 2
12)
▪ 𝑓 𝑥 =1
𝑏−𝑎
▪ cdf▪ 𝐹 𝑥 =
𝑥−𝑎
𝑏−𝑎
▪ Random variable▪ 𝑋~𝑈 𝑎, 𝑏 or 𝑋~ℎ𝑜𝑚 0, 𝜃 or 𝑋~ℎ𝑜𝑚 𝜃 etc.
EXAMPLE: UNIFORM (CONTINUOUS) DISTRIBUTION
▪ 𝑓 𝑥; 𝜇, 𝜎 =1
𝜎 2𝜋𝑒−1
2
𝑥−𝜇
𝜎
2
▪ cdf
▪ 𝐹 𝑥 = ∞−𝑥𝑓 𝑦; 𝜇, 𝜎 𝑑𝑦 =? ? ?
▪ Expected value▪ 𝐸 𝑋 = 𝜇
▪ Variance▪ var 𝑋 = 𝜎2
▪ Random variable▪ 𝑋~𝑁 𝜇, 𝜎 or 𝑋~𝑁 𝜇, 𝜎2
EXAMPLE: NORMAL (OR GAUSSIAN) DISTRIBUTION
In a concrete case indicate the
parameter’s symbol:
𝑁 12, 𝜎 = 2 or 𝑁 12, 𝜎2 = 4
Remember notation 𝜇𝑋 for expected
value and 𝜎𝑋2 for variance.
So here 𝜇𝑋 = 𝜇 and 𝜎𝑋2 = 𝜎2.
This is no coincedence!
Now, 𝜋 = 3.1415 ...
▪ Some characteristics▪ range: 𝑥 ∈ −∞,∞▪ pdf has maximum at 𝑥 = 𝜇▪ pdf is symmetric around 𝑥 = 𝜇▪ not too interesting for 𝑥 < 𝜇 − 3𝜎 and for 𝑥 > 𝜇 + 3𝜎
EXAMPLE: NORMAL (OR GAUSSIAN) DISTRIBUTION
▪ Normal distribution with 𝜇 = 0 and 𝜎 = 1▪ so a 0-parameter distribution: standard normal
▪ 𝑓 𝑥 =1
2𝜋𝑒−
1
2𝑥2
▪ cdf▪ 𝐹 𝑥 = ∞−
𝑥𝑓 𝑦 𝑑𝑦 =? ? ?= Φ 𝑥
▪ with Φ −∞ = 0, Φ ∞ = 1, Φ 0 = 0.5, 𝑑Φ
𝑑𝑥= 𝑓 𝑥
▪ Expected value▪ 𝐸 𝑋 = 0
▪ Variance▪ var 𝑋 = 1
▪ Random variable▪ 𝑋~𝑁 0,1 , we often write 𝑍~𝑁 0,1
EXAMPLE: STANDARD NORMAL DISTRIBUTION
Remember the trick:
if you don’t know
something, just give it
a name
▪ Important because any normally distributed variable can be
“standardized” to standard normal distribution
▪ Methods for determing the values of Φ 𝑥 :▪ using a graphical calculator (not at the exam)
▪ not using a formula (no formula available for Φ 𝑥 )
▪ using a table (see next slides)
▪ using Excel (see the computer tutorials)
▪ using online calculators (figure out for yourself)
EXAMPLE: STANDARD NORMAL DISTRIBUTION
▪ Calculating the value of the cdf with a table▪ 𝑃 𝑍 ≤ 1.36 = Φ 1.36▪ table C-2 (p.768): 𝑃 𝑍 ≤ 1.36 = 0.9131
EXAMPLE: STANDARD NORMAL DISTRIBUTION
Note that cdf is 𝑃 𝑍 ≤ 𝑥▪ How to find 𝑃 𝑍 < 𝑥 ?
▪ use 𝑃 𝑍 ≤ 𝑥 (why?)
▪ How to find 𝑃 𝑍 > 𝑥 ?▪ use 1 − 𝑃 𝑍 ≤ 𝑥 (why?)
▪ or use 𝑃 𝑍 > 𝑥 = 𝑃 𝑍 < −𝑥 (why?)
▪ How to find 𝑃 𝑍 ≥ 𝑥 ?▪ is easy now ...
▪ How to find 𝑃 𝑥 ≤ 𝑍 ≤ 𝑦 ?▪ use 𝑃 𝑍 ≤ 𝑦 − 𝑃 𝑍 ≤ 𝑥
▪ Etc.
EXAMPLE: STANDARD NORMAL DISTRIBUTION
= −
Scale for standard normal,
but this applies to any
continuous distribution
▪ Inverse lookup▪ 𝑃 𝑋 ≤ 𝑥 = Φ 𝑥 = 0.90▪ table C-2 (p.768): 𝑥 ≈ 1.28
EXAMPLE: STANDARD NORMAL DISTRIBUTION
No need to know this table by heart...
but two values can be convenient to know
▪ 𝑃 𝑍 ≤ 1.96 = 0.95, a 𝑧-value as large as 1.96 or
larger occurs only with 5% probability
▪ 𝑃 −1.645 ≤ 𝑍 ≤ 1.645 = 0.95, a 𝑧-value as large as
1.96 or larger or as small as −1.645 or smaller occurs
only with 5% probability
▪ so remember 1.96 and 1.645▪ (you can always look them up if you forgot or are unsure)
EXAMPLE: STANDARD NORMAL DISTRIBUTION
Note: 𝑋~𝑁 𝜇, 𝜎2 ⇔ 𝑋 − 𝜇~𝑁 0, 𝜎2 ⇔𝑋−𝜇
𝜎~𝑁 0,1
▪ Standardization
▪ 𝑥 → 𝑧 =𝑥−𝜇
𝜎and 𝑋 → 𝑍 =
𝑋−𝜇
𝜎
▪ If 𝑋~𝑁 𝜇, 𝜎2 , how to determine 𝑃 𝑋 ≤ 𝑥 ?
▪ 𝑃 𝑋 ≤ 𝑥 = 𝑃 𝑋 − 𝜇 ≤ 𝑥 − 𝜇 = 𝑃𝑋−𝜇
𝜎≤
𝑥−𝜇
𝜎= 𝑃 𝑍 ≤
𝑥−𝜇
𝜎
▪ Example▪ suppose 𝑋~𝑁 180, 𝜎2 = 25
▪ 𝑃 𝑋 ≤ 190 = 𝑃 𝑍 ≤190−180
5= 𝑃 𝑍 ≤ 2 = 0.9772
▪ 𝑃 𝑋 ≤ 𝑥 = 0.90 = 𝑃 𝑍 ≤𝑥−180
5⇒
𝑥−180
5= 1.28 ⇒ 𝑥 = 186.4
BACK TO THE NORMAL DISTRIBUTION
This is our way of doing
normalcdf and invnorm if you
don’t have a graphical calculator!
▪ What is “normal” about the normal distribution?▪ it has quite a weird pdf formula
▪ and an even weirder cdf formula
▪ But▪ it is unimodal
▪ it is symmetric
▪ very often empirical distributions “look” normal
▪ a quantity is approximately normal if it is influenced by many
additive factors, none of which is dominating
▪ several statistics (mean, sum, ...) are normally distributed
▪ You’ll learn that soon▪ when we discuss the Central Limit Theorem (CLT)
BACK TO THE NORMAL DISTRIBUTION
▪ Scaling▪ If 𝑋~𝑁 𝜇𝑋, 𝜎𝑋
2 then 𝑎𝑋 + 𝑏~𝑁 𝑎𝜇𝑋 + 𝑏, 𝑎2𝜎𝑋2
▪ Additivity▪ If 𝑋~𝑁 𝜇𝑋, 𝜎𝑋
2 and 𝑌~𝑁 𝜇𝑌, 𝜎𝑌2 and 𝑋, 𝑌 independent, then
𝑋 + 𝑌~𝑁 𝜇𝑋 + 𝜇𝑌, 𝜎𝑋2 + 𝜎𝑌
2
PROPERTIES OF THE NORMAL DISTRIBUTION
pdf of 0.825𝑋 + 11
pdf of 𝑋
Sometimes, we can approximate a “difficult” distribution by a
“simpler” one
▪ Important case: binomial normal▪ example 1: flipping a coin (𝜋 = 0.50, 𝑋 = #heads) very often
APPROXIMATIONS TO DISTRIBUTIONS
▪ But also when 𝜋 ≠ 0.50▪ example 2: flipping a biased coin (𝜋 = 0.30, 𝑋 = #heads) very
often
APPROXIMATIONS TO DISTRIBUTIONS
𝑛 = 10; 𝜋 = .30 𝑛 = 20; 𝜋 = .30 𝑛 = 40; 𝜋 = .30
▪ binomial normal▪ 𝑏𝑖𝑛 𝑛, 𝜋 𝑁 𝜇, 𝜎2
▪ using 𝜇 =? ? ? and 𝜎2 =? ? ?
We know that when 𝑋~𝑏𝑖𝑛 𝑛, 𝜋▪ 𝐸 𝑋 = 𝑛𝜋▪ var 𝑋 = 𝑛𝜋 1 − 𝜋
So, replace▪ 𝜇 = 𝑛𝜋▪ 𝜎2 = 𝑛𝜋 1 − 𝜋
So,▪ 𝑏𝑖𝑛 𝑛, 𝜋 𝑁 𝑛𝜋, 𝑛𝜋 1 − 𝜋
▪ rule: allowed when 𝑛𝜋 ≥ 5 and 𝑛 1 − 𝜋 ≥ 5
APPROXIMATIONS TO DISTRIBUTIONS
The book says ≥ 10instead of ≥ 5
▪ Example binomial normal▪ roll a die 𝑛 = 900 times
▪ study the occurrence of “sixes” (so 𝜋 =1
6)
▪ what is the probability of no more then 170 “sixes”?
▪ Exact: 𝑃𝑏𝑖𝑛 𝑛=900;𝜋=1/6 X ≤ 170 =?
▪ Two problems:▪ need to add 171 pdf-terms (𝑃 𝑋 = 0 until 𝑃 𝑋 = 170 )
▪ 900! gives an ERROR
▪ Approximation: 𝑃𝑁 𝜇=150;𝜎2=125 𝑋 ≤ 170 =
𝑃𝑍 𝑍 ≤170−150
125= Φ𝑍 1.7888 ≈ 0.9631
APPROXIMATIONS TO DISTRIBUTIONS
900 ×1
6= 150
900 ×1
6× 1 −
1
6= 125
▪ Now take 𝑋~𝑏𝑖𝑛 18,0.5▪ In a “binomial” context 𝑃 𝑋 ≤ 11 = 𝑃 𝑋 < 12▪ But in a “normal” context 𝑃 𝑋 ≤ 11 = 𝑃 𝑋 < 11
▪ So, take care about using integers
▪ Safest: go half-way: 𝑃 𝑋 ≤ 11.5 = 𝑃 𝑋 < 11.5▪ This is the continuity correction
APPROXIMATIONS TO DISTRIBUTIONS
The intuitive notion of the continuity correction▪ when approximating a discrete distribution by a continuous
distribution
APPROXIMATIONS TO DISTRIBUTIONS
𝑃𝑏𝑖𝑛 𝑋 ≤ 7 ≈ 𝑃𝑁 𝑋 ≤ 71
2𝑃𝑏𝑖𝑛 𝑋 ≥ 7 ≈ 𝑃𝑁 𝑋 ≥ 6
1
2
Improving previous result
▪ without continuity correction▪ 𝑃𝑏𝑖𝑛 𝑛=900;𝜋=1/6 X ≤ 170 = 𝑃𝑁 𝜇=150;𝜎2=125 (
)
𝑋 ≤
170 = 𝑃𝑍 𝑍 ≤170−150
125= Φ𝑍 1.788 ≈ 0.9631
▪ with continuity correction▪ 𝑃𝑏𝑖𝑛 𝑛=900;𝜋=1/6 X ≤ 170 = 𝑃𝑁 𝜇=150;𝜎2=125 (
)
𝑋 ≤
170.5 = 𝑃𝑍 𝑍 ≤170.5−150
125= Φ𝑍 1.833 ≈ 0.9664
APPROXIMATIONS TO DISTRIBUTIONS
30 June 2014, Q1d
OLD EXAM QUESTION
30 June 2014, Q1f
OLD EXAM QUESTION
Doane & Seward 5/E 6.1-6.4, 6.8, 7.1-7.5
Tutorial exercises week 1
discrete probability distributions
continuous probability distributions
expectation and variance
FURTHER STUDY