Business Statistics CONTENTS Probability distribution functions - - PowerPoint PPT Presentation

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 𝑇 as 1,2,3, … , 𝑜 or 0,1,2, … , 𝑜 − 1 ▪ Examples

▪ die 𝑦 ∈ 1,2,3,4,5,6 , so 𝑇 = 1,2,3,4,5,6 ▪ coin 𝑦 ∈ 0,1 ▪ number of broken TV sets 𝑦 ∈ 0,1,2, …

PROBABILITY DISTRIBUTION FUNCTIONS (DISCRETE)

Distribution function 𝑄 𝑦 = 𝑄 𝑌 = 𝑦 ▪ the probability that the (discrete) random variable 𝑌 assumes the value 𝑦 ▪ alternative notation: 𝑄

𝑌 𝑦

PROBABILITY DISTRIBUTION FUNCTIONS (DISCRETE)

Note our convention: capital letters (𝑌) for random variables lowercase letters (𝑦) for values

Example ▪ die: 𝑄 𝑦 =

1 6

if 𝑦 = 1

1 6

if 𝑦 = 2

1 6

if 𝑦 = 3

1 6

if 𝑦 = 4

1 6

if 𝑦 = 5

1 6

if 𝑦 = 6

• therwise

PROBABILITY DISTRIBUTION FUNCTIONS (DISCRETE)

Example: flipping a coin 3 times ▪ sample space 𝑇 = 𝐼𝐼𝐼, 𝐼𝐼𝑈, 𝐼𝑈𝐼, 𝑈𝐼𝐼, … ▪ define the random variable 𝑌 = number of heads ▪ distribution function 𝑄 𝑦 =

1 8

if 𝑦 = 0

3 8

if 𝑦 = 1

3 8

if 𝑦 = 2

1 8

if 𝑦 = 3

• therwise

▪ or: 𝑄

𝑌 0 = 1 8 , 𝑄 𝑌 1 = 3 8 , 𝑄 𝑌 2 = 3 8 , 𝑄 𝑌 3 = 1 8

PROBABILITY DISTRIBUTION FUNCTIONS (DISCRETE)

▪ 𝑄 𝑦 is a (discrete) probability distribution function (pdf or PDF) ▪ 𝑄 𝑦 = 𝑄 𝑌 = 𝑦 expresses the probability that 𝑌 = 𝑦 ▪ A random variable 𝑌 that is distributed with pdf 𝑄 is written as 𝑌~𝑄 ▪ Some properties of the pdf:

▪ 0 ≤ 𝑄 𝑦 ≤ 1 ▪ a probability is always between 0 and 1 ▪ σ𝑦∈𝑇 𝑄 𝑦 = 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 𝑛 faces

▪ 𝑄 𝑌 = 𝑦; 𝑛 = 𝑄

𝑌 𝑦; 𝑛 = 𝑄 𝑦; 𝑛 = 1 𝑛 (for 𝑦 = 1, … , 𝑛)

▪ 𝑌~𝑄 𝑛

PROBABILITY DISTRIBUTION FUNCTIONS (DISCRETE)

𝑛 = 4 𝑛 = 6 𝑛 = 8 𝑛 = 12 𝑛 = 20

In addition to the (discrete) probability distribution function (pdf) ▪ 𝑄 𝑌 = 𝑦 = 𝑄

𝑌 𝑦 = 𝑄 𝑦

we define the (discrete) cumulative distribution function (cdf or CDF) 𝐺 𝑦 = 𝐺

𝑌 𝑦 = 𝑄 𝑌 ≤ 𝑦

and therefore 𝐺 𝑦 = ෍

𝑙=−∞ 𝑦

𝑄 𝑌 = 𝑙 = ෍

𝑙=−∞ 𝑦

𝑄 𝑙 PROBABILITY DISTRIBUTION FUNCTIONS (DISCRETE)

Depending on how we count, you may also start at 𝑙 = 0 or 𝑙 = 1

Example ▪ die: 𝑄 𝑌 = 2 =

1 6, but 𝑄 𝑌 ≤ 2 = 𝑄 𝑌 = 1 +

𝑄 𝑌 = 2 =

1 3

▪ Some properties of the cdf:

▪ 𝐺 −∞ = 0 and 𝐺 ∞ = 1 ▪ monotonously increasing

PROBABILITY DISTRIBUTION FUNCTIONS (DISCRETE)

▪ pdf ▪ cdf PROBABILITY DISTRIBUTION FUNCTIONS (DISCRETE)

Expected value of 𝑌 𝐹 𝑌 = ෍

𝑗=1 𝑂

𝑦𝑗𝑄 𝑌 = 𝑦𝑗 = ෍

𝑗=1 𝑂

𝑦𝑗𝑄 𝑦𝑗 ▪ Example

▪ die with 𝑄 1 = 𝑄 2 = ⋯ = 𝑄 6 =

1 6

▪ 𝐹 𝑌 = 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: 𝜈 or 𝜈𝑌 ▪ so 𝐹 𝑌 = 𝜈𝑌

▪ Note difference between 𝜈 and the sample mean ҧ 𝑦

▪ e.g., rolling a specific die 𝑜 = 100 times may return a mean ҧ 𝑦 = 3.72 or 3.43 ▪ while 𝜈 = 7/2, always (property of die, property of “population”)

CHARACTERISTICS OF A DISCRETE DISTRIBUTION

Variance var 𝑌 = ෍

𝑗=1 𝑂

𝑦𝑗 − 𝐹 𝑌

2𝑄 𝑦𝑗

▪ Interpretation: dispersion

▪ alternative notation: 𝜏2 or 𝜏𝑌

2 or 𝑊 𝑌

▪ so var 𝑌 = 𝜏𝑌

2

▪ Note difference between 𝜏2 and the sample variance 𝑡2

▪ e.g., rolling a specific die 100 times may return a variance 𝑡2 = 2.86 or 3.04 ▪ while 𝜏2 =

35 12, always (property of die, property of “population”)

▪ And of course: standard deviation 𝜏𝑌 = var 𝑌

CHARACTERISTICS OF A DISCRETE DISTRIBUTION

Transformation rules of random variable 𝑌 and 𝑍 ▪ For means:

▪ 𝐹 𝑙 + 𝑌 = 𝑙 + 𝐹 𝑌 ▪ 𝐹 𝑏𝑌 = 𝑏𝐹 𝑌 ▪ 𝐹 𝑌 + 𝑍 = 𝐹 𝑌 + 𝐹 𝑍

▪ For variances:

▪ var 𝑙 + 𝑌 = var 𝑌 ▪ var 𝑏𝑌 = 𝑏2var 𝑌 ▪ if 𝑌 and 𝑍 independent (so if cov 𝑌, `𝑍 ): ▪ var 𝑌 + 𝑍 = var 𝑌 + var 𝑍 ▪ if 𝑌 and 𝑍 dependent: ▪ var 𝑌 + 𝑍 = var 𝑌 + 2cov 𝑌, 𝑍 + var 𝑍

CHARACTERISTICS OF A DISCRETE DISTRIBUTION

▪ Generalization of fair die:

▪ equal probability of integer outcomes from 𝑏 through 𝑐 ▪ conditions: 𝑏, 𝑐 ∈ ℤ, 𝑏 < 𝑐 ▪ zero probability elsewhere ▪ uniform discrete distribution

▪ pdf: 𝑄 𝑦; 𝑏, 𝑐 = ൝

1 𝑐−𝑏+1

𝑦 ∈ ℤ and 𝑦 ∈ 𝑏, 𝑐

• therwise

▪ Examples:

▪ coin: 𝑏 = 0, 𝑐 = 1 ▪ die: 𝑏 = 1, 𝑐 = 6

▪ Random variable:

▪ 𝑌~𝑉 𝑏, 𝑐

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 100 with equal probability and denote it by 𝑌

▪ random variable: 𝑌~𝑉 1,100 ▪ pdf: 𝑄 𝑦 = 𝑄 𝑌 = 𝑦 =

1 100 (𝑦 ∈ 1,2, … , 100 )

▪ cdf: 𝐺 𝑦 = 𝑄 𝑌 ≤ 𝑦 =

𝑦 100 (𝑦 ∈ 1,2, … , 100 )

▪ expected value: 𝐹 𝑌 = 50

1 2

▪ variance: var 𝑌 =

9999 12 ≈ 833.25

▪ Sample (𝑜 = 1000):

▪ values (e.g.): 45, 96, 33, 7, 44, 96, 20, … ▪ mean: ҧ 𝑦 = 50.92 (e.g.) ▪ variance: 𝑡𝑦

2 = 823.25 (e.g.)

EXAMPLE: UNIFORM DISTRIBUTION

Given are two dice, with outcomes 𝑌 and 𝑍.

• a. Find 𝐹 𝑌 + 𝑍
• b. Find var 𝑌 + 𝑍

EXERCISE 1

▪ Bernoulli experiment

▪ random experiment with 2 discrete outcomes (coin type) ▪ head, true, “success”, female: 𝑌 = 1 ▪ tail, false, “fail”, male: 𝑌 = 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 𝜌 (with 0 ≤ 𝜌 ≤ 1)

▪ 𝑄 1 = 𝑄 𝑌 = 1 = 𝜌 ▪ 𝑄 0 = 𝑄 𝑌 = 0 = 1 − 𝜌

▪ Random variable

▪ 𝑌~𝐶𝑓𝑠𝑜𝑝𝑣𝑚𝑚𝑗 𝜌 or 𝑌~𝑏𝑚𝑢 𝜌

EXAMPLE: BERNOULLI DISTRIBUTION

▪ Expected value

▪ 𝐹 𝑌 = 𝜌 (obviously!)

▪ Variance

▪ var 𝑌 = 𝜌 1 − 𝜌 ▪ variance zero when 𝜌 = 0 or 𝜌 = 1 (obviously!) ▪ variance maximal when 𝜌 = 1 − 𝜌 =

1 2 (obviously!)

▪ pdf: 𝑞 𝑦; 𝜌 = ቐ 𝜌 if 𝑦 = 1 1 − 𝜌 if 𝑦 = 0

• therwise

▪ cdf: (not so interesting) EXAMPLE: BERNOULLI DISTRIBUTION

▪ Repeating a Bernoulli experiment 𝑜 times

▪ 𝑌 is total number of “successes” ▪ 𝑄 𝑌 = 𝑦 is probality of 𝑦 “successes” in sample ▪ 𝑌 = 𝑌1 + 𝑌2 + ⋯ + 𝑌𝑜 ▪ where 𝑌𝑗 is the outcome of Bernoulli experiment number 𝑗 = 1,2, … , 𝑜 ▪ 𝑌 has a binomial distribution

EXAMPLE: BINOMIAL DISTRIBUTION

▪ Example

▪ flip a coin 10 times:𝑌 is number of “heads up” ▪ roll 100 dice: 𝑌 is number of “sixes” ▪ produce 1000 TV sets: 𝑌 is number of broken sets

▪ What is important?

▪ the number of repitions (𝑜) ▪ the probability of success (𝜌) per item ▪ the constancy of 𝜌 ▪ the independence of the “experiments”

EXAMPLE: BINOMIAL DISTRIBUTION

▪ Expected value

▪ 𝐹 𝑌 = 𝑜𝜌 (obviously!)

▪ Variance

▪ var 𝑌 = 𝑜𝜌 1 − 𝜌 ▪ minimum (0) when 𝜌 = 0 or 𝜌 = 1 (obviously!) ▪ maximum for given 𝑜 when 𝜌 = 1 − 𝜌 =

1 2 (obviously!)

▪ pdf:

▪ 𝑞 𝑦; 𝑜, 𝜌 =

𝑜! 𝑦! 𝑜−𝑦 ! 𝜌𝑦 1 − 𝜌 𝑜−𝑦

(𝑦 ∈ 0,1,2, … , 𝑜 )

▪ cdf:

▪ 𝐺 𝑦; 𝑜, 𝜌 = σ𝑙=0

𝑦

𝑞 𝑦; 𝑜, 𝜌

▪ Random variable:

▪ 𝑌~𝑐𝑗𝑜 𝑜, 𝜌 or 𝑌~𝑐𝑗𝑜𝑝𝑛 𝑜, 𝜌

EXAMPLE: BINOMIAL DISTRIBUTION

Recall the factorial function: 5! = 5 × 4 × 3 × 2 × 1

▪ Example:

▪ roll 10 dice: what is the distribution of 𝑌 = number of “sixes”?

▪ What is the probability model?

▪ you repeat an experiment 10 times (𝑜 = 10) ▪ with a probability 𝜌 =

1 6 of success and a probability 1 − 𝜌 = 5 6 of failure per

experiment

▪ What is the probability distribution?

▪ 𝑌~𝑐𝑗𝑜 10,

1 6

▪ where the random variable 𝑌 represents the total number of sixes ▪ so 𝑌 is not the outcome of a roll of the die!

▪ 𝐹 𝑌 = 10 ×

1 6 = 1 2 3

▪ so we expect on average 1

2 3 sixes in 10 rolls

▪ var 𝑌 = 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 𝑜 = 8, 𝜌 = 0.5

▪ what is 𝑄 3 = 𝑄 𝑌 = 3 (pdf)? ▪ what is 𝐺 3 = 𝑄 𝑌 ≤ 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

▪ 𝑄 3; 8,0.5 =

8! 3! 8−3 ! 0.53 1 − 0.5 8−3 = 0.2188

▪ or ▪ 𝑄 3; 8,0.5 =

8 3 0.53 1 − 0.5 8−3 = 0.2188

▪ using the binomial coefficient 𝑜

𝑙 = 𝑜𝐷𝑙 = 𝑜! 𝑙! 𝑜−𝑙 !

EXAMPLE: BINOMIAL DISTRIBUTION

At the exam, you can just use the tables. Much easier!

▪ pdf using the table in Appendix A

▪ 𝑄 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: 𝐺 𝑦 = 𝑄 𝑌 ≤ 𝑦 = ෍

𝑙=0 𝑦

𝑄 𝑌 = 𝑙

▪ 𝑄 𝑌 ≤ 3 = 𝑄 𝑌 = 0 + 𝑄 𝑌 = 1 + 𝑄 𝑌 = 2 + 𝑄 𝑌 = 3 ▪ use table, four times

EXAMPLE: BINOMIAL DISTRIBUTION

▪ Example

▪ 𝐺 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 𝐺 𝑦 = 𝑄 𝑌 ≤ 𝑦

▪ How to find 𝑄 𝑌 < 𝑦 ? ▪ use 𝑄 𝑌 ≤ 𝑦 = 𝑄 𝑌 ≤ 𝑦 − 1 ▪ How to find 𝑄 𝑌 > 𝑦 ? ▪ use 𝑄 X > x = 1 − 𝑄 𝑌 ≤ 𝑦 ▪ How to find 𝑄 𝑦1 < 𝑌 < 𝑦2 ? ▪ use 𝑄 𝑦1 < 𝑌 < 𝑦2 = 𝑄 𝑌 < 𝑦2 − 𝑄 𝑌 ≤ 𝑦1 ▪ Etc.

EXAMPLE: BINOMIAL DISTRIBUTION

▪ Use such rules to efficiently use the (pdf) table (𝑜 = 8)

▪ 𝑄 𝑌 ≤ 7 = 𝑄 0 + 𝑄 1 + ⋯ + 𝑄 7

▪ Much easier:

▪ 𝑄 𝑌 ≤ 7 = 1 − 𝑄 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, 𝜌uninsured = 0.20)

▪ 𝑌 is number of uninsured patients in sample ▪ 𝑄 𝑌 ≥ 2 = 𝑄 𝑌 = 2 + 𝑄 𝑌 = 3 + 𝑄 𝑌 = 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): 𝑄 𝑦 = 𝑄 𝑌 = 𝑦 ▪ probability of obtaining the value 𝑦

Continuous distributions

▪ the probability of obtaining the value 𝑦 is 0 ▪ define probability density function (pdf): 𝑔 𝑦 ▪ 𝑄 𝑏 ≤ 𝑌 ≤ 𝑐 = ׬

𝑏 𝑐 𝑔 𝑦 𝑒𝑦

▪ probability of obtaining a value between 𝑏 and 𝑐

PROBABILITY DENSITY FUNCTION (CONTINUOUS)

Compare with the probability distribution function (pdf) 𝑄 𝑌 = 𝑦 for the discrete case The red curve is the pdf, 𝑔 𝑦 The integral is the grey area under the pdf

So pdf refers to two distinct but related things:

▪ probability distribution function 𝑄 𝑦 (discrete case) ▪ probability density function 𝑔 𝑦 (continuous case)

Note also that the dimensions are different

▪ 𝑄 is a dimensionless probability ▪ example: ▪ if 𝑌 is in kg, the discrete pdf 𝑄 𝑌 is dimensionless ▪ while the continuous pdf 𝑔 𝑦 is in 1/kg

PROBABILITY DENSITY FUNCTION (CONTINUOUS)

Because ׬ 𝑔 𝑦 𝑒𝑦 should be dimensionless, and 𝑒𝑦 is in in kg

In addition to the probability density function ...

▪ 𝑄 𝑦 = 𝑄

𝑌 𝑦

... we define the cumulative distribution function (cdf or CDF) 𝐺 𝑦 = 𝑄 𝑌 ≤ 𝑦 = න

−∞ 𝑦

𝑔 𝑧 𝑒𝑧 Some properties of the cdf:

▪ 𝐺 −∞ = 0 and 𝐺 ∞ = 1 ▪ monotonously increasing

PROBABILITY DENSITY FUNCTION (CONTINUOUS)

Compare with 𝐺 𝑦 = 𝑄 𝑌 ≤ 𝑦 = ෍

𝑙=−∞ 𝑦

𝑄 𝑌 = 𝑙 for the discrete case 𝑦 𝐺 𝑦

▪ pdf ▪ cdf PROBABILITY DENSITY FUNCTION (CONTINUOUS)

𝑄 70 ≤ 𝑌 ≤ 75 = න

70 75

𝑔 𝑦 𝑒𝑦 𝑄 70 ≤ 𝑌 ≤ 75 = 𝐺 75 − 𝐺 70

▪ Expected value 𝐹 𝑌 = න

−∞ ∞

𝑦𝑔 𝑦 𝑒𝑦 ▪ Example: let 𝑔 𝑦 = 1 for 𝑦 ∈ 0,1

▪ 𝐹 𝑌 = ׬

1 𝑦𝑒𝑦 =

ቃ

1 2 𝑦2 1

=

1 2

▪ Interpretation: mean (average)

▪ alternative notation for 𝐹 𝑌 : 𝜈 or 𝜈𝑌

CHARACTERISTICS OF A CONTINUOUS DISTRIBUTION

Compare with 𝐹 𝑌 = ෍

𝑗=1 𝑜

𝑦𝑗𝑄 𝑦 for the discrete case

▪ Variance var 𝑌 = න

−∞ ∞

𝑦 − 𝐹 𝑌

2𝑔 𝑦 𝑒𝑦

▪ Interpretation: dispersion

▪ alternative notation for var 𝑌 : 𝜏2 or 𝜏𝑌

2 or V(𝑌)

CHARACTERISTICS OF A CONTINUOUS DISTRIBUTION

Compare with var 𝑌 = ෍

𝑗=1 𝑜

𝑦𝑗 − 𝐹 𝑌

2𝑄 𝑦𝑗

for the discrete case

▪ Analogy with uniform discrete distribution

▪ equal density for all outcomes between 𝑏 and 𝑐 ▪ condition: 𝑏 < 𝑐 ▪ zero probability elsewhere ▪ uniform continuous distribution

▪ pdf: 𝑔 𝑦; 𝑏, 𝑐 = ൝

1 𝑐−𝑏

𝑦 ∈ 𝑏, 𝑐

• therwise

▪ or easier: 𝑔 𝑦; 𝑏, 𝑐 =

1 𝑐−𝑏

(𝑦 ∈ 𝑏, 𝑐 ) ▪ Examples:

▪ “standard” uniform deviate: 𝑏 = 0, 𝑐 = 1

EXAMPLE: UNIFORM (CONTINUOUS) DISTRIBUTION

Example: let 𝑌 be exam grade of randomly selected student

▪ assume uniform distribution: 𝑌~𝑉 1,10 ▪ what is 𝑄 𝑌 ≥ 6.5 ?

Solution

▪ use 𝑄 𝑌 ≥ 6.5 = 1 − 𝑄 𝑌 < 6.5 = 1 − 𝑄 𝑌 ≤ 6.5 ▪ cdf: 𝑄 𝑌 ≤ 𝑦 = 𝐺 𝑦 = ׬

−∞ 𝑦 𝑔 𝑧 𝑒𝑧

▪ uniform continuous with 𝑏 = 1 and 𝑐 = 10 ▪ pdf: 𝑔 𝑦 =

1 9

(𝑦 ∈ 1,10 ) ▪ cdf: 𝑄 𝑌 ≤ 𝑦 = ׬

1 𝑦 1 9 𝑒𝑧 = 1 9 𝑦 − 1

▪ answer: 𝑄 𝑌 ≥ 6.5 = 1 −

1 9 6.5 − 1

▪ or: area of black rectangle

EXAMPLE: UNIFORM (CONTINUOUS) DISTRIBUTION

For a continuous distribution 𝑄 𝑌 < 𝑦 = 𝑄 𝑌 ≤ 𝑦 because 𝑄 𝑌 = 𝑦 = 0 1 6.5 10 1 9 𝑄 𝑌 ≥ 6.5 is the black area

▪ Expected value

▪ 𝐹 𝑌 =

𝑏+𝑐 2

▪ Variance

▪ var 𝑌 =

𝑐−𝑏 2 12

(׬

𝑏 𝑐 𝑦 − 𝑏+𝑐 2 2

×

1 𝑐−𝑏 𝑒𝑦 = 𝑐−𝑏 2 12

)

▪ pdf

▪ 𝑔 𝑦 =

1 𝑐−𝑏

▪ cdf

▪ 𝐺 𝑦 =

𝑦−𝑏 𝑐−𝑏

▪ Random variable

▪ 𝑌~𝑉 𝑏, 𝑐 or 𝑌~ℎ𝑝𝑛 0, 𝜄 or 𝑌~ℎ𝑝𝑛 𝜄 etc.

EXAMPLE: UNIFORM (CONTINUOUS) DISTRIBUTION

▪ pdf

▪ 𝑔 𝑦; 𝜈, 𝜏 =

1 𝜏 2𝜌 𝑓−1

2 𝑦−𝜈 𝜏 2

▪ cdf

▪ 𝐺 𝑦 = ׬

−∞ 𝑦 𝑔 𝑧; 𝜈, 𝜏 𝑒𝑧 =? ? ?

▪ Expected value

▪ 𝐹 𝑌 = 𝜈

▪ Variance

▪ var 𝑌 = 𝜏2

▪ Random variable

▪ 𝑌~𝑂 𝜈, 𝜏 or 𝑌~𝑂 𝜈, 𝜏2

EXAMPLE: NORMAL (OR GAUSSIAN) DISTRIBUTION

In a concrete case indicate the parameter’s symbol: 𝑂 12, 𝜏 = 2 or 𝑂 12, 𝜏2 = 4 Remember notation 𝜈𝑌 for expected value and 𝜏𝑌

2 for variance.

So here 𝜈𝑌 = 𝜈 and 𝜏𝑌

2 = 𝜏2.

This is no coincedence! Now, 𝜌 = 3.1415 ...

▪ Some characteristics

▪ range: 𝑦 ∈ −∞, ∞ ▪ pdf has maximum at 𝑦 = 𝜈 ▪ pdf is symmetric around 𝑦 = 𝜈 ▪ not too interesting for 𝑦 < 𝜈 − 3𝜏 and for 𝑦 > 𝜈 + 3𝜏

EXAMPLE: NORMAL (OR GAUSSIAN) DISTRIBUTION

▪ Normal distribution with 𝜈 = 0 and 𝜏 = 1

▪ so a 0-parameter distribution: standard normal

▪ pdf

▪ 𝑔 𝑦 =

1 2𝜌 𝑓−1

2𝑦2

▪ cdf

▪ 𝐺 𝑦 = ׬

−∞ 𝑦 𝑔 𝑧 𝑒𝑧 =? ? ? = Φ 𝑦

▪ with Φ −∞ = 0, Φ ∞ = 1, Φ 0 = 0.5,

𝑒Φ 𝑒𝑦 = 𝑔 𝑦

▪ Expected value

▪ 𝐹 𝑌 = 0

▪ Variance

▪ var 𝑌 = 1

▪ Random variable

▪ 𝑌~𝑂 0,1 , we often write 𝑎~𝑂 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 𝑦 :

▪ using a graphical calculator (not at the exam) ▪ not using a formula (no formula available for Φ 𝑦 ) ▪ 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

▪ 𝑄 𝑎 ≤ 1.36 = Φ 1.36 ▪ table C-2 (p.768): 𝑄 𝑎 ≤ 1.36 = 0.9131

EXAMPLE: STANDARD NORMAL DISTRIBUTION

Note that cdf is 𝑄 𝑎 ≤ 𝑦 ▪ How to find 𝑄 𝑎 < 𝑦 ?

▪ use 𝑄 𝑎 ≤ 𝑦 (why?)

▪ How to find 𝑄 𝑎 > 𝑦 ?

▪ use 1 − 𝑄 𝑎 ≤ 𝑦 (why?) ▪ or use 𝑄 𝑎 > 𝑦 = 𝑄 𝑎 < −𝑦 (why?)

▪ How to find 𝑄 𝑎 ≥ 𝑦 ?

▪ is easy now ...

▪ How to find 𝑄 𝑦 ≤ 𝑎 ≤ 𝑧 ?

▪ use 𝑄 𝑎 ≤ 𝑧 − 𝑄 𝑎 ≤ 𝑦

▪ Etc. EXAMPLE: STANDARD NORMAL DISTRIBUTION

= − Scale for standard normal, but this applies to any continuous distribution

▪ Inverse lookup

▪ 𝑄 𝑌 ≤ 𝑦 = Φ 𝑦 = 0.90 ▪ table C-2 (p.768): 𝑦 ≈ 1.28

EXAMPLE: STANDARD NORMAL DISTRIBUTION

No need to know this table by heart... but two values can be convenient to know ▪ 𝑄 𝑎 ≤ 1.96 = 0.95, a 𝑨-value as large as 1.96 or larger occurs only with 5% probability ▪ 𝑄 −1.645 ≤ 𝑎 ≤ 1.645 = 0.95, a 𝑨-value as large as 1.96 or larger or as small as −1.645 or smaller occurs

• nly 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: 𝑌~𝑂 𝜈, 𝜏2 ⇔ 𝑌 − 𝜈~𝑂 0, 𝜏2 ⇔

𝑌−𝜈 𝜏 ~𝑂 0,1

▪ Standardization

▪ 𝑦 → 𝑨 =

𝑦−𝜈 𝜏 and 𝑌 → 𝑎 = 𝑌−𝜈 𝜏

▪ If 𝑌~𝑂 𝜈, 𝜏2 , how to determine 𝑄 𝑌 ≤ 𝑦 ?

▪ 𝑄 𝑌 ≤ 𝑦 = 𝑄 𝑌 − 𝜈 ≤ 𝑦 − 𝜈 = 𝑄

𝑌−𝜈 𝜏

≤

𝑦−𝜈 𝜏

= 𝑄 𝑎 ≤

𝑦−𝜈 𝜏

▪ Example

▪ suppose 𝑌~𝑂 180, 𝜏2 = 25 ▪ 𝑄 𝑌 ≤ 190 = 𝑄 𝑎 ≤

190−180 5

= 𝑄 𝑎 ≤ 2 = 0.9772 ▪ 𝑄 𝑌 ≤ 𝑦 = 0.90 = 𝑄 𝑎 ≤

𝑦−180 5

⇒

𝑦−180 5

= 1.28 ⇒ 𝑦 = 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 𝑌~𝑂 𝜈𝑌, 𝜏𝑌

2 then 𝑏𝑌 + 𝑐~𝑂 𝑏𝜈𝑌 + 𝑐, 𝑏2𝜏𝑌 2

▪ Additivity

▪ If 𝑌~𝑂 𝜈𝑌, 𝜏𝑌

2 and 𝑍~𝑂 𝜈𝑍, 𝜏𝑍 2 and 𝑌, 𝑍 independent, then

𝑌 + 𝑍~𝑂 𝜈𝑌 + 𝜈𝑍, 𝜏𝑌

2 + 𝜏𝑍 2

PROPERTIES OF THE NORMAL DISTRIBUTION

pdf of 0.825𝑌 + 11 pdf of 𝑌

Sometimes, we can approximate a “difficult” distribution by a “simpler” one ▪ Important case: binomial  normal

▪ example 1: flipping a coin (𝜌 = 0.50, 𝑌 = #heads) very often

APPROXIMATIONS TO DISTRIBUTIONS

▪ But also when 𝜌 ≠ 0.50

▪ example 2: flipping a biased coin (𝜌 = 0.30, 𝑌 = #heads) very

• ften

APPROXIMATIONS TO DISTRIBUTIONS

𝑜 = 10; 𝜌 = .30 𝑜 = 20; 𝜌 = .30 𝑜 = 40; 𝜌 = .30

▪ binomial  normal

▪ 𝑐𝑗𝑜 𝑜, 𝜌  𝑂 𝜈, 𝜏2 ▪ using 𝜈 =? ? ? and 𝜏2 =? ? ?

We know that when 𝑌~𝑐𝑗𝑜 𝑜, 𝜌

▪ 𝐹 𝑌 = 𝑜𝜌 ▪ var 𝑌 = 𝑜𝜌 1 − 𝜌

So, replace

▪ 𝜈 = 𝑜𝜌 ▪ 𝜏2 = 𝑜𝜌 1 − 𝜌

So,

▪ 𝑐𝑗𝑜 𝑜, 𝜌  𝑂 𝑜𝜌, 𝑜𝜌 1 − 𝜌 ▪ rule: allowed when 𝑜𝜌 ≥ 5 and 𝑜 1 − 𝜌 ≥ 5

APPROXIMATIONS TO DISTRIBUTIONS

The book says ≥ 10 instead of ≥ 5

▪ Example binomial  normal

▪ roll a die 𝑜 = 900 times ▪ study the occurrence of “sixes” (so 𝜌 =

1 6)

▪ what is the probability of no more then 170 “sixes”?

▪ Exact: 𝑄𝑐𝑗𝑜 𝑜=900;𝜌=1/6 X ≤ 170 =? ▪ Two problems:

▪ need to add 171 pdf-terms (𝑄 𝑌 = 0 until 𝑄 𝑌 = 170 ) ▪ 900! gives an ERROR

▪ Approximation: 𝑄𝑂 𝜈=150;𝜏2=125 𝑌 ≤ 170 = 𝑄𝑎 𝑎 ≤

170−150 125

= Φ𝑎 1.7888 ≈ 0.9631 APPROXIMATIONS TO DISTRIBUTIONS

900 × 1 6 = 150 900 × 1 6 × 1 − 1 6 = 125

▪ Now take 𝑌~𝑐𝑗𝑜 18,0.5

▪ In a “binomial” context 𝑄 𝑌 ≤ 11 = 𝑄 𝑌 < 12 ▪ But in a “normal” context 𝑄 𝑌 ≤ 11 = 𝑄 𝑌 < 11

▪ So, take care about using integers ▪ Safest: go half-way: 𝑄 𝑌 ≤ 11.5 = 𝑄 𝑌 < 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

𝑄𝑐𝑗𝑜 𝑌 ≤ 7 ≈ 𝑄𝑂 𝑌 ≤ 7 1 2 𝑄𝑐𝑗𝑜 𝑌 ≥ 7 ≈ 𝑄𝑂 𝑌 ≥ 6 1 2

Improving previous result ▪ without continuity correction

▪ 𝑄𝑐𝑗𝑜 𝑜=900;𝜌=1/6 X ≤ 170 = 𝑄𝑂 𝜈=150;𝜏2=125 ( ) 𝑌 ≤ 170 = 𝑄𝑎 𝑎 ≤

170−150 125

= Φ𝑎 1.788 ≈ 0.9631

▪ with continuity correction

▪ 𝑄𝑐𝑗𝑜 𝑜=900;𝜌=1/6 X ≤ 170 = 𝑄𝑂 𝜈=150;𝜏2=125 ( ) 𝑌 ≤ 170.5 = 𝑄𝑎 𝑎 ≤

170.5−150 125

= Φ𝑎 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