Today’s lesson

• Probability calculations with the standard normal distribution.

• Making predictions based on the specification of a normal distribution.

Chapter Ten: The Normal Distribution

• Definition of Normal Distribution

• Using tables of the standard normal distribution.

• Solving basic problems with the standard normal distribution.

• Central limit theorem for sums and averages

Definition of the normal distribution

• Familiar bell-shaped curve

• Continuous distribution, unimodal, symmetric, rapid fall-off of probability for values far from mean.

• Probability density function, φ(z)

• Cumulative distribution function, Ф(z)

Example Normal Distributions

• IQ scores are set to be normal with mean 100 and standard deviation 10 or 15, depending on the form.

• ETS examination score results are normal distributed with mean 500 and standard deviation 100.

Standard Normal Distribution

• I always use Z to denote a standard normal.

• E(Z)=0

• var(Z)=1

• Φ(z)=Pr{Z<=z}

• Appendix D gives right and two-sided tail areas of standard normal (page 549).

• I recommend using cdf tables (Ф(z)).

General Normal Distribution

• Solve problems about any normal distribution by converting to standard normal.

• STANDARDIZE the problem:

• standard units=(value-expected value)/standard deviation.

• Find probability.

Today’s Example Scenario

• The winnings W in one play of a game of a game of chance is a normally distributed random variable with expected value -$200 and standard deviation $1000.

• Advice: always sketch the distribution you are working with.

What is the probability that a gambler will win money in one

play of this game of chance?

• To win money means that W>0.

• Must find Pr{W>0}.

• Standardize both sides:

• Pr{(W-EW)/σW > (0-(-200))/1000}= Pr{Z>0.2}=1-Ф(0.2)=1-0.5793=0.4207.

• Answer is 0.4207. Does it make sense?

Prediction Intervals

• ASS-U-ME quantity to be predicted Y has a normal distribution with known mean E(Y) and known variance σ2.

• 95% prediction interval for Y is the interval between E(Y)-1.960σ and E(Y)+1.960σ.

• 99% prediction interval for Y is the interval between E(Y)-2.576σ and E(Y)+2.576σ.

Differences between Prediction Intervals and Confidence

Intervals

• Forms are very similar.

• A prediction interval contains an observable future value with specified probability. It is thus easy to know when a prediction interval is incorrect.

• A confidence interval contains an unknown parameter with specified “confidence”.

What is the 99% prediction interval for the winnings in the

next play of the game of chance?

• The left end-point is E(W)-2.576σ.

• Here, -$200-2.576(1000)=-$200-$2576.

• There is a 0.005 probability that the gambler will lose $2776 or more.

• The right end-point is E(W)+2.576σ=$2376.

• There is a 0.005 probability that the gambler will win $2376 or more.

Central Limit Theorem for Sums

• ASS-U-ME n independent identically distributed observations (usually called a random sample).

• Focus on the sum of the n observations:

• Sn=W1+…+Wn

Central Limit Theorem for Sums

• E(Sn)=nE(W)

• The “merry-go-round” principle.

• Var(Sn)=nvar(W)

• Note that sd(Sn)=n0.5sd(W)

• The distribution of Sn is asymptotically normal.

What are the expected total winnings after 400 independent plays of this game of chance?

• E(S400)=400E(W).

• E(S400)=400(-$200)=-$80000.

• Notice how quickly the losses mount.

Second standard problem

• What is the standard deviation of the total winnings after 400 independent plays of this game of chance?

Solution

• Sd(Sn)=n0.5sd(W)

• Sd(S400)=4000.5(1000)=$20,000

Third Standard Problem

• What is the symmetric 99% prediction interval for S400?

• Solution:

• Left endpoint is E(S400)-2.576sd(S400)

• This is -$80000-2.576($20000)=-$131,520.

• That is, there is a 0.005 probability that the gambler will lose $131,520 or more.

Third Standard Problem

• Right endpoint is E(S400)+2.576sd(S400)

• This is -$80000+2.576($20000)=$-28480.• That is, there is a 0.005 probability that the

gambler will lose $28,480 or less.• The answer is that the 99% prediction interval is

the interval between -$131,520 and -$28,480.• The gambler is very sure to lose a lot of money!

Fourth Standard Problem

• What is the probability that a gambler will have total winnings that are greater than zero after 400 independent plays of this game of chance?

Solution

• Standardize

• Pr{S400>0}=

• Pr{[(S400-E(S400))/sd(S400)]

• (0-(-80000))/20000=4.

• That is, =Pr{Z>4}=1-Φ(4)=0.00003.

• The gambler has almost no chance of winning money after 400 independent plays.

Discussion of previous problems

• The quantities sought are standard approaches to understanding the level of risk involved in a betting (insurance) strategy.

• Realistic problems may require more advanced mathematics or simulation techniques.

Central Limit Theorem for Averages

• ASS-U-ME n independent identically distributed observations (usually called a random sample).

• Focus on the average of the n observations:

• Mean=Sn/n=(W1+…+Wn)/n

Central Limit Theorem for Averages

• E(Mean)=E(Sn)/n=(nE(W))/n=E(W)

• The expected value of the mean is the expected value of the random variable that was sampled

• Var(Mean)=(nvar(W))/n2=var(W)/n.• Note that sd(mean)=sd(W)/n0.5

• The distribution of Sn is asymptotically normal.

Major points covered

• Definition of the normal distribution.

• Use of the normal distribution tables.

• Risk management example problems using the normal distribution.

• Central limit theorem for sums.

• Central limit theorem for averages.