#### Transcript Lecture 5

```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
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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?
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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
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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
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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
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.
```