Math 3680

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Math 3680 Lecture 12 The Central Limit
Theorem for Sample Sums and Sample Averages
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• In the previous lecture, we introduced the
  central limit theorem when drawing from a box
  containing exclusively 0s and 1s.
• We now generalize this technique to other kinds
  of populations heights of college students,
  incomes, or anything else which is not
  dichotomous.

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• Linear Combinations of
• Random Variables

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• Theorem. For any random variables X and Y
  and any constants a, b, and c, we have
• E( a X b Y c ) a E( X ) b E ( Y ) c.
• Note X and Y do not have to be independent.

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• Proof. We show the discrete case the continuous
  case is complete analogous.

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• Definition Random variables X and Y are
  independent if for all constants a and b,
• P(X a and Y b) P(X a) P(Y b)
• For discrete random variables, this is the same
  as saying that for all x and all y,
• P(Xx and Yy) P(X x) P(Yy)
• Theorem. If X and Y are independent, then
• E( X Y ) E( X ) E ( Y ).

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• Proof. We show the discrete case the continuous
  case is analogous (using integrals instead of
  summations).

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• Theorem. For any random variable X and any
  constant a, we have
• Var( a X ) a2 Var( X ),
• SD( a X ) a SD( X ).
• (Remember this from earlier?)

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• Proof.

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• Theorem. For any independent random variables X
  and Y, we have
• Var( X Y ) Var( X ) Var ( Y ).
• Proof.

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• Note The assumption of independence is critical
  in the last theorem. For example, if X Y, then
• Var( X X ) Var( 2 X ) 4 Var( X )
• 
  ? Var( X ) Var( X )

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• Theorem. For any independent random variables X
  and Y and any constants a, b, and c, we have
• Var( a X b Y c ) a2 Var( X ) b2 Var ( Y
  ).

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• Example. Let X and Y be independent r.v.s
  with X Binomial(8, 0.4) and Y Binomial(8,
  0.4). Find E( X 2 ) and E( X Y ).

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• Example. Let S be the sum of 5 thrown dice.
  Find E( S ) and SD( S ).

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• The Central Limit Theorem
• (or the Law of Averages
• for the Sample Sum)

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• The normal approximation may be used for other
  random variables beside binomial ones.
• Theorem. Suppose random variables X1, X2, , Xn
  are drawn with replacement from a large
  population with mean m and standard deviation
  s. Let
• SUM X1 X2 Xn.
• Then E(SUM) n m and SD(SUM)
  (Why?)
• Furthermore, if n is large, then we may
  accurately approximate probabilities of SUM by
  converting to standard units and using the normal
  curve.

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• The normal approximation may be used for other
  random variables beside binomial ones.
• Theorem. Suppose random variables X1, X2, , Xn
  are drawn without replacement from a population
  of size N with mean m and standard deviation
  s. Let
• SUM X1 X2 Xn.
• Then E(SUM) n m and SD(SUM)
  (Why?)
• Furthermore, if n is large, then we may
  accurately approximate probabilities of SUM by
  converting to standard units and using the normal
  curve.

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• Question How large is large enough for the
  normal curve to be applicable? The answer is, It
  depends. If the box itself follows the normal
  distribution exactly, then so will SUM, no matter
  what the value of n is. However, this trivial
  case rarely happens in practice.
• For a more typical example, lets look at
• P(X 1) 1/3
• P(X 2) 1/3
• P(X 3) 1/3

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• Another example suppose
• P(X 1) 1/7
• P(X 2) 1/7
• P(X 5) 3/7
• P(X 9) 1/7
• P(X 20) 1/7

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• Example. Two hundred tickets are drawn at random
  with replacement from the following box of
  tickets
• What is the smallest possible sum? The biggest?
• What is the expected sum?
• Find the probability that the sum of the tickets
  is more than 630.

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• Example Thirty-six jets wait to take off from
  an airport. The average taxi and take-off time
  for each jet is 8.5 minutes, with an SD of 2.5
  minutes. What is the probability that the total
  taxi and take-off time for the 36 jets is less
  than 320 minutes?

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• Example. A gambler makes 1000 column bets at
  roulette. The chance of winning on any one play
  is 12/38. The gambler can either win 2 or lose
  1 on each play. Find the probability that, in
  total, the gambler wins at least 0.

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• Example. A gambler makes 10,000 column bets at
  roulette. The chance of winning on any one play
  is 12/38. The gambler can either win 2 or lose
  1 on each play. Find the probability that, in
  total, the gambler wins at least 0.

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• The Central Limit Theorem
• (or the Law of Averages)
• for the Sample Mean

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• Theorem. Suppose random variables X1, X2, , Xn
  are drawn with replacement from a large
  population with mean m and standard deviation
  s. Let
• Then E( ) m and SD( )
  (Why?)
• Furthermore, if n is large, then we may
  accurately approximate probabilities of by
  converting to standard units and using the normal
  curve. (As before, this is exact if the original
  population follows the normal curve.)

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• Theorem. Suppose random variables X1, X2, , Xn
  are drawn without replacement from a population
  of size N with mean m and standard deviation
  s. Let
• Then E( ) m and SD( )
  (Why?)
• Furthermore, if n is large, then we may
  accurately approximate probabilities of by
  converting to standard units and using the normal
  curve. (As before, this is exact if the original
  population follows the normal curve.)

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• Example The cookie machine at Chips Ahoy adds
  a random number of chips to each cookie. The
  number of chips is a random number with average
  28.5 and SD 5.3. Find the probability that, in a
  bag of 50 cookies, the average number of chips
  per cookie is at least 30.

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• Example A computer generates 100 random
  numbers between 0 and 1 (presumably evenly). Find
  the probability that the average of these numbers
  is between 0.48 and 0.49.

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• Review of Law of Averages
• Population Statistic Use the Law of
• (Parameter) Averages for
• Sample...
• K Count
• Dichotomous (p)
• (0-1 box)
  p Proportion
• SUM Sum
• Quantitative (m)
• Mean

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• COUNT (0/1 box)
• E(K) n p
• SD(K)

PROPORTION (0/1 box) E(P) p SD(P)

SUM E(SUM) n m SD(SUM)
AVERAGE E( ) m SD( )
For all four types of problems, multiply the SD
by the finite population correction factor if
drawing without replacement.