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Math 1710 Class 11 V1cu Normal Approximation Making Math 1710 Class 11 Normal Approximation Accurate Prop CLT Dr. Allen Back Failure Picture A Random Variable Problem Sep. 19, 2016 CLT Suppose 70% approve the President . . . Math

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Math 1710 Class 11 V1cu Normal Approximation Making Normal Approximation Accurate Prop CLT Failure Picture A Random Variable Problem CLT

Math 1710 Class 11

  • Dr. Allen Back
  • Sep. 19, 2016
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Math 1710 Class 11 V1cu Normal Approximation Making Normal Approximation Accurate Prop CLT Failure Picture A Random Variable Problem CLT

Suppose 70% approve the President . . .

You poll 100 people. What is the probability that between 60 and 65 report approval? (including 60 and 65.)

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Math 1710 Class 11 V1cu Normal Approximation Making Normal Approximation Accurate Prop CLT Failure Picture A Random Variable Problem CLT

Suppose 70% approve the President . . .

You poll 100 people. What is the probability that between 60 and 65 report approval? (including 60 and 65.) Solution: Let X = Binomial(100, .7)

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Math 1710 Class 11 V1cu Normal Approximation Making Normal Approximation Accurate Prop CLT Failure Picture A Random Variable Problem CLT

Suppose 70% approve the President . . .

You poll 100 people. What is the probability that between 60 and 65 report approval? (including 60 and 65.) Solution: Let X = Binomial(100, .7) X has mean µ = (100)(.7) = 70 and σ =

  • (100)(.7)(.3) =

√ 21 = 4.58.

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SLIDE 5

Math 1710 Class 11 V1cu Normal Approximation Making Normal Approximation Accurate Prop CLT Failure Picture A Random Variable Problem CLT

Suppose 70% approve the President . . .

You poll 100 people. What is the probability that between 60 and 65 report approval? (including 60 and 65.) Solution: Let X = Binomial(100, .7) X has mean µ = (100)(.7) = 70 and σ =

  • (100)(.7)(.3) =

√ 21 = 4.58. X will then be approximated by a normal distribution Y with the same mean and standard deviation.

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SLIDE 6

Math 1710 Class 11 V1cu Normal Approximation Making Normal Approximation Accurate Prop CLT Failure Picture A Random Variable Problem CLT

Suppose 70% approve the President . . .

You poll 100 people. What is the probability that between 60 and 65 report approval? (including 60 and 65.) Solution: Let X = Binomial(100, .7) X has mean µ = (100)(.7) = 70 and σ =

  • (100)(.7)(.3) =

√ 21 = 4.58. X will then be approximated by a normal distribution Y with the same mean and standard deviation. Y = N(70, 4.58).

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Math 1710 Class 11 V1cu Normal Approximation Making Normal Approximation Accurate Prop CLT Failure Picture A Random Variable Problem CLT

Suppose 70% approve the President . . .

You poll 100 people. What is the probability that between 60 and 65 report approval? (including 60 and 65.) Solution: Let X = Binomial(100, .7) X has mean µ = (100)(.7) = 70 and σ =

  • (100)(.7)(.3) =

√ 21 = 4.58. X will then be approximated by a normal distribution Y with the same mean and standard deviation. Y = N(70, 4.58). We can approximate P(60 ≤ X ≤ 65) by P(60 ≤ Y ≤ 65).

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Math 1710 Class 11 V1cu Normal Approximation Making Normal Approximation Accurate Prop CLT Failure Picture A Random Variable Problem CLT

Suppose 70% approve the President . . .

You poll 100 people. What is the probability that between 60 and 65 report approval? (including 60 and 65.) X will then be approximated by a normal distribution Y with the same mean and standard deviation. Y = N(70, 4.58). We can approximate P(60 ≤ X ≤ 65) by P(60 ≤ Y ≤ 65). The Z score of 60 is −10

4.58 = −2.18.

The Z score of 65 is

−5 4.58 = −1.09.

So P(60 ≤ Y ≤ 65) = P(Z < −1.09) − P(Z < −2.18) = .1379 − .0146 = .1233.

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SLIDE 9

Math 1710 Class 11 V1cu Normal Approximation Making Normal Approximation Accurate Prop CLT Failure Picture A Random Variable Problem CLT

Suppose 70% approve the President . . .

You poll 100 people. What is the probability that between 60 and 65 report approval? (including 60 and 65.) We can approximate P(60 ≤ X ≤ 65) by P(60 ≤ Y ≤ 65). The Z score of 60 is −10

4.58 = −2.18.

The Z score of 65 is

−5 4.58 = −1.09.

So P(60 ≤ Y ≤ 65) = P(Z < −1.09) − P(Z < −2.18) = .1379 − .0146 = .1233. Actually P(60 ≤ X ≤ 65) = .15036. So .1233 is not that good an approximation.

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Math 1710 Class 11 V1cu Normal Approximation Making Normal Approximation Accurate Prop CLT Failure Picture A Random Variable Problem CLT

Suppose 70% approve the President . . .

You poll 100 people. What is the probability that between 60 and 65 report approval? (including 60 and 65.) We can approximate P(60 ≤ X ≤ 65) by P(60 ≤ Y ≤ 65). The Z score of 60 is −10

4.58 = −2.18.

The Z score of 65 is

−5 4.58 = −1.09.

So P(60 ≤ Y ≤ 65) = P(Z < −1.09) − P(Z < −2.18) = .1379 − .0146 = .1233. Actually P(60 ≤ X ≤ 65) = .15036. So .1233 is not that good an approximation. P(60 < X < 65) = .09509. would also be approximated by .1233!

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Math 1710 Class 11 V1cu Normal Approximation Making Normal Approximation Accurate Prop CLT Failure Picture A Random Variable Problem CLT

X = Binomial(100, .7) and Y = N(70, 4.58).

P(60 ≤ X ≤ 65) = .15036 P(60 < X < 65) = .09509

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SLIDE 12

Math 1710 Class 11 V1cu Normal Approximation Making Normal Approximation Accurate Prop CLT Failure Picture A Random Variable Problem CLT

X = Binomial(100, .7) and Y = N(70, 4.58).

P(60 ≤ X ≤ 65) = .15036 P(60 < X < 65) = .09509 P(60 ≤ Y ≤ 65) = .1233 is not that good an approximation to either.

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Math 1710 Class 11 V1cu Normal Approximation Making Normal Approximation Accurate Prop CLT Failure Picture A Random Variable Problem CLT

X = Binomial(100, .7) and Y = N(70, 4.58).

P(60 ≤ X ≤ 65) = .15036 P(60 < X < 65) = .09509 P(60 ≤ Y ≤ 65) = .1233 is not that good an approximation to either. The problem is that for a continuous model like Y , P(Y = 65) = 0. But P(X = 65) = .04678.

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Math 1710 Class 11 V1cu Normal Approximation Making Normal Approximation Accurate Prop CLT Failure Picture A Random Variable Problem CLT

X = Binomial(100, .7) and Y = N(70, 4.58).

P(60 ≤ X ≤ 65) = .15036 P(60 < X < 65) = .09509 P(60 ≤ Y ≤ 65) = .1233 So Y doesn’t care about < vs. ≤ . But X does!

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Math 1710 Class 11 V1cu Normal Approximation Making Normal Approximation Accurate Prop CLT Failure Picture A Random Variable Problem CLT

X = Binomial(100, .7) and Y = N(70, 4.58).

P(60 ≤ X ≤ 65) = .15036 P(60 < X < 65) = .09509 P(60 ≤ Y ≤ 65) = .1233 Looking closely at the picture provides the key:

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Math 1710 Class 11 V1cu Normal Approximation Making Normal Approximation Accurate Prop CLT Failure Picture A Random Variable Problem CLT

X = Binomial(100, .7) and Y = N(70, 4.58).

P(60 ≤ X ≤ 65) = .15036 P(60 < X < 65) = .09509 P(60 ≤ Y ≤ 65) = .1233

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Math 1710 Class 11 V1cu Normal Approximation Making Normal Approximation Accurate Prop CLT Failure Picture A Random Variable Problem CLT

X = Binomial(100, .7) and Y = N(70, 4.58).

P(60 ≤ X ≤ 65) = .15036 P(60 < X < 65) = .09509 P(60 ≤ Y ≤ 65) = .1233 It is really the area under the normal curve between 64.5 and 65.5 which approximates P(X=65).

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Math 1710 Class 11 V1cu Normal Approximation Making Normal Approximation Accurate Prop CLT Failure Picture A Random Variable Problem CLT

X = Binomial(100, .7) and Y = N(70, 4.58).

P(60 ≤ X ≤ 65) = .15036 P(60 < X < 65) = .09509 P(60 ≤ Y ≤ 65) = .1233

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SLIDE 19

Math 1710 Class 11 V1cu Normal Approximation Making Normal Approximation Accurate Prop CLT Failure Picture A Random Variable Problem CLT

X = Binomial(100, .7) and Y = N(70, 4.58).

P(60 ≤ X ≤ 65) = .15036 P(60 < X < 65) = .09509 P(60 ≤ Y ≤ 65) = .1233 So P(60 ≤ X ≤ 65) = .15036 should be approximated by P(59.5 ≤ Y ≤ 65.5).

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Math 1710 Class 11 V1cu Normal Approximation Making Normal Approximation Accurate Prop CLT Failure Picture A Random Variable Problem CLT

X = Binomial(100, .7) and Y = N(70, 4.58).

P(60 ≤ X ≤ 65) = .15036 P(60 < X < 65) = .09509 P(60 ≤ Y ≤ 65) = .1233 So P(60 ≤ X ≤ 65) = .15036 should be approximated by P(59.5 ≤ Y ≤ 65.5). The z-scores of 59.5 and 65.5 are

−10.5 4.58 = −2.29 and −4.5 4.58 = −.98 resp.

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Math 1710 Class 11 V1cu Normal Approximation Making Normal Approximation Accurate Prop CLT Failure Picture A Random Variable Problem CLT

X = Binomial(100, .7) and Y = N(70, 4.58).

P(60 ≤ X ≤ 65) = .15036 P(60 < X < 65) = .09509 P(60 ≤ Y ≤ 65) = .1233 So P(60 ≤ X ≤ 65) = .15036 should be approximated by P(59.5 ≤ Y ≤ 65.5). The z-scores of 59.5 and 65.5 are

−10.5 4.58 = −2.29 and −4.5 4.58 = −.98 resp.

So P(59.5 ≤ Y ≤ 65.5) = P(−2.29 ≤ Z ≤ −.98) = .1635 − .0110 = .1525. Much closer to P(60 ≤ X ≤ 65) = .15036!

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Math 1710 Class 11 V1cu Normal Approximation Making Normal Approximation Accurate Prop CLT Failure Picture A Random Variable Problem CLT

X = Binomial(100, .7) and Y = N(70, 4.58).

P(60 ≤ X ≤ 65) = .15036 P(60 < X < 65) = .09509 P(60 ≤ Y ≤ 65) = .1233 So P(60 ≤ X ≤ 65) = .15036 should be approximated by P(59.5 ≤ Y ≤ 65.5). Similarly P(60.5 ≤ Y ≤ 64.5) = P(Z < −1.20) − P(Z < −2.07) = .1151 − .0192 = .0959 is a great approximation to P(60 < X < 65) = .09509.

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Math 1710 Class 11 V1cu Normal Approximation Making Normal Approximation Accurate Prop CLT Failure Picture A Random Variable Problem CLT

X = Binomial(100, .7) and Y = N(70, 4.58).

P(60 ≤ X ≤ 65) = .15036 P(60 < X < 65) = .09509 P(60 ≤ Y ≤ 65) = .1233 So P(60 ≤ X ≤ 65) = .15036 should be approximated by P(59.5 ≤ Y ≤ 65.5). This is called the continuity approximation. You needn’t use it on exams or homework! (We suggest you not.)

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Math 1710 Class 11 V1cu Normal Approximation Making Normal Approximation Accurate Prop CLT Failure Picture A Random Variable Problem CLT

Normal Approximation of X=Binomial(30,.1)

The binomial distribution may be well approximated by a normal distribution with the same mean and standard deviation as long as the expected number of successes (np) and the expected number of failures (n(1 − p)) are both at least 10. Some people call this the “success/failure” condition. The suitabilty of normal approximation when it holds is “the weak central limit theorem.”

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Math 1710 Class 11 V1cu Normal Approximation Making Normal Approximation Accurate Prop CLT Failure Picture A Random Variable Problem CLT

Normal Approximation of X=Binomial(30,.1)

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Math 1710 Class 11 V1cu Normal Approximation Making Normal Approximation Accurate Prop CLT Failure Picture A Random Variable Problem CLT

Normal Approximation of X=Binomial(30,.1)

µ = 3 and σ =

  • 30(.1)(.9 = 1.64.
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Math 1710 Class 11 V1cu Normal Approximation Making Normal Approximation Accurate Prop CLT Failure Picture A Random Variable Problem CLT

Normal Approximation of X=Binomial(30,.1)

µ = 3 and σ =

  • 30(.1)(.9 = 1.64.

Two standard deviations to the left of the mean and our normal model is reporting negative values for the number of successes!

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Math 1710 Class 11 V1cu Normal Approximation Making Normal Approximation Accurate Prop CLT Failure Picture A Random Variable Problem CLT

Normal Approximation of X=Binomial(30,.1)

µ = 3 and σ =

  • 30(.1)(.9 = 1.64.

This is typical of what happens when success/failure is not satisfied.

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Math 1710 Class 11 V1cu Normal Approximation Making Normal Approximation Accurate Prop CLT Failure Picture A Random Variable Problem CLT

RV Review Problem #5

In problem #3, suppose the amounts of cereal in both size bowls are normally distributed. (N(1.5, .3) and N(2.5, .4).) You pour a small and a large bowl. (a) What is the probability you poured out more than 4.5

  • unces of cereal in the two bowls together.

(b) What is the probability the small bowl contains more cereal than the large one?

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Math 1710 Class 11 V1cu Normal Approximation Making Normal Approximation Accurate Prop CLT Failure Picture A Random Variable Problem CLT

RV Review Problem #5

small bowls N(1.5, .3) large bowls N(2.5, .4)

  • ne small, one large bowl

P(total > 4.5)?

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Math 1710 Class 11 V1cu Normal Approximation Making Normal Approximation Accurate Prop CLT Failure Picture A Random Variable Problem CLT

RV Review Problem #5

small bowls N(1.5, .3) large bowls N(2.5, .4)

  • ne small, one large bowl

P(total > 4.5)? Solution: First define some RV’s:

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Math 1710 Class 11 V1cu Normal Approximation Making Normal Approximation Accurate Prop CLT Failure Picture A Random Variable Problem CLT

RV Review Problem #5

small bowls N(1.5, .3) large bowls N(2.5, .4)

  • ne small, one large bowl

P(total > 4.5)? Solution: First define some RV’s: Let S be an RV describing amt. of cereal in a small bowl.

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Math 1710 Class 11 V1cu Normal Approximation Making Normal Approximation Accurate Prop CLT Failure Picture A Random Variable Problem CLT

RV Review Problem #5

small bowls N(1.5, .3) large bowls N(2.5, .4)

  • ne small, one large bowl

P(total > 4.5)? Solution: First define some RV’s: Let S be an RV describing amt. of cereal in a small bowl. Let L be an RV describing amt. of cereal in a lg. bowl.

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Math 1710 Class 11 V1cu Normal Approximation Making Normal Approximation Accurate Prop CLT Failure Picture A Random Variable Problem CLT

RV Review Problem #5

small bowls N(1.5, .3) large bowls N(2.5, .4)

  • ne small, one large bowl

P(total > 4.5)? Solution: First define some RV’s: Let S be an RV describing amt. of cereal in a small bowl. Let L be an RV describing amt. of cereal in a lg. bowl. Let T be an RV describing total amt. of cereal.

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Math 1710 Class 11 V1cu Normal Approximation Making Normal Approximation Accurate Prop CLT Failure Picture A Random Variable Problem CLT

RV Review Problem #5

small bowls N(1.5, .3) large bowls N(2.5, .4)

  • ne small, one large bowl

P(total > 4.5)? Let S be an RV describing amt. of cereal in a small bowl. Let L be an RV describing amt. of cereal in a lg. bowl. Let T be an RV describing total amt. of cereal. T = L + S

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Math 1710 Class 11 V1cu Normal Approximation Making Normal Approximation Accurate Prop CLT Failure Picture A Random Variable Problem CLT

RV Review Problem #5

small bowls N(1.5, .3) large bowls N(2.5, .4)

  • ne small, one large bowl

P(total > 4.5)? T = L + S The sum of (indep.) normal RV’s is again normal.

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Math 1710 Class 11 V1cu Normal Approximation Making Normal Approximation Accurate Prop CLT Failure Picture A Random Variable Problem CLT

RV Review Problem #5

small bowls N(1.5, .3) large bowls N(2.5, .4)

  • ne small, one large bowl

P(total > 4.5)? T = L + S The sum of (indep.) normal RV’s is again normal. So what we need are: E(T)? Var(T)? σT?

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SLIDE 38

Math 1710 Class 11 V1cu Normal Approximation Making Normal Approximation Accurate Prop CLT Failure Picture A Random Variable Problem CLT

RV Review Problem #5

small bowls N(1.5, .3) large bowls N(2.5, .4)

  • ne small, one large bowl

P(total > 4.5)? T = L + S The sum of (indep.) normal RV’s is again normal. So what we need are: E(T)? Var(T)? σT? E(T) = E(L) + E(S) = 1.5 + 2.5 = 4.

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SLIDE 39

Math 1710 Class 11 V1cu Normal Approximation Making Normal Approximation Accurate Prop CLT Failure Picture A Random Variable Problem CLT

RV Review Problem #5

small bowls N(1.5, .3) large bowls N(2.5, .4)

  • ne small, one large bowl

P(total > 4.5)? T = L + S The sum of (indep.) normal RV’s is again normal. So what we need are: E(T)? Var(T)? σT? E(T) = E(L) + E(S) = 1.5 + 2.5 = 4. Assuming independence Var(T) = Var(L) + Var(S) = .42 + .32 = .25.

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Math 1710 Class 11 V1cu Normal Approximation Making Normal Approximation Accurate Prop CLT Failure Picture A Random Variable Problem CLT

RV Review Problem #5

small bowls N(1.5, .3) large bowls N(2.5, .4)

  • ne small, one large bowl

P(total > 4.5)? T = L + S The sum of (indep.) normal RV’s is again normal. So what we need are: E(T)? Var(T)? σT? E(T) = E(L) + E(S) = 1.5 + 2.5 = 4. Assuming independence Var(T) = Var(L) + Var(S) = .42 + .32 = .25. σT = √ .25 = .5

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Math 1710 Class 11 V1cu Normal Approximation Making Normal Approximation Accurate Prop CLT Failure Picture A Random Variable Problem CLT

RV Review Problem #5

small bowls N(1.5, .3) large bowls N(2.5, .4)

  • ne small, one large bowl

P(total > 4.5)? T = L + S The sum of (indep.) normal RV’s is again normal. So what we need are: E(T)? Var(T)? σT? E(T) = E(L) + E(S) = 1.5 + 2.5 = 4. Assuming independence Var(T) = Var(L) + Var(S) = .42 + .32 = .25. σT = √ .25 = .5 Thus T ∼ N(4, .5).

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Math 1710 Class 11 V1cu Normal Approximation Making Normal Approximation Accurate Prop CLT Failure Picture A Random Variable Problem CLT

RV Review Problem #5

small bowls N(1.5, .3) large bowls N(2.5, .4)

  • ne small, one large bowl

P(total > 4.5)? Thus T ∼ N(4, .5). We need P(T > 4.5):

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Math 1710 Class 11 V1cu Normal Approximation Making Normal Approximation Accurate Prop CLT Failure Picture A Random Variable Problem CLT

RV Review Problem #5

Thus T ∼ N(4, .5). We need P(T > 4.5): N(4, .5)

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Math 1710 Class 11 V1cu Normal Approximation Making Normal Approximation Accurate Prop CLT Failure Picture A Random Variable Problem CLT

RV Review Problem #5

Thus T ∼ N(4, .5). We need P(T > 4.5): The z score is 1 so N(0, 1)

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Math 1710 Class 11 V1cu Normal Approximation Making Normal Approximation Accurate Prop CLT Failure Picture A Random Variable Problem CLT

RV Review Problem #5

The z score is 1 so N(0, 1) And P(Z > 1) = P(Z < −1) = .1587.

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Math 1710 Class 11 V1cu Normal Approximation Making Normal Approximation Accurate Prop CLT Failure Picture A Random Variable Problem CLT

RV Review Problem #5

In problem #3, suppose the amounts of cereal in both size bowls are normally distributed. (N(1.5, .3) and N(2.5, .4).) You pour a small and a large bowl. (b) What is the probability the small bowl contains more cereal than the large one?

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SLIDE 47

Math 1710 Class 11 V1cu Normal Approximation Making Normal Approximation Accurate Prop CLT Failure Picture A Random Variable Problem CLT

RV Review Problem #5

In problem #3, suppose the amounts of cereal in both size bowls are normally distributed. (N(1.5, .3) and N(2.5, .4).) You pour a small and a large bowl. (b) What is the probability the small bowl contains more cereal than the large one? small bowls N(1.5, .3) large bowls N(2.5, .4)

  • ne small, one large bowl

P(S > L)?

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Math 1710 Class 11 V1cu Normal Approximation Making Normal Approximation Accurate Prop CLT Failure Picture A Random Variable Problem CLT

RV Review Problem #5

small bowls N(1.5, .3) large bowls N(2.5, .4)

  • ne small, one large bowl

P(S > L)? At first glance this is hard! Two curves, how do you relate them?

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Math 1710 Class 11 V1cu Normal Approximation Making Normal Approximation Accurate Prop CLT Failure Picture A Random Variable Problem CLT

RV Review Problem #5

small bowls N(1.5, .3) large bowls N(2.5, .4)

  • ne small, one large bowl

P(S > L)? Trick: Translate to P(L − S < 0).

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Math 1710 Class 11 V1cu Normal Approximation Making Normal Approximation Accurate Prop CLT Failure Picture A Random Variable Problem CLT

RV Review Problem #5

small bowls N(1.5, .3) large bowls N(2.5, .4)

  • ne small, one large bowl

P(S > L)? Trick: Translate to P(L − S < 0). As before L − S has mean 1 and std. dev. .5.

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Math 1710 Class 11 V1cu Normal Approximation Making Normal Approximation Accurate Prop CLT Failure Picture A Random Variable Problem CLT

RV Review Problem #5

P(S > L)? Trick: Translate to P(L − S < 0). N(1, .5)

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SLIDE 52

Math 1710 Class 11 V1cu Normal Approximation Making Normal Approximation Accurate Prop CLT Failure Picture A Random Variable Problem CLT

RV Review Problem #5

P(S > L)? Trick: Translate to P(L − S < 0). The z score is -2 so N(0, 1)

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SLIDE 53

Math 1710 Class 11 V1cu Normal Approximation Making Normal Approximation Accurate Prop CLT Failure Picture A Random Variable Problem CLT

RV Review Problem #5

P(S > L)? Trick: Translate to P(L − S < 0). The z score is -2 so N(0, 1) P(Z < −2) = .0228.

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Math 1710 Class 11 V1cu Normal Approximation Making Normal Approximation Accurate Prop CLT Failure Picture A Random Variable Problem CLT

Three CLT problems

(1) 80% of all cars on the interstate speed. Randomly sample 50. What proportion of speeders might we see?

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Math 1710 Class 11 V1cu Normal Approximation Making Normal Approximation Accurate Prop CLT Failure Picture A Random Variable Problem CLT

Three CLT problems

(1) 80% of all cars on the interstate speed. Randomly sample 50. What proportion of speeders might we see? (e.g.: A policeman has a quota of at least 35 speeders to ticket. What is the chance he’ll meet his quota in such a sample?)

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Math 1710 Class 11 V1cu Normal Approximation Making Normal Approximation Accurate Prop CLT Failure Picture A Random Variable Problem CLT

Three CLT problems

(2) At birth, babies average 7.8 pounds with a standard deviation of 2.1 pounds. 34 Babies born near a large possibly polluting factory average 7.2 pounds.

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SLIDE 57

Math 1710 Class 11 V1cu Normal Approximation Making Normal Approximation Accurate Prop CLT Failure Picture A Random Variable Problem CLT

Three CLT problems

(2) At birth, babies average 7.8 pounds with a standard deviation of 2.1 pounds. 34 Babies born near a large possibly polluting factory average 7.2 pounds. Is that unusually low?

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Math 1710 Class 11 V1cu Normal Approximation Making Normal Approximation Accurate Prop CLT Failure Picture A Random Variable Problem CLT

Three CLT problems

(3) Suppose SAT’s have a mean of 500, standard deviation of 100 and are approximately normal. Form means of samples of 20 students.

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Math 1710 Class 11 V1cu Normal Approximation Making Normal Approximation Accurate Prop CLT Failure Picture A Random Variable Problem CLT

Three CLT problems

(3) Suppose SAT’s have a mean of 500, standard deviation of 100 and are approximately normal. Form means of samples of 20 students. What distribution do these follow? What is the chance of a sample averaging over 600? Over 550?

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Math 1710 Class 11 V1cu Normal Approximation Making Normal Approximation Accurate Prop CLT Failure Picture A Random Variable Problem CLT

Proportion Case of the CLT

X ∼ Binomial(n,p) X approximated by normal RV Y ∼ N(np, √npq). Suppose we keep track of the observed proportion ˆ p = X n here.

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Math 1710 Class 11 V1cu Normal Approximation Making Normal Approximation Accurate Prop CLT Failure Picture A Random Variable Problem CLT

Proportion Case of the CLT

X ∼ Binomial(n,p) X approximated by normal RV Y ∼ N(np, √npq). Suppose we keep track of the observed proportion ˆ p = X n here. Then ˆ p is an RV approximated by Y /n ∼ N(p, SD(ˆ p)). where SD(ˆ p) = pq n .

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Math 1710 Class 11 V1cu Normal Approximation Making Normal Approximation Accurate Prop CLT Failure Picture A Random Variable Problem CLT

Proportion Case of the CLT

X ∼ Binomial(n,p) X approximated by normal RV Y ∼ N(np, √npq). Suppose we keep track of the observed proportion ˆ p = X n here. Then ˆ p is an RV approximated by Y /n ∼ N(p, SD(ˆ p)). where SD(ˆ p) = pq n . Thus N(p,

  • pq

n ) is the sampling distribution of ˆ

p.

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Math 1710 Class 11 V1cu Normal Approximation Making Normal Approximation Accurate Prop CLT Failure Picture A Random Variable Problem CLT

Proportion Case of the CLT

Some books uses the notation SD(ˆ p) = pq n .

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Math 1710 Class 11 V1cu Normal Approximation Making Normal Approximation Accurate Prop CLT Failure Picture A Random Variable Problem CLT

Proportion Case of the CLT

Some books uses the notation SD(ˆ p) = pq n . SD(ˆ p) stands for Standard Deviation of the Sampling Distribution of ˆ p.

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Math 1710 Class 11 V1cu Normal Approximation Making Normal Approximation Accurate Prop CLT Failure Picture A Random Variable Problem CLT

Proportion Case of the CLT

Some books uses the notation SD(ˆ p) = pq n . SD(ˆ p) stands for Standard Deviation of the Sampling Distribution of ˆ p. And the notation SE(ˆ p) =

  • ˆ

pˆ q n . SE(ˆ p) stands for the Standard Error of ˆ p.

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Math 1710 Class 11 V1cu Normal Approximation Making Normal Approximation Accurate Prop CLT Failure Picture A Random Variable Problem CLT

Proportion Case of the CLT

Others, including your textbook, I believe, are happy to refer to both of these as standard errors of ˆ

  • p. One is exact (but hard to

know in practice) while the other is only an estimate, but feasible to produce.