Normal Binomial Lecture 23 - Purdue Universityhuang251/slides23.pdfBinomial Review of Normal Distribution Normal approximation 23.8 Normal approximation of Binomial Distribution We

Normalapproximation to

Binomial

Review of NormalDistribution

Normal approximation

23.1

Lecture 23Normal approximation to BinomialSTAT 225 Introduction to Probability ModelsApril 4, 2014

Whitney HuangPurdue University


Binomial



23.2

Agenda

1 Review of Normal Distribution

2 Normal approximation


Binomial



23.3

Normal Distribution

Characteristics of the Normal random variable:Let X be a Normal r.v.

The support for X : (−∞,∞)

Its parameter(s) and definition(s): µ : mean and σ2 :variance

The probability density function (pdf): 1√2πσ

e−(x−µ)2

2σ2 for−∞ < x <∞The cumulative distribution function (cdf): No explicit form,look at the value Φ( x−µ

σ ) for −∞ < x <∞ from standardnormal tableThe expected value: E[X ] = µ

The variance: Var(X ) = σ2


Binomial



23.3

Normal Distribution





e−(x−µ)2





Binomial



23.3

Normal Distribution





e−(x−µ)2

2σ2 for−∞ < x <∞

The cumulative distribution function (cdf): No explicit form,look at the value Φ( x−µ




Binomial



23.3

Normal Distribution





e−(x−µ)2


σ ) for −∞ < x <∞ from standardnormal table

The expected value: E[X ] = µ



Binomial



23.3

Normal Distribution





e−(x−µ)2





Binomial



23.3

Normal Distribution





e−(x−µ)2





Binomial



23.4

Standard Normal Z ∼ N(µ = 0, σ2 = 1)

Normal random variable X with mean µ and standarddeviation σ can convert to standard normal Z by thefollowing :

Z =X − µσ

The cdf of the standard normal, denoted by Φ(z), can befound from the standard normal tableThe probability P(a ≤ X ≤ b) where X ∼ N(µ, σ2) can becompute

P(a ≤ X ≤ b) = P(a− µσ≤ Z ≤ b − µ

σ)

= Φ(b − µσ

)− Φ(a− µσ

)


Binomial



23.4



Z =X − µσ

The cdf of the standard normal, denoted by Φ(z), can befound from the standard normal table

The probability P(a ≤ X ≤ b) where X ∼ N(µ, σ2) can becompute

P(a ≤ X ≤ b) = P(a− µσ≤ Z ≤ b − µ

σ)

= Φ(b − µσ

)− Φ(a− µσ

)


Binomial



23.4



Z =X − µσ

The cdf of the standard normal, denoted by Φ(z), can befound from the standard normal tableThe probability P(a ≤ X ≤ b) where X ∼ N(µ, σ2) can becompute

P(a ≤ X ≤ b) = P(a− µσ≤ Z ≤ b − µ

σ)

= Φ(b − µσ

)− Φ(a− µσ

)


Binomial



23.5

Properties of Φ

Φ(0) = .50⇒ Mean and Median (50th percentile) forstandard normal are both 0

Φ(−z) = 1− Φ(z)

P(Z > z) = 1− Φ(z) = Φ(−z)


Binomial



23.5

Properties of Φ

Φ(0) = .50⇒ Mean and Median (50th percentile) forstandard normal are both 0Φ(−z) = 1− Φ(z)

P(Z > z) = 1− Φ(z) = Φ(−z)


Binomial



23.5

Properties of Φ

Φ(0) = .50⇒ Mean and Median (50th percentile) forstandard normal are both 0Φ(−z) = 1− Φ(z)

P(Z > z) = 1− Φ(z) = Φ(−z)


Binomial



23.6

Sums of Normal Random Variables

If Xi 1 ≤ i ≤ n are independent normal random variables withmean µi are variance σ2

i , respectively.Let Sn =

∑ni=1 Xi then Sn ∼ N(

∑ni=1 µi ,

∑ni=1 σ

2i )

This can be applied for any integer n


Binomial



23.6

Sums of Normal Random Variables

If Xi 1 ≤ i ≤ n are independent normal random variables withmean µi are variance σ2

i , respectively.Let Sn =

∑ni=1 Xi then Sn ∼ N(

∑ni=1 µi ,

∑ni=1 σ

2i )

This can be applied for any integer n


Binomial



23.7

The Empirical Rules

The Empirical Rules are a way to approximate certainprobabilities for the Normal Distribution as the following table:

Interval Percentage with intervalµ± σ 68%µ± 2σ 95%µ± 3σ 99.7%


Binomial



23.8

Normal approximation of Binomial Distribution

We can use a Normal Distribution to approximate aBinomial Distribution if n is large and p is close to .5

Rule of thumb for this approximation to be valid (in thisclass) is np > 5 and n(1− p) > 5If X ∼ Binomial(n,p) with np > 5 and n(1− p) > 5 thenwe can use X ∗ ∼ N(µ = np, σ2 = np(1− p)) toapproximate XNotice that Binomial is a discrete distribution but normal isa continuous distribution so that P(X ∗ = x) = 0 ∀xBy using continuity correction we useP(x − 0.5 ≤ X ∗ ≤ x + 0.5) to approximate P(X = x)


Binomial



23.8


We can use a Normal Distribution to approximate aBinomial Distribution if n is large and p is close to .5Rule of thumb for this approximation to be valid (in thisclass) is np > 5 and n(1− p) > 5

If X ∼ Binomial(n,p) with np > 5 and n(1− p) > 5 thenwe can use X ∗ ∼ N(µ = np, σ2 = np(1− p)) toapproximate XNotice that Binomial is a discrete distribution but normal isa continuous distribution so that P(X ∗ = x) = 0 ∀xBy using continuity correction we useP(x − 0.5 ≤ X ∗ ≤ x + 0.5) to approximate P(X = x)


Binomial



23.8


We can use a Normal Distribution to approximate aBinomial Distribution if n is large and p is close to .5Rule of thumb for this approximation to be valid (in thisclass) is np > 5 and n(1− p) > 5If X ∼ Binomial(n,p) with np > 5 and n(1− p) > 5 thenwe can use X ∗ ∼ N(µ = np, σ2 = np(1− p)) toapproximate X

Notice that Binomial is a discrete distribution but normal isa continuous distribution so that P(X ∗ = x) = 0 ∀xBy using continuity correction we useP(x − 0.5 ≤ X ∗ ≤ x + 0.5) to approximate P(X = x)


Binomial



23.8


We can use a Normal Distribution to approximate aBinomial Distribution if n is large and p is close to .5Rule of thumb for this approximation to be valid (in thisclass) is np > 5 and n(1− p) > 5If X ∼ Binomial(n,p) with np > 5 and n(1− p) > 5 thenwe can use X ∗ ∼ N(µ = np, σ2 = np(1− p)) toapproximate XNotice that Binomial is a discrete distribution but normal isa continuous distribution so that P(X ∗ = x) = 0 ∀x

By using continuity correction we useP(x − 0.5 ≤ X ∗ ≤ x + 0.5) to approximate P(X = x)


Binomial



23.8


We can use a Normal Distribution to approximate aBinomial Distribution if n is large and p is close to .5Rule of thumb for this approximation to be valid (in thisclass) is np > 5 and n(1− p) > 5If X ∼ Binomial(n,p) with np > 5 and n(1− p) > 5 thenwe can use X ∗ ∼ N(µ = np, σ2 = np(1− p)) toapproximate XNotice that Binomial is a discrete distribution but normal isa continuous distribution so that P(X ∗ = x) = 0 ∀xBy using continuity correction we useP(x − 0.5 ≤ X ∗ ≤ x + 0.5) to approximate P(X = x)


Binomial



23.9

Example 57

For this entire problem, please use the empirical rules. Thenumber of pairs of shoes in an adult female’s closet is Normalwith a mean of 58 and a standard deviation of 5

1 What interval contains the middle 68% of the distribution?

2 Find the value such that 2.5% are lower than that value

3 What is the probability an adult female’s closet hasbetween 48 and 63 pairs of shoes?


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23.10

Example 57 cont’d

Solution.

1 (58− 5,58 + 5) = (53,63)

2 58− 2× 5 = 483 68% + 95%−68%

2 = 81.5%


Binomial



23.11

Example 58

Suppose a class has 400 students (to begin with), that eachstudent drops independently of any other student with aprobability of .07. Let X be the number of students that finishthis course

1 Find the probability that X is between 370 and 373inclusive

2 Is an approximation appropriate for the number of studentsthat finish the course?

3 If so, what is this distribution and what are the value(s) ofits parameter(s)?

4 Find the probability that is between 370 and 373 inclusiveby using the approximation (if an approximationappropriate)


Binomial



23.12

Example 58 cont’d

Solution.

X ∼ Binomial(n = 400, .93)

1 P(370 ≤ X ≤ 373) = .3010

2 Yes, since np = 372 > 5 and n(1− p) = 28 > 5. Thenormal approximation for binomial is appropriate

3 X ∗ ∼ N(µ = 372, σ2 = 400(.93)(.07) = 26.04)

4 P(369.5 ≤ X ∗ ≤ 373.5) = P(−.49 ≤ Z ≤ .29) =.6141− .3121 = .3020


Binomial



23.12

Example 58 cont’d

Solution.

X ∼ Binomial(n = 400, .93)

1 P(370 ≤ X ≤ 373) = .30102 Yes, since np = 372 > 5 and n(1− p) = 28 > 5. The

normal approximation for binomial is appropriate

3 X ∗ ∼ N(µ = 372, σ2 = 400(.93)(.07) = 26.04)

4 P(369.5 ≤ X ∗ ≤ 373.5) = P(−.49 ≤ Z ≤ .29) =.6141− .3121 = .3020


Binomial



23.12

Example 58 cont’d

Solution.

X ∼ Binomial(n = 400, .93)


normal approximation for binomial is appropriate3 X ∗ ∼ N(µ = 372, σ2 = 400(.93)(.07) = 26.04)

4 P(369.5 ≤ X ∗ ≤ 373.5) = P(−.49 ≤ Z ≤ .29) =.6141− .3121 = .3020


Binomial



23.12

Example 58 cont’d

Solution.

X ∼ Binomial(n = 400, .93)


normal approximation for binomial is appropriate3 X ∗ ∼ N(µ = 372, σ2 = 400(.93)(.07) = 26.04)

4 P(369.5 ≤ X ∗ ≤ 373.5) = P(−.49 ≤ Z ≤ .29) =.6141− .3121 = .3020

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Normal Binomial Lecture 23 - Purdue Universityhuang251/slides23.pdfBinomial Review of Normal Distribution Normal approximation 23.8 Normal approximation of Binomial Distribution We