Lecture 13: Some Important Continuous Probability Distributions (Part 2)
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1 Lecture 13: Some Important Continuous Probability Distributions (Part 2) Kateřina Staňková Statistics (MAT1003) May 14, 2012
2 Outline 1 Erlang distribution Formulation Application of Erlang distribution Gamma distribution 2 Various Exercises 3 Chi-squared distribution Basics Applications Examples book: Sections ,6.6
3 And now... 1 Erlang distribution Formulation Application of Erlang distribution Gamma distribution 2 Various Exercises 3 Chi-squared distribution Basics Applications Examples
4 Formulation
5 Formulation What is Erlang PDF? Let X 1, X 2,... X n be IID exponentially distributed RV s with parameter λ. Then the RV X def = X 1 + X X n has an Erlang distribution with parameters n and λ, X Erl(n, λ),
6 Formulation What is Erlang PDF? Let X 1, X 2,... X n be IID exponentially distributed RV s with parameter λ. Then the RV X def = X 1 + X X n has an Erlang distribution with parameters n and λ, X Erl(n, λ), { λ n x n 1 exp( λ x), x > 0, f (x) = (n 1)! 0, elsewhere
7 Formulation What is Erlang PDF? Let X 1, X 2,... X n be IID exponentially distributed RV s with parameter λ. Then the RV X def = X 1 + X X n has an Erlang distribution with parameters n and λ, X Erl(n, λ), { λ n x n 1 exp( λ x), x > 0, f (x) = (n 1)! 0, elsewhere
8 Application of Erlang distribution
9 Application of Erlang distribution In a Poisson process the sum of n interarrival times has an Erlang distribution with parameters n and λ
10 Application of Erlang distribution In a Poisson process the sum of n interarrival times has an Erlang distribution with parameters n and λ Example 5(c) (from before) Suppose on average 6 people call some service number per minute. What is the probability that it takes at least one minute for 3 people to call?
11 Application of Erlang distribution In a Poisson process the sum of n interarrival times has an Erlang distribution with parameters n and λ Example 5(c) (from before) Suppose on average 6 people call some service number per minute. What is the probability that it takes at least one minute for 3 people to call? Z = 3 interarrival times
12 Application of Erlang distribution In a Poisson process the sum of n interarrival times has an Erlang distribution with parameters n and λ Example 5(c) (from before) Suppose on average 6 people call some service number per minute. What is the probability that it takes at least one minute for 3 people to call? Z = 3 interarrival times Z Erl(3, 18)
13 Application of Erlang distribution In a Poisson process the sum of n interarrival times has an Erlang distribution with parameters n and λ Example 5(c) (from before) Suppose on average 6 people call some service number per minute. What is the probability that it takes at least one minute for 3 people to call? Z = 3 interarrival times Z Erl(3, 18) (λ = 18 for 3 minute time-unit)
14 Application of Erlang distribution In a Poisson process the sum of n interarrival times has an Erlang distribution with parameters n and λ Example 5(c) (from before) Suppose on average 6 people call some service number per minute. What is the probability that it takes at least one minute for 3 people to call? Z = 3 interarrival times Z Erl(3, 18) (λ = 18 for 3 minute time-unit) P(Z 1 3 )
15 Application of Erlang distribution In a Poisson process the sum of n interarrival times has an Erlang distribution with parameters n and λ Example 5(c) (from before) Suppose on average 6 people call some service number per minute. What is the probability that it takes at least one minute for 3 people to call? Z = 3 interarrival times Z Erl(3, 18) (λ = 18 for 3 minute time-unit) P(Z 1 3 ) = λ n x n 1 exp( λ x) dx 1 (n 1)! 3
16 Application of Erlang distribution In a Poisson process the sum of n interarrival times has an Erlang distribution with parameters n and λ Example 5(c) (from before) Suppose on average 6 people call some service number per minute. What is the probability that it takes at least one minute for 3 people to call? Z = 3 interarrival times Z Erl(3, 18) (λ = 18 for 3 minute time-unit) P(Z 1 3 ) = λ n x n 1 exp( λ x) dx = 1 (n 1)! x 2 exp( 18 x) dx 2!
17 Application of Erlang distribution In a Poisson process the sum of n interarrival times has an Erlang distribution with parameters n and λ Example 5(c) (from before) Suppose on average 6 people call some service number per minute. What is the probability that it takes at least one minute for 3 people to call? Z = 3 interarrival times Z Erl(3, 18) (λ = 18 for 3 minute time-unit) P(Z 1 3 ) = λ n x n 1 exp( λ x) dx = 1 (n 1)! 3 [ ( 18 3 = 2 1 ) x 2 exp( 18 x) ] x= x 2 exp( 18 x) dx 2! x exp( 18 x) dx = [ 162 x 2 exp( 18 x) ] x= 1 3
18 Application of Erlang distribution In a Poisson process the sum of n interarrival times has an Erlang distribution with parameters n and λ Example 5(c) (from before) Suppose on average 6 people call some service number per minute. What is the probability that it takes at least one minute for 3 people to call? Z = 3 interarrival times Z Erl(3, 18) (λ = 18 for 3 minute time-unit) P(Z 1 3 ) = λ n x n 1 exp( λ x) dx = 1 (n 1)! 3 [ ( 18 3 = 2 1 ) x 2 exp( 18 x) ] x= x 2 exp( 18 x) dx 2! x exp( 18 x) dx = [ 162 x 2 exp( 18 x) [ 324 ( 1 ) x exp( 18 x) 18 ] x= ] x= 1 3 exp( 18 x) dx
19 Application of Erlang distribution In a Poisson process the sum of n interarrival times has an Erlang distribution with parameters n and λ Example 5(c) (from before) Suppose on average 6 people call some service number per minute. What is the probability that it takes at least one minute for 3 people to call? Z = 3 interarrival times Z Erl(3, 18) (λ = 18 for 3 minute time-unit) P(Z 1 3 ) = λ n x n 1 exp( λ x) dx = 1 (n 1)! 3 [ ( 18 3 = 2 1 ) x 2 exp( 18 x) ] x= x 2 exp( 18 x) dx 2! x exp( 18 x) dx = [ 162 x 2 exp( 18 x) [ 324 ( 1 ) x exp( 18 x) 18 ] x= = [18 exp( 6) + 6 exp( 6) exp( 18 x)] x= ] x= 1 3 exp( 18 x) dx
20 Application of Erlang distribution In a Poisson process the sum of n interarrival times has an Erlang distribution with parameters n and λ Example 5(c) (from before) Suppose on average 6 people call some service number per minute. What is the probability that it takes at least one minute for 3 people to call? Z = 3 interarrival times Z Erl(3, 18) (λ = 18 for 3 minute time-unit) P(Z 1 3 ) = λ n x n 1 exp( λ x) dx = 1 (n 1)! 3 [ ( 18 3 = 2 1 ) x 2 exp( 18 x) ] x= x 2 exp( 18 x) dx 2! x exp( 18 x) dx = [ 162 x 2 exp( 18 x) [ 324 ( 1 ) x exp( 18 x) 18 ] x= ] x= 1 3 exp( 18 x) dx = [18 exp( 6) + 6 exp( 6) exp( 18 x)] x=
21 Gamma distribution Derivation
22 Gamma distribution Derivation One can prove that (n 1)!
23 Gamma distribution Derivation One can prove that (n 1)! = Define the so-called Gamma-function: 0 t n 1 exp( t) d t
24 Gamma distribution Derivation One can prove that (n 1)! = Define the so-called Gamma-function: 0 t n 1 exp( t) d t Γ(α) def = 0 t α 1 exp( t) dt for α R
25 Gamma distribution Derivation One can prove that (n 1)! = Define the so-called Gamma-function: 0 t n 1 exp( t) d t Γ(α) def = Thenf (x) = λα x α 1 Γ(α) 0 t α 1 exp( t) dt exp( λ x) for α R for x > 0 and 0 elsewhere
26 Gamma distribution Derivation One can prove that (n 1)! = Define the so-called Gamma-function: 0 t n 1 exp( t) d t Γ(α) def = Thenf (x) = λα x α 1 Γ(α) 0 t α 1 exp( t) dt exp( λ x) for α R for x > 0 and 0 elsewhere is a PDF. An RV with such a PDF has so-called Gamma distribution with parameters α and λ, X Γ(λ, α)
27 Gamma distribution Derivation One can prove that (n 1)! = Define the so-called Gamma-function: 0 t n 1 exp( t) d t Γ(α) def = Thenf (x) = λα x α 1 Γ(α) 0 t α 1 exp( t) dt exp( λ x) for α R for x > 0 and 0 elsewhere is a PDF. An RV with such a PDF has so-called Gamma distribution with parameters α and λ, X Γ(λ, α) (Plug in n = 1 or α = 1 to obtain the exponential distribution)
28 And now... 1 Erlang distribution Formulation Application of Erlang distribution Gamma distribution 2 Various Exercises 3 Chi-squared distribution Basics Applications Examples
29 Example 6 (from last Thursday)
30 Example 6 (from last Thursday) The net weight of a pack of coffee (500 grams) is a normally distributed RV with parameters µ = 505 grams and σ = 5 grams.
31 Example 6 (from last Thursday) The net weight of a pack of coffee (500 grams) is a normally distributed RV with parameters µ = 505 grams and σ = 5 grams. (a) What is the probability that the net weight of a pack is at least 500 grams?
32 Example 6 (from last Thursday) The net weight of a pack of coffee (500 grams) is a normally distributed RV with parameters µ = 505 grams and σ = 5 grams. (a) What is the probability that the net weight of a pack is at least 500 grams?
33 Example 6 (from last Thursday) The net weight of a pack of coffee (500 grams) is a normally distributed RV with parameters µ = 505 grams and σ = 5 grams. (a) What is the probability that the net weight of a pack is at least 500 grams? P(X 500) = P(Z 500 µ = = 1) σ 5
34 Example 6 (from last Thursday) The net weight of a pack of coffee (500 grams) is a normally distributed RV with parameters µ = 505 grams and σ = 5 grams. (a) What is the probability that the net weight of a pack is at least 500 grams? P(X 500) = P(Z 500 µ = = 1) σ 5 = 1 P(Z 1) = =
35 Example 6 (from last Thursday) The net weight of a pack of coffee (500 grams) is a normally distributed RV with parameters µ = 505 grams and σ = 5 grams. (a) What is the probability that the net weight of a pack is at least 500 grams? P(X 500) = P(Z 500 µ = = 1) σ 5 = 1 P(Z 1) = = (b) What is the probability that the gross weight of a pack is between 510 and 520 grams?
36 Example 6 (from last Thursday) The net weight of a pack of coffee (500 grams) is a normally distributed RV with parameters µ = 505 grams and σ = 5 grams. (a) What is the probability that the net weight of a pack is at least 500 grams? P(X 500) = P(Z 500 µ = = 1) σ 5 = 1 P(Z 1) = = (b) What is the probability that the gross weight of a pack is between 510 and 520 grams? A : gross weight of a pack
37 Example 6 (from last Thursday) The net weight of a pack of coffee (500 grams) is a normally distributed RV with parameters µ = 505 grams and σ = 5 grams. (a) What is the probability that the net weight of a pack is at least 500 grams? P(X 500) = P(Z 500 µ = = 1) σ 5 = 1 P(Z 1) = = (b) What is the probability that the gross weight of a pack is between 510 and 520 grams? A : gross weight of a pack A = X + Y
38 Example 6 (from last Thursday) The net weight of a pack of coffee (500 grams) is a normally distributed RV with parameters µ = 505 grams and σ = 5 grams. (a) What is the probability that the net weight of a pack is at least 500 grams? P(X 500) = P(Z 500 µ = = 1) σ 5 = 1 P(Z 1) = = (b) What is the probability that the gross weight of a pack is between 510 and 520 grams? A : gross weight of a pack ( ) A = X + Y N µ X + µ Y, σx 2 + σ2 Y
39 Example 6 (from last Thursday) The net weight of a pack of coffee (500 grams) is a normally distributed RV with parameters µ = 505 grams and σ = 5 grams. (a) What is the probability that the net weight of a pack is at least 500 grams? P(X 500) = P(Z 500 µ = = 1) σ 5 = 1 P(Z 1) = = (b) What is the probability that the gross weight of a pack is between 510 and 520 grams? A : gross weight of a pack ( ) A = X + Y N µ X + µ Y, σx 2 + σ2 Y = N ( , )
40 Example 6 (from last Thursday) The net weight of a pack of coffee (500 grams) is a normally distributed RV with parameters µ = 505 grams and σ = 5 grams. (a) What is the probability that the net weight of a pack is at least 500 grams? P(X 500) = P(Z 500 µ = = 1) σ 5 = 1 P(Z 1) = = (b) What is the probability that the gross weight of a pack is between 510 and 520 grams? A : gross weight of a pack ( ) A = X + Y N µ X + µ Y, σx 2 + σ2 Y = N ( , ) = N ( 515, ) 26
41 Example 6 (from last Thursday) The net weight of a pack of coffee (500 grams) is a normally distributed RV with parameters µ = 505 grams and σ = 5 grams. (a) What is the probability that the net weight of a pack is at least 500 grams? P(X 500) = P(Z 500 µ = = 1) σ 5 = 1 P(Z 1) = = (b) What is the probability that the gross weight of a pack is between 510 and 520 grams? A : gross weight of a pack ( ) A = X + Y N µ X + µ Y, σx 2 + σ2 Y = N ( , ) = N ( 515, ) 26 Z = A 515 N (0, 1) P( Z ) 26 26
42 Example 6 (from last Thursday) The net weight of a pack of coffee (500 grams) is a normally distributed RV with parameters µ = 505 grams and σ = 5 grams. (a) What is the probability that the net weight of a pack is at least 500 grams? P(X 500) = P(Z 500 µ = = 1) σ 5 = 1 P(Z 1) = = (b) What is the probability that the gross weight of a pack is between 510 and 520 grams? A : gross weight of a pack ( ) A = X + Y N µ X + µ Y, σx 2 + σ2 Y = N ( , ) = N ( 515, ) 26 Z = A 515 N (0, 1) P( Z ) = P(Z 0.98) P(Z 0.98) 26
43 Example 6 (from last Thursday) The net weight of a pack of coffee (500 grams) is a normally distributed RV with parameters µ = 505 grams and σ = 5 grams. (a) What is the probability that the net weight of a pack is at least 500 grams? P(X 500) = P(Z 500 µ = = 1) σ 5 = 1 P(Z 1) = = (b) What is the probability that the gross weight of a pack is between 510 and 520 grams? A : gross weight of a pack ( ) A = X + Y N µ X + µ Y, σx 2 + σ2 Y = N ( , ) = N ( 515, ) 26 Z = A 515 N (0, 1) P( Z ) = P(Z 0.98) P(Z 0.98) 26 =
44 Example 6 (from last Thursday)
45 Example 6 (from last Thursday) The net weight of a pack of coffee (500 grams) is a normally distributed RV with parameters µ = 505 grams and σ = 5 grams.
46 Example 6 (from last Thursday) The net weight of a pack of coffee (500 grams) is a normally distributed RV with parameters µ = 505 grams and σ = 5 grams. (c) What is the probability that the gross weight of 10 packs is between 5120 and 5170 grams?
47 Example 6 (from last Thursday) The net weight of a pack of coffee (500 grams) is a normally distributed RV with parameters µ = 505 grams and σ = 5 grams. (c) What is the probability that the gross weight of 10 packs is between 5120 and 5170 grams? B : Gross weight of 10 packs,
48 Example 6 (from last Thursday) The net weight of a pack of coffee (500 grams) is a normally distributed RV with parameters µ = 505 grams and σ = 5 grams. (c) What is the probability that the gross weight of 10 packs is between 5120 and 5170 grams? B : Gross weight of 10 packs, B = 10 i=1 (X i + Y i ), with X i N (505, 5), Y i N (10, 1)
49 Example 6 (from last Thursday) The net weight of a pack of coffee (500 grams) is a normally distributed RV with parameters µ = 505 grams and σ = 5 grams. (c) What is the probability that the gross weight of 10 packs is between 5120 and 5170 grams? B : Gross weight of 10 packs, B = 10 i=1 (X i + Y i ), with X i N (505, 5), Y i N (10, 1) B N (10 µ X + 10 µ Y, 10 σx σ2 Y ) = N (5150, 260)
50 Example 6 (from last Thursday) The net weight of a pack of coffee (500 grams) is a normally distributed RV with parameters µ = 505 grams and σ = 5 grams. (c) What is the probability that the gross weight of 10 packs is between 5120 and 5170 grams? B : Gross weight of 10 packs, B = 10 i=1 (X i + Y i ), with X i N (505, 5), Y i N (10, 1) B N (10 µ X + 10 µ Y, 10 σx σ2 Y ) = N (5150, 260) Z = B N (0, 1)
51 Example 6 (from last Thursday) The net weight of a pack of coffee (500 grams) is a normally distributed RV with parameters µ = 505 grams and σ = 5 grams. (c) What is the probability that the gross weight of 10 packs is between 5120 and 5170 grams? B : Gross weight of 10 packs, B = 10 i=1 (X i + Y i ), with X i N (505, 5), Y i N (10, 1) B N (10 µ X + 10 µ Y, 10 σx σ2 Y ) = N (5150, 260) Z = B N (0, 1)
52 Example 6 (from last Thursday) The net weight of a pack of coffee (500 grams) is a normally distributed RV with parameters µ = 505 grams and σ = 5 grams. (c) What is the probability that the gross weight of 10 packs is between 5120 and 5170 grams? B : Gross weight of 10 packs, B = 10 i=1 (X i + Y i ), with X i N (505, 5), Y i N (10, 1) B N (10 µ X + 10 µ Y, 10 σx σ2 Y ) = N (5150, 260) Z = B N (0, 1) P(5120 B 5170)
53 Example 6 (from last Thursday) The net weight of a pack of coffee (500 grams) is a normally distributed RV with parameters µ = 505 grams and σ = 5 grams. (c) What is the probability that the gross weight of 10 packs is between 5120 and 5170 grams? B : Gross weight of 10 packs, B = 10 i=1 (X i + Y i ), with X i N (505, 5), Y i N (10, 1) B N (10 µ X + 10 µ Y, 10 σx σ2 Y ) = N (5150, 260) Z = B N (0, 1) ( ) P(5120 B 5170) = P Z
54 Example 6 (from last Thursday) The net weight of a pack of coffee (500 grams) is a normally distributed RV with parameters µ = 505 grams and σ = 5 grams. (c) What is the probability that the gross weight of 10 packs is between 5120 and 5170 grams? B : Gross weight of 10 packs, B = 10 i=1 (X i + Y i ), with X i N (505, 5), Y i N (10, 1) B N (10 µ X + 10 µ Y, 10 σx σ2 Y ) = N (5150, 260) Z = B N (0, 1) ( ) P(5120 B 5170) = P Z = P ( 1.86 Z 1.24)
55 Example 6 (from last Thursday) The net weight of a pack of coffee (500 grams) is a normally distributed RV with parameters µ = 505 grams and σ = 5 grams. (c) What is the probability that the gross weight of 10 packs is between 5120 and 5170 grams? B : Gross weight of 10 packs, B = 10 i=1 (X i + Y i ), with X i N (505, 5), Y i N (10, 1) B N (10 µ X + 10 µ Y, 10 σx σ2 Y ) = N (5150, 260) Z = B N (0, 1) ( ) P(5120 B 5170) = P Z = P ( 1.86 Z 1.24) = P (Z 1.24) P (Z 1.86)
56 Example 6 (from last Thursday) The net weight of a pack of coffee (500 grams) is a normally distributed RV with parameters µ = 505 grams and σ = 5 grams. (c) What is the probability that the gross weight of 10 packs is between 5120 and 5170 grams? B : Gross weight of 10 packs, B = 10 i=1 (X i + Y i ), with X i N (505, 5), Y i N (10, 1) B N (10 µ X + 10 µ Y, 10 σx σ2 Y ) = N (5150, 260) Z = B N (0, 1) ( ) P(5120 B 5170) = P Z = P ( 1.86 Z 1.24) = P (Z 1.24) P (Z 1.86) = =
57 Example 6 (from last Thursday)
58 Example 6 (from last Thursday) The net weight of a pack of coffee (500 grams) is a normally distributed RV with parameters µ = 505 grams and σ = 5 grams.
59 Example 6 (from last Thursday) The net weight of a pack of coffee (500 grams) is a normally distributed RV with parameters µ = 505 grams and σ = 5 grams. (d) 80 % of the gross weight of the packs is heavier than a particular weight. What is that weight?
60 Example 6 (from last Thursday) The net weight of a pack of coffee (500 grams) is a normally distributed RV with parameters µ = 505 grams and σ = 5 grams. (d) 80 % of the gross weight of the packs is heavier than a particular weight. What is that weight? A N (515, 26)
61 Example 6 (from last Thursday) The net weight of a pack of coffee (500 grams) is a normally distributed RV with parameters µ = 505 grams and σ = 5 grams. (d) 80 % of the gross weight of the packs is heavier than a particular weight. What is that weight? A N (515, 26) P(A a) = 0.8. What is a?
62 Example 6 (from last Thursday) The net weight of a pack of coffee (500 grams) is a normally distributed RV with parameters µ = 505 grams and σ = 5 grams. (d) 80 % of the gross weight of the packs is heavier than a particular weight. What is that weight? A N (515, 26) P(A a) = 0.8. What is a? Consider Z = A N (0, 1)
63 Example 6 (from last Thursday) The net weight of a pack of coffee (500 grams) is a normally distributed RV with parameters µ = 505 grams and σ = 5 grams. (d) 80 % of the gross weight of the packs is heavier than a particular weight. What is that weight? A N (515, 26) P(A a) = 0.8. What is a? Consider Z = A N (0, 1) P(A a)
64 Example 6 (from last Thursday) The net weight of a pack of coffee (500 grams) is a normally distributed RV with parameters µ = 505 grams and σ = 5 grams. (d) 80 % of the gross weight of the packs is heavier than a particular weight. What is that weight? A N (515, 26) P(A a) = 0.8. What is a? Consider Z = A N (0, 1) ) ( P(A a) = P Z a = 0.8
65 Example 6 (from last Thursday) The net weight of a pack of coffee (500 grams) is a normally distributed RV with parameters µ = 505 grams and σ = 5 grams. (d) 80 % of the gross weight of the packs is heavier than a particular weight. What is that weight? A N (515, 26) P(A a) = 0.8. What is a? Consider Z = A N (0, 1) ( P(A a) = P Z a 515 ) = ( P Z a 515 ) =
66 Example 6 (from last Thursday) The net weight of a pack of coffee (500 grams) is a normally distributed RV with parameters µ = 505 grams and σ = 5 grams. (d) 80 % of the gross weight of the packs is heavier than a particular weight. What is that weight? A N (515, 26) P(A a) = 0.8. What is a? Consider Z = A N (0, 1) ( P(A a) = P Z a 515 ) = ( P Z a 515 ) = But P (Z z) = 0.2 for z = 0.84
67 Example 6 (from last Thursday) The net weight of a pack of coffee (500 grams) is a normally distributed RV with parameters µ = 505 grams and σ = 5 grams. (d) 80 % of the gross weight of the packs is heavier than a particular weight. What is that weight? A N (515, 26) P(A a) = 0.8. What is a? Consider Z = A N (0, 1) ( P(A a) = P Z a 515 ) = ( P Z a 515 ) = But P (Z z) = 0.2 for z = 0.84 Apparently a = 0.84 a = = 510.7
68 Example 6 (from last Thursday) The net weight of a pack of coffee (500 grams) is a normally distributed RV with parameters µ = 505 grams and σ = 5 grams. (d) 80 % of the gross weight of the packs is heavier than a particular weight. What is that weight? A N (515, 26) P(A a) = 0.8. What is a? Consider Z = A N (0, 1) ( P(A a) = P Z a 515 ) = ( P Z a 515 ) = But P (Z z) = 0.2 for z = 0.84 Apparently a = 0.84 a = = In general...
69 Example 6 (from last Thursday) The net weight of a pack of coffee (500 grams) is a normally distributed RV with parameters µ = 505 grams and σ = 5 grams. (d) 80 % of the gross weight of the packs is heavier than a particular weight. What is that weight? A N (515, 26) P(A a) = 0.8. What is a? Consider Z = A N (0, 1) ( P(A a) = P Z a 515 ) = ( P Z a 515 ) = But P (Z z) = 0.2 for z = 0.84 Apparently a = 0.84 a = = In general... X N (µ, σ) and P(X x) = p, P ( ) Z z = x µ σ = p
70 Example 6 (from last Thursday) The net weight of a pack of coffee (500 grams) is a normally distributed RV with parameters µ = 505 grams and σ = 5 grams. (d) 80 % of the gross weight of the packs is heavier than a particular weight. What is that weight? A N (515, 26) P(A a) = 0.8. What is a? Consider Z = A N (0, 1) ( P(A a) = P Z a 515 ) = ( P Z a 515 ) = But P (Z z) = 0.2 for z = 0.84 Apparently a = 0.84 a = = In general... X N (µ, σ) and P(X x) = p, P ( ) Z z = x µ σ = p Look up z in table A3 and then calculate x = σ x + µ
71 Example 6 (from last Thursday)
72 Example 6 (from last Thursday) The net weight of a pack of coffee (500 grams) is a normally distributed RV with parameters µ = 505 grams and σ = 5 grams.
73 Example 6 (from last Thursday) The net weight of a pack of coffee (500 grams) is a normally distributed RV with parameters µ = 505 grams and σ = 5 grams. (e) What is the probability that the difference of net weights of two packs is lower or equal than two grams?
74 Example 6 (from last Thursday) The net weight of a pack of coffee (500 grams) is a normally distributed RV with parameters µ = 505 grams and σ = 5 grams. (e) What is the probability that the difference of net weights of two packs is lower or equal than two grams? X 1, X 2 N (505, 5)
75 Example 6 (from last Thursday) The net weight of a pack of coffee (500 grams) is a normally distributed RV with parameters µ = 505 grams and σ = 5 grams. (e) What is the probability that the difference of net weights of two packs is lower or equal than two grams? X 1, X 2 N (505, 5) Q : P( X 1 X 2 2) = P( 2 X 1 X 2 2)
76 Example 6 (from last Thursday) The net weight of a pack of coffee (500 grams) is a normally distributed RV with parameters µ = 505 grams and σ = 5 grams. (e) What is the probability that the difference of net weights of two packs is lower or equal than two grams? X 1, X 2 N (505, 5) Q : P( X 1 X 2 2) = P( 2 X 1 X 2 2) X 2 N (µ X1 µ X2 ), σ X1 X 2 = V (X 1 X 2 )
77 Example 6 (from last Thursday) The net weight of a pack of coffee (500 grams) is a normally distributed RV with parameters µ = 505 grams and σ = 5 grams. (e) What is the probability that the difference of net weights of two packs is lower or equal than two grams? X 1, X 2 N (505, 5) Q : P( X 1 X 2 2) = P( 2 X 1 X 2 2) X 2 N (µ X1 µ X2 ), σ X1 X 2 = V (X 1 X 2 ) V (X 1 X 2 ) = V (X 1 ) + V ( X 2 ) = 1 2 V (X 1 ) + ( 1) 2 V (X 2 ) = V (X 1 ) + V (X 2 )
78 Example 6 (from last Thursday) The net weight of a pack of coffee (500 grams) is a normally distributed RV with parameters µ = 505 grams and σ = 5 grams. (e) What is the probability that the difference of net weights of two packs is lower or equal than two grams? X 1, X 2 N (505, 5) Q : P( X 1 X 2 2) = P( 2 X 1 X 2 2) X 2 N (µ X1 µ X2 ), σ X1 X 2 = V (X 1 X 2 ) V (X 1 X 2 ) = V (X 1 ) + V ( X 2 ) = 1 2 V (X 1 ) + ( 1) 2 V (X 2 ) = V (X 1 ) + V (X 2 ) Hence X 1 X 2 N (0, σx σx 2 2 )
79 Example 6 (from last Thursday) The net weight of a pack of coffee (500 grams) is a normally distributed RV with parameters µ = 505 grams and σ = 5 grams. (e) What is the probability that the difference of net weights of two packs is lower or equal than two grams? X 1, X 2 N (505, 5) Q : P( X 1 X 2 2) = P( 2 X 1 X 2 2) X 2 N (µ X1 µ X2 ), σ X1 X 2 = V (X 1 X 2 ) V (X 1 X 2 ) = V (X 1 ) + V ( X 2 ) = 1 2 V (X 1 ) + ( 1) 2 V (X 2 ) = V (X 1 ) + V (X 2 ) ( Hence X 1 X 2 N (0, σx σx 2 2 ) = N 0, ) ( = N 0, ) 50
80 Example 6 (from last Thursday) The net weight of a pack of coffee (500 grams) is a normally distributed RV with parameters µ = 505 grams and σ = 5 grams. (e) What is the probability that the difference of net weights of two packs is lower or equal than two grams? X 1, X 2 N (505, 5) Q : P( X 1 X 2 2) = P( 2 X 1 X 2 2) X 2 N (µ X1 µ X2 ), σ X1 X 2 = V (X 1 X 2 ) V (X 1 X 2 ) = V (X 1 ) + V ( X 2 ) = 1 2 V (X 1 ) + ( 1) 2 V (X 2 ) = V (X 1 ) + V (X 2 ) ( Hence X 1 X 2 N (0, σx σx 2 2 ) = N 0, ) ( = N 0, ) 50 Z = (x 1 x 2 ) 0 50 N (0, 1)
81 Example 6 (from last Thursday) The net weight of a pack of coffee (500 grams) is a normally distributed RV with parameters µ = 505 grams and σ = 5 grams. (e) What is the probability that the difference of net weights of two packs is lower or equal than two grams? X 1, X 2 N (505, 5) Q : P( X 1 X 2 2) = P( 2 X 1 X 2 2) X 2 N (µ X1 µ X2 ), σ X1 X 2 = V (X 1 X 2 ) V (X 1 X 2 ) = V (X 1 ) + V ( X 2 ) = 1 2 V (X 1 ) + ( 1) 2 V (X 2 ) = V (X 1 ) + V (X 2 ) ( Hence X 1 X 2 N (0, σx σx 2 2 ) = N 0, ) ( = N 0, ) 50 Z = (x 1 x 2 ) 0 50 N (0, 1) P ( 2 X 1 X 2 2) = P(X 1 X 2 2) P(X 1 X 2 2)
82 Example 6 (from last Thursday) The net weight of a pack of coffee (500 grams) is a normally distributed RV with parameters µ = 505 grams and σ = 5 grams. (e) What is the probability that the difference of net weights of two packs is lower or equal than two grams? X 1, X 2 N (505, 5) Q : P( X 1 X 2 2) = P( 2 X 1 X 2 2) X 2 N (µ X1 µ X2 ), σ X1 X 2 = V (X 1 X 2 ) V (X 1 X 2 ) = V (X 1 ) + V ( X 2 ) = 1 2 V (X 1 ) + ( 1) 2 V (X 2 ) = V (X 1 ) + V (X 2 ) ( Hence X 1 X 2 N (0, σx σx 2 2 ) = N 0, ) ( = N 0, ) 50 Z = (x 1 x 2 ) 0 50 N (0, 1) P ( 2 ( X 1 ) X 2 ( 2) = P(X 1 ) X 2 2) P(X 1 X 2 2) P Z P Z
83 Example 6 (from last Thursday) The net weight of a pack of coffee (500 grams) is a normally distributed RV with parameters µ = 505 grams and σ = 5 grams. (e) What is the probability that the difference of net weights of two packs is lower or equal than two grams? X 1, X 2 N (505, 5) Q : P( X 1 X 2 2) = P( 2 X 1 X 2 2) X 2 N (µ X1 µ X2 ), σ X1 X 2 = V (X 1 X 2 ) V (X 1 X 2 ) = V (X 1 ) + V ( X 2 ) = 1 2 V (X 1 ) + ( 1) 2 V (X 2 ) = V (X 1 ) + V (X 2 ) ( Hence X 1 X 2 N (0, σx σx 2 2 ) = N 0, ) ( = N 0, ) 50 Z = (x 1 x 2 ) 0 50 N (0, 1) P ( 2 ( X 1 ) X 2 ( 2) = P(X 1 ) X 2 2) P(X 1 X 2 2) P Z P Z P (Z 0.28) P (Z 0.28) = =
84 Observation
85 Observation Let Z N (0, 1). For cumulative density of the RV X = Z 2 we have:
86 Observation Let Z N (0, 1). For cumulative density of the RV X = Z 2 we have: F(x) = P(X x) = P(Z 2 x) = P( z x) = P( x Z x) = G( x) G( x)
87 Observation Let Z N (0, 1). For cumulative density of the RV X = Z 2 we have: F(x) = P(X x) = P(Z 2 x) = P( z x) = P( x Z x) = G( x) G( x) where G is the cumulative density of Z. So
88 Observation Let Z N (0, 1). For cumulative density of the RV X = Z 2 we have: F(x) = P(X x) = P(Z 2 x) = P( z x) = P( x Z x) = G( x) G( x) where G is the cumulative density of Z. So f (x) = F (x) = 1 2 x 1 2 G ( x) ( 1 2 )x 1 2 G ( x)
89 Observation Let Z N (0, 1). For cumulative density of the RV X = Z 2 we have: F(x) = P(X x) = P(Z 2 x) = P( z x) = P( x Z x) = G( x) G( x) where G is the cumulative density of Z. So f (x) = F (x) = 1 2 x 1 2 G ( x) ( 1 2 )x 1 2 G ( x) = 1 2 x π exp( x ) + 1 x π exp( x ) 2
90 Observation Let Z N (0, 1). For cumulative density of the RV X = Z 2 we have: F(x) = P(X x) = P(Z 2 x) = P( z x) = P( x Z x) = G( x) G( x) where G is the cumulative density of Z. So f (x) = F (x) = 1 2 x 1 2 G ( x) ( 1 2 )x 1 2 G ( x) = 1 2 x π exp( x ) + 1 x π exp( x )= x 2 exp( x 2 ) 2 4 π
91 Observation Let Z N (0, 1). For cumulative density of the RV X = Z 2 we have: F(x) = P(X x) = P(Z 2 x) = P( z x) = P( x Z x) = G( x) G( x) where G is the cumulative density of Z. So f (x) = F (x) = 1 2 x 1 2 G ( x) ( 1 2 )x 1 2 G ( x) = 1 2 x π exp( x ) + 1 x π exp( x )= x 2 exp( x 2 ) 2 4 π Apparently: Γ( 1 2 ) = π and X = Z 2 has a Gamma-distribution with parameters λ = 1 2 (β = 2) and α = 1 2
92 And now... 1 Erlang distribution Formulation Application of Erlang distribution Gamma distribution 2 Various Exercises 3 Chi-squared distribution Basics Applications Examples
93 Basics Formulation
94 Basics Formulation If X 1, X 2,..., X n are IID, X i N (0, 1), then χ 2 def = X X X 2 n
95 Basics Formulation If X 1, X 2,..., X n are IID, X i N (0, 1), then χ 2 def = X X X 2 n has a Gamma-distribution with parameters α = n/2 and λ = 1/2 (β = 2)
96 Basics Formulation If X 1, X 2,..., X n are IID, X i N (0, 1), then χ 2 def = X X X 2 n has a Gamma-distribution with parameters α = n/2 and λ = 1/2 (β = 2) This distribution is also called a Chi-squared distribution with n degrees of freedom
97 Basics Formulation If X 1, X 2,..., X n are IID, X i N (0, 1), then χ 2 def = X X X 2 n has a Gamma-distribution with parameters α = n/2 and λ = 1/2 (β = 2) This distribution is also called a Chi-squared distribution with n degrees of freedom Notation χ 2 χ 2 n
98 Basics Formulation If X 1, X 2,..., X n are IID, X i N (0, 1), then χ 2 def = X X X 2 n has a Gamma-distribution with parameters α = n/2 and λ = 1/2 (β = 2) This distribution is also called a Chi-squared distribution with n degrees of freedom Notation χ 2 χ 2 n Book: Table A5: X s.t. P(χ 2 x) = α for several values of v ( degrees of freedom) and α
99 Applications Estimation of σ
100 Applications Estimation of σ Let X 1, X 2,..., X n N (µ, σ) IDD with known µ and unknown σ
101 Applications Estimation of σ Let X 1, X 2,..., X n N (µ, σ) IDD with known µ and unknown σ Then Z i = X i µ σ N (0, 1)
102 Applications Estimation of σ Let X 1, X 2,..., X n N (µ, σ) IDD with known µ and unknown σ Then Z i = X i µ N (0, 1) σ Zi 2 = (X i µ) 2 χ 2 σ 2 i
103 Applications Estimation of σ Let X 1, X 2,..., X n N (µ, σ) IDD with known µ and unknown σ Then Z i = X i µ N (0, 1) σ Zi 2 = (X i µ) 2 χ 2 σ 2 i χ 2 def = n i=1 Z i 2 = 1 n σ 2 i=1 (X i µ) 2 χ 2 n
104 Examples Example 7
105 Examples Example 7 Let X 1, X 2,..., X n N (µ, σ) IDD with µ = 5 and unknown σ. The realizations for X 1,..., X 5 are 3.5, 5.7, 1.2, 6.8, and 7.1. What values of σ are reasonable?
106 Examples Example 7 Let X 1, X 2,..., X n N (µ, σ) IDD with µ = 5 and unknown σ. The realizations for X 1,..., X 5 are 3.5, 5.7, 1.2, 6.8, and 7.1. What values of σ are reasonable? Let χ 2 def = 1 5 σ 2 i=1 (X i µ) 2
107 Examples Example 7 Let X 1, X 2,..., X n N (µ, σ) IDD with µ = 5 and unknown σ. The realizations for X 1,..., X 5 are 3.5, 5.7, 1.2, 6.8, and 7.1. What values of σ are reasonable? Let χ 2 def = 1 5 σ 2 i=1 (X i µ) 2 = 1 (( 1.5) 2 + (0.7) 2 + ( 3.8) 2 + (1.8) 2 + (2.1) 2 ) = σ 2 σ 2
108 Examples Example 7 Let X 1, X 2,..., X n N (µ, σ) IDD with µ = 5 and unknown σ. The realizations for X 1,..., X 5 are 3.5, 5.7, 1.2, 6.8, and 7.1. What values of σ are reasonable? Let χ 2 def = 1 5 σ 2 i=1 (X i µ) 2 = 1 (( 1.5) 2 + (0.7) 2 + ( 3.8) 2 + (1.8) 2 + (2.1) 2 ) = σ 2 σ 2 Table A5: P(χ ) = and P(χ ) = 0.975
109 Examples Example 7 Let X 1, X 2,..., X n N (µ, σ) IDD with µ = 5 and unknown σ. The realizations for X 1,..., X 5 are 3.5, 5.7, 1.2, 6.8, and 7.1. What values of σ are reasonable? Let χ 2 def = 1 5 σ 2 i=1 (X i µ) 2 = 1 (( 1.5) 2 + (0.7) 2 + ( 3.8) 2 + (1.8) 2 + (2.1) 2 ) = σ 2 σ 2 Table A5: P(χ ) = and P(χ ) = So (fill the realization for χ 2 :) P( σ ) = = 0.95 P(σ 2 [1.935, 29.88]) = 0.95
110 Examples Example 7 Let X 1, X 2,..., X n N (µ, σ) IDD with µ = 5 and unknown σ. The realizations for X 1,..., X 5 are 3.5, 5.7, 1.2, 6.8, and 7.1. What values of σ are reasonable? Let χ 2 def = 1 5 σ 2 i=1 (X i µ) 2 = 1 (( 1.5) 2 + (0.7) 2 + ( 3.8) 2 + (1.8) 2 + (2.1) 2 ) = σ 2 σ 2 Table A5: P(χ ) = and P(χ ) = So (fill the realization for χ 2 :) P( σ ) = = 0.95 P(σ 2 [1.935, 29.88]) = 0.95 P(σ [1.391, 5.466]) = 0.95
111 Examples Example 7 Let X 1, X 2,..., X n N (µ, σ) IDD with µ = 5 and unknown σ. The realizations for X 1,..., X 5 are 3.5, 5.7, 1.2, 6.8, and 7.1. What values of σ are reasonable? Let χ 2 def = 1 5 σ 2 i=1 (X i µ) 2 = 1 (( 1.5) 2 + (0.7) 2 + ( 3.8) 2 + (1.8) 2 + (2.1) 2 ) = σ 2 σ 2 Table A5: P(χ ) = and P(χ ) = So (fill the realization for χ 2 :) P( σ ) = = 0.95 P(σ 2 [1.935, 29.88]) = 0.95 P(σ [1.391, 5.466]) = 0.95 Apparently reasonable values for σ are between and We call [1.391, 5.466] a 95 % confidence interval for σ
112 Examples Example 7 Let X 1, X 2,..., X n N (µ, σ) IDD with µ = 5 and unknown σ. The realizations for X 1,..., X 5 are 3.5, 5.7, 1.2, 6.8, and 7.1. What values of σ are reasonable? Let χ 2 def = 1 5 σ 2 i=1 (X i µ) 2 = 1 (( 1.5) 2 + (0.7) 2 + ( 3.8) 2 + (1.8) 2 + (2.1) 2 ) = σ 2 σ 2 Table A5: P(χ ) = and P(χ ) = So (fill the realization for χ 2 :) P( σ ) = = 0.95 P(σ 2 [1.935, 29.88]) = 0.95 P(σ [1.391, 5.466]) = 0.95 Apparently reasonable values for σ are between and We call [1.391, 5.466] a 95 % confidence interval for σ Similarly, [1.935, 29.88] is a 95 % confidence interval for σ 2
113 Examples Example 8: What if µ is unknown as well?
114 Examples Example 8: What if µ is unknown as well? Data: X 1, X 2,..., X n N (µ, σ) IDD with µ, σ unknown.
115 Examples Example 8: What if µ is unknown as well? Data: X 1, X 2,..., X n N (µ, σ) IDD with µ, σ unknown. Some calculation show: n (X i µ) 2 i=1
116 Examples Example 8: What if µ is unknown as well? Data: X 1, X 2,..., X n N (µ, σ) IDD with µ, σ unknown. Some calculation show: n (X i µ) 2 = i=1 n ((x i x) + ( x µ)) 2 i=1
117 Examples Example 8: What if µ is unknown as well? Data: X 1, X 2,..., X n N (µ, σ) IDD with µ, σ unknown. Some calculation show: n (X i µ) 2 = i=1 = n ((x i x) + ( x µ)) 2 i=1 n n (x i x) (x i x)( x µ) + i=1 i=1 n ( x µ) 2 i=1
118 Examples Example 8: What if µ is unknown as well? Data: X 1, X 2,..., X n N (µ, σ) IDD with µ, σ unknown. Some calculation show: n (X i µ) 2 = i=1 = = n ((x i x) + ( x µ)) 2 i=1 n n (x i x) (x i x)( x µ) + i=1 i=1 n (x i x) 2 + n ( x µ) 2 i=1 n ( x µ) 2 i=1
119 Examples Example 8: What if µ is unknown as well? Data: X 1, X 2,..., X n N (µ, σ) IDD with µ, σ unknown. Some calculation show: n (X i µ) 2 = i=1 = = n ((x i x) + ( x µ)) 2 i=1 n n (x i x) (x i x)( x µ) + i=1 i=1 n (x i x) 2 + n ( x µ) 2 i=1 Now, since s 2 = 1 n 1 n i=1 (x i x) 2, we have n ( x µ) 2 i=1
120 Examples Example 8: What if µ is unknown as well? Data: X 1, X 2,..., X n N (µ, σ) IDD with µ, σ unknown. Some calculation show: n (X i µ) 2 = i=1 = = n ((x i x) + ( x µ)) 2 i=1 n n (x i x) (x i x)( x µ) + i=1 i=1 n (x i x) 2 + n ( x µ) 2 i=1 Now, since s 2 = 1 n n 1 i=1 (x i x) 2, we have 1 n (x i µ) 2 σ 2 i=1 }{{} χ 2 n = (n 1) s2 σ 2 + ( x µ)2 σ 2 /n n ( x µ) 2 i=1
121 Examples Example 8: What if µ is unknown as well? Data: X 1, X 2,..., X n N (µ, σ) IDD with µ, σ unknown. Some calculation show: n (X i µ) 2 = i=1 = = n ((x i x) + ( x µ)) 2 i=1 n n (x i x) (x i x)( x µ) + i=1 i=1 n (x i x) 2 + n ( x µ) 2 i=1 Now, since s 2 = 1 n n 1 i=1 (x i x) 2, we have 1 n (x i µ) 2 X = 1 n n i=1 x i N (µ, σ 2 i=1 }{{} χ 2 n = (n 1) s2 σ 2 + ( x µ)2 σ 2 /n n ( x µ) 2 i=1 σ n )
122 Examples Example 8: What if µ is unknown as well? Data: X 1, X 2,..., X n N (µ, σ) IDD with µ, σ unknown. Some calculation show: n (X i µ) 2 = i=1 = = n ((x i x) + ( x µ)) 2 i=1 n n (x i x) (x i x)( x µ) + i=1 i=1 n (x i x) 2 + n ( x µ) 2 i=1 Now, since s 2 = 1 n n 1 i=1 (x i x) 2, we have 1 n (x i µ) 2 X = 1 n n i=1 x i N (µ, σ 2 i=1 }{{} χ 2 n = (n 1) s2 σ 2 + ( x µ)2 σ 2 /n n ( x µ) 2 i=1 σ n )
123 Examples Example 8: What if µ is unknown as well? Data: X 1, X 2,..., X n N (µ, σ) IDD with µ, σ unknown. Some calculation show: n (X i µ) 2 = i=1 = = n ((x i x) + ( x µ)) 2 i=1 n n (x i x) (x i x)( x µ) + i=1 i=1 n (x i x) 2 + n ( x µ) 2 i=1 Now, since s 2 = 1 n n 1 i=1 (x i x) 2, we have 1 n (x i µ) 2 X = 1 n n i=1 x i N (µ, σ 2 i=1 }{{} = (n 1) s2 σ 2 + ( x µ)2 σ 2 /n n ( x µ) 2 i=1 σ n ) χ 2 n ( X µ) 2 σ 2 / n = ( X µ σ/ n ) 2 χ 2 1
124 Examples Example 8: What if µ is unknown as well? Data: X 1, X 2,..., X n N (µ, σ) IDD with µ, σ unknown. Some calculation show: n (X i µ) 2 = i=1 = = n ((x i x) + ( x µ)) 2 i=1 n n (x i x) (x i x)( x µ) + i=1 i=1 n (x i x) 2 + n ( x µ) 2 i=1 Now, since s 2 = 1 n n 1 i=1 (x i x) 2, we have 1 n (x i µ) 2 X = 1 n n i=1 x i N (µ, σ 2 i=1 }{{} = (n 1) s2 σ 2 + ( x µ)2 σ 2 /n n ( x µ) 2 i=1 σ n ) χ 2 n ( X µ) 2 σ 2 / n = ( X µ σ/ n ) 2 χ 2 1 the statistics
125 Examples Example 8: What if µ is unknown as well? Data: X 1, X 2,..., X n N (µ, σ) IDD with µ, σ unknown. Some calculation show: n (X i µ) 2 = i=1 = = n ((x i x) + ( x µ)) 2 i=1 n n (x i x) (x i x)( x µ) + i=1 i=1 n (x i x) 2 + n ( x µ) 2 i=1 Now, since s 2 = 1 n n 1 i=1 (x i x) 2, we have 1 n (x i µ) 2 X = 1 n n i=1 x i N (µ, σ 2 i=1 }{{} = (n 1) s2 σ 2 + ( x µ)2 σ 2 /n n ( x µ) 2 i=1 σ n ) χ 2 n ( X µ) 2 σ 2 / n χ 2 = (n 1) s2 σ 2 = ( X µ σ/ n ) 2 χ 2 1 the statistics = n (x i x) 2 i=1 χ 2 σ 2 n 1
126 Examples Example 7 (b)
127 Examples Example 7 (b) Let X 1, X 2,..., X n N (µ, σ) IDD with µ = 5 and unknown σ. The realizations for X 1,..., X 5 are 3.5, 5.7, 1.2, 6.8, and 7.1. What values of σ are reasonable
128 Examples Example 7 (b) Let X 1, X 2,..., X n N (µ, σ) IDD with µ = 5 and unknown σ. The realizations for X 1,..., X 5 are 3.5, 5.7, 1.2, 6.8, and 7.1. What values of σ are reasonable Let χ 2 def = (n 1)s2 χ 2 σ 2 n 1 = χ 2 4
129 Examples Example 7 (b) Let X 1, X 2,..., X n N (µ, σ) IDD with µ = 5 and unknown σ. The realizations for X 1,..., X 5 are 3.5, 5.7, 1.2, 6.8, and 7.1. What values of σ are reasonable Let χ 2 def = (n 1)s2 χ 2 σ 2 n 1 = χ 2 4 X = 1 ( ) =
130 Examples Example 7 (b) Let X 1, X 2,..., X n N (µ, σ) IDD with µ = 5 and unknown σ. The realizations for X 1,..., X 5 are 3.5, 5.7, 1.2, 6.8, and 7.1. What values of σ are reasonable Let χ 2 def = (n 1)s2 χ 2 σ 2 n 1 = χ 2 4 X = 1 ( ) = (n 1)s 2 = n i=1 (x i x) 2 =
131 Examples Example 7 (b) Let X 1, X 2,..., X n N (µ, σ) IDD with µ = 5 and unknown σ. The realizations for X 1,..., X 5 are 3.5, 5.7, 1.2, 6.8, and 7.1. What values of σ are reasonable Let χ 2 def = (n 1)s2 χ 2 σ 2 n 1 = χ 2 4 X = 1 ( ) = (n 1)s 2 = n i=1 (x i x) 2 = χ 2 = σ 2
132 Examples Example 7 (b) Let X 1, X 2,..., X n N (µ, σ) IDD with µ = 5 and unknown σ. The realizations for X 1,..., X 5 are 3.5, 5.7, 1.2, 6.8, and 7.1. What values of σ are reasonable Let χ 2 def = (n 1)s2 χ 2 σ 2 n 1 = χ 2 4 X = 1 ( ) = (n 1)s 2 = n i=1 (x i x) 2 = χ 2 = σ 2 Construct a 95% confidence interval for σ :
133 Examples Example 7 (b) Let X 1, X 2,..., X n N (µ, σ) IDD with µ = 5 and unknown σ. The realizations for X 1,..., X 5 are 3.5, 5.7, 1.2, 6.8, and 7.1. What values of σ are reasonable Let χ 2 def = (n 1)s2 χ 2 σ 2 n 1 = χ 2 4 X = 1 ( ) = (n 1)s 2 = n i=1 (x i x) 2 = χ 2 = σ 2 Construct a 95% confidence interval for σ : Table A5: P(χ ) = and P(χ ) = 0.975
134 Examples Example 7 (b) Let X 1, X 2,..., X n N (µ, σ) IDD with µ = 5 and unknown σ. The realizations for X 1,..., X 5 are 3.5, 5.7, 1.2, 6.8, and 7.1. What values of σ are reasonable Let χ 2 def = (n 1)s2 χ 2 σ 2 n 1 = χ 2 4 X = 1 ( ) = (n 1)s 2 = n i=1 (x i x) 2 = χ 2 = σ 2 Construct a 95% confidence interval for σ : Table A5: P(χ ) = and P(χ ) = P( /σ ) = 0.95
135 Examples Example 7 (b) Let X 1, X 2,..., X n N (µ, σ) IDD with µ = 5 and unknown σ. The realizations for X 1,..., X 5 are 3.5, 5.7, 1.2, 6.8, and 7.1. What values of σ are reasonable Let χ 2 def = (n 1)s2 χ 2 σ 2 n 1 = χ 2 4 X = 1 ( ) = (n 1)s 2 = n i=1 (x i x) 2 = χ 2 = σ 2 Construct a 95% confidence interval for σ : Table A5: P(χ ) = and P(χ ) = P( /σ ) = 0.95 P(24.732/ σ /0.484) = 0.95
136 Examples Example 7 (b) Let X 1, X 2,..., X n N (µ, σ) IDD with µ = 5 and unknown σ. The realizations for X 1,..., X 5 are 3.5, 5.7, 1.2, 6.8, and 7.1. What values of σ are reasonable Let χ 2 def = (n 1)s2 χ 2 σ 2 n 1 = χ 2 4 X = 1 ( ) = (n 1)s 2 = n i=1 (x i x) 2 = χ 2 = σ 2 Construct a 95% confidence interval for σ : Table A5: P(χ ) = and P(χ ) = P( /σ ) = 0.95 P(24.732/ σ /0.484) = % CI for σ 2 : [24.732/11.143, /0.484] = [0.220, ]
137 Examples Example 7 (b) Let X 1, X 2,..., X n N (µ, σ) IDD with µ = 5 and unknown σ. The realizations for X 1,..., X 5 are 3.5, 5.7, 1.2, 6.8, and 7.1. What values of σ are reasonable Let χ 2 def = (n 1)s2 χ 2 σ 2 n 1 = χ 2 4 X = 1 ( ) = (n 1)s 2 = n i=1 (x i x) 2 = χ 2 = σ 2 Construct a 95% confidence interval for σ : Table A5: P(χ ) = and P(χ ) = P( /σ ) = 0.95 P(24.732/ σ /0.484) = % CI for σ 2 : [24.732/11.143, /0.484] = [0.220, ] 95% CI for σ: [1.490, 7.148]
138 Examples Example 7 (b) Let X 1, X 2,..., X n N (µ, σ) IDD with µ = 5 and unknown σ. The realizations for X 1,..., X 5 are 3.5, 5.7, 1.2, 6.8, and 7.1. What values of σ are reasonable Let χ 2 def = (n 1)s2 χ 2 σ 2 n 1 = χ 2 4 X = 1 ( ) = (n 1)s 2 = n i=1 (x i x) 2 = χ 2 = σ 2 Construct a 95% confidence interval for σ : Table A5: P(χ ) = and P(χ ) = P( /σ ) = 0.95 P(24.732/ σ /0.484) = % CI for σ 2 : [24.732/11.143, /0.484] = [0.220, ] 95% CI for σ: [1.490, 7.148] Important!: These calculations use the fact that X i s are normally distributed. If it is not the case, we cannot use χ 2 -distributions!
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