The Normal Distribution

Page 1

The Normal distribution A random variable Z is said to follow the standard Normal distribution if it has pdf f (z) =

The Normal distribution A random variable Z is said to follow the standard Normal distribution if it has pdf f (z) =

1 √ exp − z 2 2 2Ď€

The Normal distribution A random variable Z is said to follow the standard Normal distribution if it has pdf f (z) =

1 √ exp − z 2 2 2Ď€

−∞<z <∞

The Normal distribution A random variable Z is said to follow the standard Normal distribution if it has pdf f (z) =

1 √ exp − z 2 2 2Ď€

−∞<z <∞

The standard Normal pdf (see handout) is unimodal, bell-shaped, and symmetrical about z = 0.

The Normal distribution A random variable Z is said to follow the standard Normal distribution if it has pdf f (z) =

1 √ exp − z 2 2 2Ď€

−∞<z <∞

The standard Normal pdf (see handout) is unimodal, bell-shaped, and symmetrical about z = 0. From symmetry:

E[Z] =

The Normal distribution A random variable Z is said to follow the standard Normal distribution if it has pdf f (z) =

1 √ exp − z 2 2 2Ď€

−∞<z <∞

The standard Normal pdf (see handout) is unimodal, bell-shaped, and symmetrical about z = 0. From symmetry:

E[Z] = 0

The Normal distribution A random variable Z is said to follow the standard Normal distribution if it has pdf f (z) =

1 √ exp − z 2 2 2Ď€

−∞<z <∞

The standard Normal pdf (see handout) is unimodal, bell-shaped, and symmetrical about z = 0. From symmetry: Can also show:

E[Z] = 0 Variance[Z] =

The Normal distribution A random variable Z is said to follow the standard Normal distribution if it has pdf f (z) =

1 √ exp − z 2 2 2Ď€

−∞<z <∞

The standard Normal pdf (see handout) is unimodal, bell-shaped, and symmetrical about z = 0. From symmetry: Can also show:

E[Z] = 0 Variance[Z] = 1

The Normal distribution More generally, the Normal probability density function with mean µ and standard deviation σ is f (x) =

for − ∞ < x < ∞

The Normal distribution More generally, the Normal probability density function with mean µ and standard deviation σ is ! 1 1 x−µ 2 √ exp − for − ∞ < x < ∞ f (x) = 2 σ σ 2π

The Normal distribution More generally, the Normal probability density function with mean µ and standard deviation σ is ! 1 1 x−µ 2 √ exp − for − ∞ < x < ∞ f (x) = 2 σ σ 2π Notation: If X has the above pdf, we write

The Normal distribution More generally, the Normal probability density function with mean µ and standard deviation σ is ! 1 1 x−µ 2 √ exp − for − ∞ < x < ∞ f (x) = 2 σ σ 2π Notation: If X has the above pdf, we write X ∼ N µ, σ 2

The Normal distribution More generally, the Normal probability density function with mean µ and standard deviation σ is ! 1 1 x−µ 2 √ exp − for − ∞ < x < ∞ f (x) = 2 σ σ 2π Notation: If X has the above pdf, we write X ∼ N µ, σ 2

There are two parameters,

13

The Normal distribution More generally, the Normal probability density function with mean µ and standard deviation σ is ! 1 1 x−µ 2 √ exp − for − ∞ < x < ∞ f (x) = 2 σ σ 2π Notation: If X has the above pdf, we write X ∼ N µ, σ 2

There are two parameters, the mean µ with −∞ < µ < ∞

14

The Normal distribution More generally, the Normal probability density function with mean µ and standard deviation σ is ! 1 1 x−µ 2 √ exp − for − ∞ < x < ∞ f (x) = 2 σ σ 2π Notation: If X has the above pdf, we write X ∼ N µ, σ 2

There are two parameters, the mean µ with −∞ < µ < ∞ and the standard deviation σ with σ > 0

15

The Normal distribution More generally, the Normal probability density function with mean µ and standard deviation σ is ! 1 1 x−µ 2 √ exp − for − ∞ < x < ∞ f (x) = 2 σ σ 2π Notation: If X has the above pdf, we write X ∼ N µ, σ 2

There are two parameters, the mean µ with −∞ < µ < ∞ and the standard deviation σ with σ > 0 The standard Normal distribution is denoted

16

The Normal distribution More generally, the Normal probability density function with mean µ and standard deviation σ is ! 1 1 x−µ 2 √ exp − for − ∞ < x < ∞ f (x) = 2 σ σ 2π Notation: If X has the above pdf, we write X ∼ N µ, σ 2

There are two parameters, the mean µ with −∞ < µ < ∞ and the standard deviation σ with σ > 0 The standard Normal distribution is denoted N (0, 1)

17

The Normal distribution

The distribution function of the standard Normal distribution is denoted by ÎŚ. That is,

The Normal distribution

The distribution function of the standard Normal distribution is denoted by Φ. That is, Z z Φ(z) = du −∞

The Normal distribution

The distribution function of the standard Normal distribution is denoted by Φ. That is, Z z 1 √ exp −u2 2 du Φ(z) = 2Ï€ −∞

The Normal distribution

The distribution function of the standard Normal distribution is denoted by ÎŚ. That is, Z z 1 √ exp −u2 2 du ÎŚ(z) = 2Ď€ −∞

Values of ÎŚ(z) are obtained numerically using tables (handout).

21

The Normal distribution

The distribution function of the standard Normal distribution is denoted by ÎŚ. That is, Z z 1 √ exp −u2 2 du ÎŚ(z) = 2Ď€ −∞

Values of ÎŚ(z) are obtained numerically using tables (handout). To use the tables for Normal distributions other than N (0, 1), need 2 to convert from N Âľ, Ďƒ to N (0, 1) as follows.

22

The Normal distribution If X ∼ N µ, σ 2 , and we define

23

The Normal distribution If X ∼ N µ, σ 2 , and we define Z =

X−µ σ ,

24

The Normal distribution If X ∼ N µ, σ 2 , and we define Z =

X−µ σ ,

then Z ∼ N (0, 1).

25

The Normal distribution If X ∼ N µ, σ 2 , and we define Z = Hence P (X ≤ x) =

X−µ σ ,

then Z ∼ N (0, 1).

The Normal distribution If X ∼ N µ, σ 2 , and we define Z = X−µ σ , then Z ∼ N (0, 1). Hence x−µ P (X ≤ x) = P Z ≤ = σ

The Normal distribution If X ∼ N µ, σ 2 , and we define Z = X−µ σ , then Z ∼ N (0, 1). Hence x−µ x−µ P (X ≤ x) = P Z ≤ =Φ σ σ

The Normal distribution If X ∼ N µ, σ 2 , and we define Z = X−µ σ , then Z ∼ N (0, 1). Hence x−µ x−µ P (X ≤ x) = P Z ≤ =Φ σ σ Note:

29

The Normal distribution If X ∼ N µ, σ 2 , and we define Z = X−µ σ , then Z ∼ N (0, 1). Hence x−µ x−µ P (X ≤ x) = P Z ≤ =Φ σ σ Note: z =

x−µ σ

is often called a z-score.

30

The Normal distribution If X ∼ N µ, σ 2 , and we define Z = X−µ σ , then Z ∼ N (0, 1). Hence x−µ x−µ P (X ≤ x) = P Z ≤ =Φ σ σ Note: z =

x−µ σ

is often called a z-score.

Example: The heights of female students are Normally distributed with mean 169cm and standard deviation 9cm. Find the probability that a randomly chosen female student is over 187cm tall.

The Normal distribution Solution: Let X denote the height of the chosen student, so

The Normal distribution Solution: Let X denote the height of the chosen student, so X âˆź N 169, 92

33

The Normal distribution Solution: Let X denote the height of the chosen student, so X âˆź N 169, 92 Defining Z =

, then Z âˆź N (0, 1)

The Normal distribution Solution: Let X denote the height of the chosen student, so X ∼ N 169, 92 Defining Z =

X−169 , 9

then Z ∼ N (0, 1)

The Normal distribution Solution: Let X denote the height of the chosen student, so X ∼ N 169, 92 Defining Z =

X−169 , 9

then Z ∼ N (0, 1)

P (X > 187) = = = = = =

The Normal distribution Solution: Let X denote the height of the chosen student, so X ∼ N 169, 92 X−169 , 9

then Z ∼ N (0, 1) 187 − 169 P (X > 187) = P Z > 9 =

Defining Z =

= = = =

The Normal distribution Solution: Let X denote the height of the chosen student, so X ∼ N 169, 92 X−169 , 9

then Z ∼ N (0, 1) 187 − 169 P (X > 187) = P Z > 9 = P (Z > 2)

Defining Z =

= = = =

The Normal distribution Solution: Let X denote the height of the chosen student, so X ∼ N 169, 92 X−169 , 9

then Z ∼ N (0, 1) 187 − 169 P (X > 187) = P Z > 9 = P (Z > 2)

Defining Z =

= 1 − P (Z ≤ 2) = = =

The Normal distribution Solution: Let X denote the height of the chosen student, so X ∼ N 169, 92 X−169 , 9

then Z ∼ N (0, 1) 187 − 169 P (X > 187) = P Z > 9 = P (Z > 2)

Defining Z =

= 1 − P (Z ≤ 2) = 1 − Φ(2) = =

The Normal distribution Solution: Let X denote the height of the chosen student, so X ∼ N 169, 92 X−169 , 9

then Z ∼ N (0, 1) 187 − 169 P (X > 187) = P Z > 9 = P (Z > 2)

Defining Z =

= 1 − P (Z ≤ 2) = 1 − Φ(2) = 1 − 0.9772 =

The Normal distribution Solution: Let X denote the height of the chosen student, so X ∼ N 169, 92 X−169 , 9

then Z ∼ N (0, 1) 187 − 169 P (X > 187) = P Z > 9 = P (Z > 2)

Defining Z =

= 1 − P (Z ≤ 2) = 1 − Φ(2) = 1 − 0.9772 = 0.0228

The Normal distribution

Since the standard Normal pdf is symmetric about zero, values of Φ(z) only need to be tabulated for z ≥ 0. For negative argument, the value of Φ is obtained by symmetry as Φ(−a) =

The Normal distribution

Since the standard Normal pdf is symmetric about zero, values of Φ(z) only need to be tabulated for z ≥ 0. For negative argument, the value of Φ is obtained by symmetry as Φ(−a) = 1 − Φ(a)

The Normal distribution Example: Packets of breakfast cereal nominally contain 500 g. However, actual contents vary approximately Normally. If average contents are 508g and the SD is 5g, what is the probability of an underweight packet?

45

The Normal distribution Example: Packets of breakfast cereal nominally contain 500 g. However, actual contents vary approximately Normally. If average contents are 508g and the SD is 5g, what is the probability of an underweight packet? Solution: Write W for weight of packet, and take that Z âˆź N (0, 1)

, so

46

The Normal distribution Example: Packets of breakfast cereal nominally contain 500 g. However, actual contents vary approximately Normally. If average contents are 508g and the SD is 5g, what is the probability of an underweight packet? Solution: Write W for weight of packet, and take Z = that Z âˆź N (0, 1)

W −508 , 5

so

47

The Normal distribution Example: Packets of breakfast cereal nominally contain 500 g. However, actual contents vary approximately Normally. If average contents are 508g and the SD is 5g, what is the probability of an underweight packet? Solution: Write W for weight of packet, and take Z = that Z âˆź N (0, 1) P (W < 500) = = = = = =

W −508 , 5

so

The Normal distribution Example: Packets of breakfast cereal nominally contain 500 g. However, actual contents vary approximately Normally. If average contents are 508g and the SD is 5g, what is the probability of an underweight packet? Solution: Write W for weight of packet, and take Z = that Z ∼ N (0, 1) 500 − 508 P (W < 500) = P Z < 5 = = = = =

W −508 , 5

so

The Normal distribution Example: Packets of breakfast cereal nominally contain 500 g. However, actual contents vary approximately Normally. If average contents are 508g and the SD is 5g, what is the probability of an underweight packet? Solution: Write W for weight of packet, and take Z = that Z ∼ N (0, 1) 500 − 508 P (W < 500) = P Z < 5 = P (Z < −1.6) = = = =

W −508 , 5

so

The Normal distribution Example: Packets of breakfast cereal nominally contain 500 g. However, actual contents vary approximately Normally. If average contents are 508g and the SD is 5g, what is the probability of an underweight packet? Solution: Write W for weight of packet, and take Z = that Z ∼ N (0, 1) 500 − 508 P (W < 500) = P Z < 5 = P (Z < −1.6) = P (Z > 1.6) = = =

W −508 , 5

so

The Normal distribution Example: Packets of breakfast cereal nominally contain 500 g. However, actual contents vary approximately Normally. If average contents are 508g and the SD is 5g, what is the probability of an underweight packet? Solution: Write W for weight of packet, and take Z = that Z ∼ N (0, 1) 500 − 508 P (W < 500) = P Z < 5 = P (Z < −1.6) = P (Z > 1.6) = 1 − P (Z ≤ 1.6) = =

W −508 , 5

so

The Normal distribution Example: Packets of breakfast cereal nominally contain 500 g. However, actual contents vary approximately Normally. If average contents are 508g and the SD is 5g, what is the probability of an underweight packet? Solution: Write W for weight of packet, and take Z = that Z ∼ N (0, 1) 500 − 508 P (W < 500) = P Z < 5 = P (Z < −1.6) = P (Z > 1.6) = 1 − P (Z ≤ 1.6) = 1 − Φ(1.6) =

W −508 , 5

so

The Normal distribution Example: Packets of breakfast cereal nominally contain 500 g. However, actual contents vary approximately Normally. If average contents are 508g and the SD is 5g, what is the probability of an underweight packet? Solution: Write W for weight of packet, and take Z = that Z ∼ N (0, 1) 500 − 508 P (W < 500) = P Z < 5 = P (Z < −1.6) = P (Z > 1.6) = 1 − P (Z ≤ 1.6) = 1 − Φ(1.6) = 1 − 0.9452

W −508 , 5

so

The Normal distribution Example: Packets of breakfast cereal nominally contain 500 g. However, actual contents vary approximately Normally. If average contents are 508g and the SD is 5g, what is the probability of an underweight packet? Solution: Write W for weight of packet, and take Z = that Z ∼ N (0, 1) 500 − 508 P (W < 500) = P Z < 5 = P (Z < −1.6) = P (Z > 1.6) = 1 − P (Z ≤ 1.6) = 1 − Φ(1.6) = 1 − 0.9452 = 0.0548

W −508 , 5

so

Two-sided situations

Example: The diameter X of a steel bearing is approximately Normally distributed with Âľ = 3.76cm and Ďƒ = 0.02cm. Acceptable tolerances are that diameter should lie in the range 3.73cm to 3.79cm. What is the probability that a randomly selected bearing will be OK?

Two-sided situations

Example: The diameter X of a steel bearing is approximately Normally distributed with µ = 3.76cm and σ = 0.02cm. Acceptable tolerances are that diameter should lie in the range 3.73cm to 3.79cm. What is the probability that a randomly selected bearing will be OK? Solution: Let Z =

, so Z ∼ N (0, 1)

Two-sided situations

Example: The diameter X of a steel bearing is approximately Normally distributed with µ = 3.76cm and σ = 0.02cm. Acceptable tolerances are that diameter should lie in the range 3.73cm to 3.79cm. What is the probability that a randomly selected bearing will be OK? Solution: Let Z =

X −µ X − 3.76 = , so Z ∼ N (0, 1) σ 0.02

Two-sided situations Probability that bearing is OK is

59

Two-sided situations Probability that bearing is OK is P (3.73 < X < 3.79) =

60

Two-sided situations Probability that bearing is OK is 3.73 − 3.76 3.79 − 3.76 P (3.73 < X < 3.79) = P <Z< 0.02 0.02 =

61

Two-sided situations Probability that bearing is OK is 3.73 − 3.76 3.79 − 3.76 P (3.73 < X < 3.79) = P <Z< 0.02 0.02 = P (−1.5 < Z < 1.5) =

62

Two-sided situations Probability that bearing is OK is 3.73 − 3.76 3.79 − 3.76 P (3.73 < X < 3.79) = P <Z< 0.02 0.02 = P (−1.5 < Z < 1.5) = P (Z < 1.5) − P (Z < −1.5) =

63

Two-sided situations Probability that bearing is OK is 3.73 − 3.76 3.79 − 3.76 P (3.73 < X < 3.79) = P <Z< 0.02 0.02 = P (−1.5 < Z < 1.5) = P (Z < 1.5) − P (Z < −1.5) = P (Z < 1.5) − (1 − P (Z < 1.5)) =

64

Two-sided situations Probability that bearing is OK is 3.73 − 3.76 3.79 − 3.76 P (3.73 < X < 3.79) = P <Z< 0.02 0.02 = P (−1.5 < Z < 1.5) = P (Z < 1.5) − P (Z < −1.5) = P (Z < 1.5) − (1 − P (Z < 1.5)) = 2 × Φ(1.5) − 1 =

65

Two-sided situations Probability that bearing is OK is 3.73 − 3.76 3.79 − 3.76 P (3.73 < X < 3.79) = P <Z< 0.02 0.02 = P (−1.5 < Z < 1.5) = P (Z < 1.5) − P (Z < −1.5) = P (Z < 1.5) − (1 − P (Z < 1.5)) = 2 × Φ(1.5) − 1 = 2 × 0.9332 − 1 =

66

Two-sided situations Probability that bearing is OK is 3.73 − 3.76 3.79 − 3.76 P (3.73 < X < 3.79) = P <Z< 0.02 0.02 = P (−1.5 < Z < 1.5) = P (Z < 1.5) − P (Z < −1.5) = P (Z < 1.5) − (1 − P (Z < 1.5)) = 2 × Φ(1.5) − 1 = 2 × 0.9332 − 1 = 0.8664

67

Two-sided situations In general,

Two-sided situations In general, P (b ≤ Z ≤ c) = P (Z ≤ c) − P (Z ≤ b)

Two-sided situations In general, P (b ≤ Z ≤ c) = P (Z ≤ c) − P (Z ≤ b) Note: Because the Normal distribution is continuous,

Two-sided situations In general, P (b ≤ Z ≤ c) = P (Z ≤ c) − P (Z ≤ b) Note: Because the Normal distribution is continuous, P (Z < b) = P (Z ≤ b) = Φ(b)

71

Two-sided situations In general, P (b ≤ Z ≤ c) = P (Z ≤ c) − P (Z ≤ b) Note: Because the Normal distribution is continuous, P (Z < b) = P (Z ≤ b) = Φ(b)

Example: For Z ∼ N (0, 1) and a > 0, find P ({Z < −a} ∪ {Z > a})

Two-sided situations In general, P (b ≤ Z ≤ c) = P (Z ≤ c) − P (Z ≤ b) Note: Because the Normal distribution is continuous, P (Z < b) = P (Z ≤ b) = Φ(b)

Example: For Z ∼ N (0, 1) and a > 0, find P ({Z < −a} ∪ {Z > a}) Solution: P ({Z < −a} ∪ {Z > a}) = = = 73

Two-sided situations In general, P (b ≤ Z ≤ c) = P (Z ≤ c) − P (Z ≤ b) Note: Because the Normal distribution is continuous, P (Z < b) = P (Z ≤ b) = Φ(b)

Example: For Z ∼ N (0, 1) and a > 0, find P ({Z < −a} ∪ {Z > a}) Solution: P ({Z < −a} ∪ {Z > a}) = P (Z < −a) + P (Z > a) = = 74

Two-sided situations In general, P (b ≤ Z ≤ c) = P (Z ≤ c) − P (Z ≤ b) Note: Because the Normal distribution is continuous, P (Z < b) = P (Z ≤ b) = Φ(b)

Example: For Z ∼ N (0, 1) and a > 0, find P ({Z < −a} ∪ {Z > a}) Solution: P ({Z < −a} ∪ {Z > a}) = P (Z < −a) + P (Z > a) = 2P (Z > a)

(by symmetry)

= 75

Two-sided situations In general, P (b ≤ Z ≤ c) = P (Z ≤ c) − P (Z ≤ b) Note: Because the Normal distribution is continuous, P (Z < b) = P (Z ≤ b) = Φ(b)

Example: For Z ∼ N (0, 1) and a > 0, find P ({Z < −a} ∪ {Z > a}) Solution: P ({Z < −a} ∪ {Z > a}) = P (Z < −a) + P (Z > a) = 2P (Z > a)

(by symmetry)

= 2(1 − Φ(a)) 76

The Normal distribution

Note: The standard Normal distribution is in terms of ‘number of standard deviations from the mean’.

77

The Normal distribution

Note: The standard Normal distribution is in terms of ‘number of standard deviations from the mean’. So if X ∼ N µ, σ 2 , then Φ(c) gives the probability of being not more than c standard deviations above the mean.

78

The Normal distribution

Note: The standard Normal distribution is in terms of ‘number of standard deviations from the mean’. So if X ∼ N µ, σ 2 , then Φ(c) gives the probability of being not more than c standard deviations above the mean. (And by symmetry, Φ(c) also gives the probability of being not more than c standard deviations below the mean.)

79

The Normal distribution Φ(1) = 0.8413

Φ(2) = 0.9772

Φ(3) = 0.9987

80

The Normal distribution Φ(1) = 0.8413

16% chance of being more than 1 SD above the mean

Φ(2) = 0.9772

Φ(3) = 0.9987

81

The Normal distribution Φ(1) = 0.8413

16% chance of being more than 1 SD above the mean

Φ(2) = 0.9772

2.3% chance of being more than 2 SDs above the mean

Φ(3) = 0.9987

82

The Normal distribution Φ(1) = 0.8413

16% chance of being more than 1 SD above the mean

Φ(2) = 0.9772

2.3% chance of being more than 2 SDs above the mean

Φ(3) = 0.9987

0.13% chance of being more than 3 SDs above the mean

83

The Normal distribution Example: The diameter X of a steel bearing is Normally distributed with Âľ = 3.76cm and Ďƒ = 0.02cm. Find the value of c such that 97.5% of steel bearings have diameters less than c

84

The Normal distribution Example: The diameter X of a steel bearing is Normally distributed with µ = 3.76cm and σ = 0.02cm. Find the value of c such that 97.5% of steel bearings have diameters less than c Solution: Taking Z = require

, so that Z ∼ N (0, 1), we

85

The Normal distribution Example: The diameter X of a steel bearing is Normally distributed with µ = 3.76cm and σ = 0.02cm. Find the value of c such that 97.5% of steel bearings have diameters less than c Solution: Taking Z = (X − 3.76)/0.02, so that Z ∼ N (0, 1), we require

86

The Normal distribution Example: The diameter X of a steel bearing is Normally distributed with µ = 3.76cm and σ = 0.02cm. Find the value of c such that 97.5% of steel bearings have diameters less than c Solution: Taking Z = (X − 3.76)/0.02, so that Z ∼ N (0, 1), we require P (X < c) = 0.975 = = = 87

The Normal distribution Example: The diameter X of a steel bearing is Normally distributed with µ = 3.76cm and σ = 0.02cm. Find the value of c such that 97.5% of steel bearings have diameters less than c Solution: Taking Z = (X − 3.76)/0.02, so that Z ∼ N (0, 1), we require P (X < c) = 0.975 ⇔P

c − 3.76 Z< = 0.975 0.02 = = 88

The Normal distribution Example: The diameter X of a steel bearing is Normally distributed with µ = 3.76cm and σ = 0.02cm. Find the value of c such that 97.5% of steel bearings have diameters less than c Solution: Taking Z = (X − 3.76)/0.02, so that Z ∼ N (0, 1), we require P (X < c) = 0.975 ⇔P

c − 3.76 Z< = 0.975 0.02 ⇔

c − 3.76 0.02

=

(from tables)

= 89

The Normal distribution Example: The diameter X of a steel bearing is Normally distributed with µ = 3.76cm and σ = 0.02cm. Find the value of c such that 97.5% of steel bearings have diameters less than c Solution: Taking Z = (X − 3.76)/0.02, so that Z ∼ N (0, 1), we require P (X < c) = 0.975 ⇔P

c − 3.76 Z< = 0.975 0.02 ⇔

c − 3.76 0.02

= 1.96 (from tables) = 90

The Normal distribution Example: The diameter X of a steel bearing is Normally distributed with µ = 3.76cm and σ = 0.02cm. Find the value of c such that 97.5% of steel bearings have diameters less than c Solution: Taking Z = (X − 3.76)/0.02, so that Z ∼ N (0, 1), we require P (X < c) = 0.975 ⇔P

c − 3.76 Z< = 0.975 0.02 ⇔

c − 3.76 0.02

= 1.96 (from tables)

⇔c = 91

The Normal distribution Example: The diameter X of a steel bearing is Normally distributed with µ = 3.76cm and σ = 0.02cm. Find the value of c such that 97.5% of steel bearings have diameters less than c Solution: Taking Z = (X − 3.76)/0.02, so that Z ∼ N (0, 1), we require P (X < c) = 0.975 ⇔P

c − 3.76 Z< = 0.975 0.02 ⇔

c − 3.76 0.02

= 1.96 (from tables)

⇔ c = 1.96 × 0.02 + 3.76 = 3.7992

The Normal distribution Example: The diameter X of a steel bearing is Normally distributed with µ = 3.76cm and σ = 0.02cm. Find the value of c such that 97.5% of steel bearings have diameters less than c Solution: Taking Z = (X − 3.76)/0.02, so that Z ∼ N (0, 1), we require P (X < c) = 0.975 ⇔P

c − 3.76 Z< = 0.975 0.02 ⇔

c − 3.76 0.02

= 1.96 (from tables)

⇔ c = 1.96 × 0.02 + 3.76 = 3.7992cm

Distribution of the sample mean Suppose we take a sample of n observations X1 , . . . , Xn from a Normal population.

94

Distribution of the sample mean Suppose we take a sample of n observations X1 , . . . , Xn from a Normal population. That is, X1 , . . . , Xn are independent and identically distributed, with Xi ∼ N µ, σ 2

95

Distribution of the sample mean Suppose we take a sample of n observations X1 , . . . , Xn from a Normal population. That is, X1 , . . . , Xn are independent and identically distributed, with Xi ∼ N µ, σ 2

Then the sample mean X̄ is itself a random variable; if we took a different sample, we’d get a different value of X̄. So what is the distribution of X̄?

96

Distribution of the sample mean First, work out the expectation: E XĚ„ =

97

Distribution of the sample mean First, work out the expectation: # " E XĚ„ = E

98

Distribution of the sample mean First, work out the expectation: # " n 1X Xi E XÌ„ = E n i=1

99

Distribution of the sample mean First, work out the expectation: # " n n 1X 1X = Xi E [Xi ] E XÌ„ = E n n i=1

i=1

100

Distribution of the sample mean First, work out the expectation: # " n n 1X 1X 1 = Xi E [Xi ] = (nµ) E X̄ = E n n n i=1

i=1

101

Distribution of the sample mean First, work out the expectation: # " n n 1X 1X 1 = Xi E [Xi ] = (nµ) = µ E X̄ = E n n n i=1

i=1

102

Distribution of the sample mean First, work out the expectation: # " n n 1X 1X 1 = Xi E [Xi ] = (nµ) = µ E X̄ = E n n n i=1

i=1

For the variance, a similar but more complicated calculation shows that Variance X̄ =

103

Distribution of the sample mean First, work out the expectation: # " n n 1X 1X 1 = Xi E [Xi ] = (nµ) = µ E X̄ = E n n n i=1

i=1

For the variance, a similar but more complicated calculation shows that Variance X̄ =

σ2 n

104

Distribution of the sample mean First, work out the expectation: # " n n 1X 1X 1 = Xi E [Xi ] = (nµ) = µ E X̄ = E n n n i=1

i=1

For the variance, a similar but more complicated calculation shows that Variance X̄ = so that SD X̄ =

σ2 n

Distribution of the sample mean First, work out the expectation: # " n n 1X 1X 1 = Xi E [Xi ] = (nµ) = µ E X̄ = E n n n i=1

i=1

For the variance, a similar but more complicated calculation shows that Variance X̄ =

σ2 n

so that SD X̄ =

σ √ n

Distribution of the sample mean

Finally, it can be shown that

Distribution of the sample mean

Finally, it can be shown that if the individual values X1 , . . . , Xn are Normally distributed then the sample mean XĚ„ is also Normally distributed,

Distribution of the sample mean

Finally, it can be shown that if the individual values X1 , . . . , Xn are Normally distributed then the sample mean XĚ„ is also Normally distributed, and so we have

Distribution of the sample mean

Finally, it can be shown that if the individual values X1 , . . . , Xn are Normally distributed then the sample mean X̄ is also Normally distributed, and so we have σ2 X̄ ∼ N µ, n

Distribution of the sample mean

Finally, it can be shown that if the individual values X1 , . . . , Xn are Normally distributed then the sample mean X̄ is also Normally distributed, and so we have σ2 X̄ ∼ N µ, n Note: The distribution of a sample statistic, such as the sample mean, is often referred to as its

Distribution of the sample mean

Finally, it can be shown that if the individual values X1 , . . . , Xn are Normally distributed then the sample mean X̄ is also Normally distributed, and so we have σ2 X̄ ∼ N µ, n Note: The distribution of a sample statistic, such as the sample mean, is often referred to as its sampling distribution.

Distribution of the sample mean Example: Weights of packets of breakfast cereal are Normally distributed with mean 508g, standard deviation 5g. A sample of 100 packets is taken at random and weighed.

Distribution of the sample mean Example: Weights of packets of breakfast cereal are Normally distributed with mean 508g, standard deviation 5g. A sample of 100 packets is taken at random and weighed. The mean weight of the packets in the sample, XĚ„, is

Distribution of the sample mean Example: Weights of packets of breakfast cereal are Normally distributed with mean 508g, standard deviation 5g. A sample of 100 packets is taken at random and weighed. The mean weight of the packets in the sample, XĚ„, is Normally distributed

Distribution of the sample mean Example: Weights of packets of breakfast cereal are Normally distributed with mean 508g, standard deviation 5g. A sample of 100 packets is taken at random and weighed. The mean weight of the packets in the sample, XĚ„, is Normally distributed with E XĚ„ =

Distribution of the sample mean Example: Weights of packets of breakfast cereal are Normally distributed with mean 508g, standard deviation 5g. A sample of 100 packets is taken at random and weighed. The mean weight of the packets in the sample, XĚ„, is Normally distributed with E XĚ„ = E[X] = 508g

Distribution of the sample mean Example: Weights of packets of breakfast cereal are Normally distributed with mean 508g, standard deviation 5g. A sample of 100 packets is taken at random and weighed. The mean weight of the packets in the sample, X̄, is Normally distributed with E X̄ = E[X] = 508g Variance X̄ =

118

Distribution of the sample mean Example: Weights of packets of breakfast cereal are Normally distributed with mean 508g, standard deviation 5g. A sample of 100 packets is taken at random and weighed. The mean weight of the packets in the sample, X̄, is Normally distributed with E X̄ = E[X] = 508g Variance X̄ =

Variance[X] 52 = = 0.25 n 100

119

Distribution of the sample mean Example: Weights of packets of breakfast cereal are Normally distributed with mean 508g, standard deviation 5g. A sample of 100 packets is taken at random and weighed. The mean weight of the packets in the sample, X̄, is Normally distributed with E X̄ = E[X] = 508g Variance X̄ =

that is,

Variance[X] 52 = = 0.25 n 100

Distribution of the sample mean Example: Weights of packets of breakfast cereal are Normally distributed with mean 508g, standard deviation 5g. A sample of 100 packets is taken at random and weighed. The mean weight of the packets in the sample, X̄, is Normally distributed with E X̄ = E[X] = 508g Variance X̄ = that is, X̄ ∼ N 508, 0.52

Variance[X] 52 = = 0.25 n 100

121

Turn static files into dynamic content formats.

Create a flipbook
Issuu converts static files into: digital portfolios, online yearbooks, online catalogs, digital photo albums and more. Sign up and create your flipbook.