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Normal Distribution

“Bell Curve”

19/04/23Characteristics of a normal distribution…

Example:

The heights of males and females in the UK .

Few very short people Few very tall people

Most people around the average height

19/04/23Characteristics of a normal distribution…

Example:

X is the r.v. “The heights, in cm, of males and females in the UK”

The probability distribution for the random variable X

μ = 178 cm μ = 163 cm X

P(X = x) P(X = x)

Height is continuous, so X is a CONTINUOUS RANDOM VARIABLE

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Properties of a normal distribution

For a continuous random variable, X:

Symmetrical distribution

Mean = mode = median

- < X < + asymptote

Total area under curve = 1Since P(X = x) = 1

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Parameters of a normal distribution

P(X = x) depends ONLY on two things: mean (μ) and variance (σ2)

μ + 3 σ μ – 3σ

Nearly ALL values of X lie within ± 3 standard deviations away from the mean

X ~ N (μ, σ2)

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Shape of a normal distribution

The values of the parameters (μ and σ2) change the shape of the curve.

Example: Heights (cm) of adult males and females in the UKX ~ N (μ, σ2)

μMμF X

XF ~ N (163, 252) XM ~ N (178, 252)

Different means distributions have different location

Equal variances distributions have same dispersion

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Shape of a normal distribution

The values of the parameters (μ and σ2) change the shape of the curve.

Example: Length (cm) of hair in adult males and females in the 1970s

X ~ N (μ, σ2)

μF = μM X

XF ~ N (30, 42) XM ~ N (30, 102)

Equal means distributions have same location

Different variances distributions have different dispersion

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Calculating probabilities

All probabilities depend on the values of the mean and variance. Normal distributions for different variables will differ.

a

P(V < a)

V b

P(W > b)

W

c d

P(c < X < d)

X

The probabilities are VERY difficult to calculate!

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The standard normal distributionAny variable can be transformed (coded) so it has a mean of zero and a variance of one:

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The standard normal distributionAny variable can be transformed (coded) so it has a mean of zero and a variance of one:

Z is the standard normal distribution, and all its associated probabilities can be found using statistical tables.

Z ~ N(0, 12)

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Transforming your variableAny variable can be transformed (coded) so it has a mean of zero and a variance of one.

Example:

X ~ N(100, 152)

X Z

Z ~ N(0, 12)

X

Z

15100

X

ZEach value of X has a corresponding value of Z after standardising

When X = 115, 115100115

Z

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Probabilities 1

Example 1.

X ~ N(10, 42)

Z

Z ~ N(0, 12)

X

Z

410

X

Z

Find P(X < 12)

Transform X using

0.5)P(Z4

10-12ZP 12) P(X

Find P(X < 14)

Similarly… 1)P(Z4

10-14ZP 14) P(X

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Using tables

X

Z

Your statistical tables use a notation that you need to become familiar with…

Φ(0.5)

means

P(Z < 0.5)

Z

Z ~ N(0, 12)

0.5P(Z < 0.5) = Φ(0.5) = 0.6915

Φ(z) = P(Z < z)

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Using tables

X

Z

Your statistical tables use a notation that you need to become familiar with…

Φ(1)

means

P(Z < 1)

Z

Z ~ N(0, 12)

1P(Z < 1) = Φ(0.5) = 0.8413

Φ(z) = P(Z < z)

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Using tables

X

Z

Your statistical tables use a notation that you need to become familiar with…

Z

Z ~ N(0, 12)

1

P(0.5 < Z < 1)

Φ(z) = P(Z < z)

0.5

= Φ(1) – Φ(0.5) = 0.8413 – 0.6915 = 0.1498

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When z is negativeFind P(Z < -1)

Z

Z ~ N(0, 12)

-1

Therefore Φ(-1) = 1 – Φ(1) = 1 – 0.8413 = 0.1587

Use SYMMETRY properties

1

P(Z < -1) = P(Z > 1)

BUT tables only give you probabilities LESS THAN z!

P(Z < -1) = 1 – P(Z < 1)

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Find these probabilities…

(a) 0.1587(b) 0.0228(c) 0.6247

Z

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Find these probabilities…

(a) 0.1151(b) 0.3159(c) 0.2638

Y

Z Classwork:

Exercise 9A Q1 – 6