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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
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