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Measures of Dispersion
CUMULATIVE FREQUENCIES
INTER-QUARTILE RANGE
RANGE MEAN DEVIATION
VARIANCE andSTANDARD DEVIATION
STATISTICS: DESCRIBING VARIABILITY
Variability = Uncertainty
Probabilities
STATISTICS: PROBABILITIES
CALCULATING PROBABILITIES
NORMAL DISTRIBUTION
What are probabilities? z-DISTRIBUTION
t-DISTRIBUTION
Probabilities
Probability
= 0.167
The chance of NOT throwing a = 5/6 = 0.833
= 1 – 0.167
The chance of throwing a = 1/6
The probability of an event A, symbolized by P(A), is a number between 0 and 1
The higher the P value the more probable the event
The Total Number Of Possible Outcomes
The Number Of Ways Event A Can Occur P(A) =
What is the probability of picking a student of 1.65 m high from the class?
0
2
4
6
8
10
12
14
16
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9 2
2.1
2.2
Height (m)
Fre
quen
cy
Height (m) Frequency1.2 1
1.25 11.3 1
1.35 21.4 2
1.45 31.5 5
1.55 61.6 8
1.65 101.7 12
1.75 151.8 11
1.85 91.9 7
1.95 52 3
2.05 22.1 1
2.15 12.2 1
P(NOT 1.65) = 1 – P(1.65) = 1 – 0.094 = 0.906
Depends on how the data are
distributed
STATISTICS: PROBABILITY
P(1.65) = 10/106 = 0.094
The probability is a number between 0 and 1
Class has 106 students (n=106)
10 are 1.65m tall
Frequency Distributions
Height (m)
0
2
4
6
8
10
12
14
16
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9 2
2.1
2.2
Height (m)
Fre
quen
cy (
%)
STATISTICS: PROBABILITY
Area under graph = total number of observations
Can display frequency distributions as % or proportion
0
2
4
6
8
10
12
14
16
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9 2
2.1
2.2N
o. o
f ob
serv
atio
ns
Area under graph = 1.0
(for proportion)
= 100 (for percentage)
12 people
10 people
0.113
0.094
0
2
4
6
8
10
12
14
16
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9 2
2.1
2.2
Height (m)
Fre
qu
ency
(%
)STATISTICS: PROBABILITY
Normal Distributio
n
• Data clustered around the mean • Therefore good chance (high probability) of
picking (at random) a student with a height close to the mean
• Small chance (low probability) of picking (at random) a student who is either very tall or very short
Tails
POPULATION DYNAMICS POPULATION DYNAMICS Required background knowledge:
• Data and variability concepts
Data collection
• Measures of central tendency (mean, median, mode, variance, stdev)
• Normal distribution and Standard Error
• Student’s t-test and 95% confidence intervals
• Chi-Square tests
• MS Excel
Properties of a normal distribution:•The mean, median and mode are the same•The frequency distribution is completely symmetrical either side of the mean
•The area under the curve is proportional to number of observations
Height (mm)
Fre
qu
ency
(%
)
02468
1012
0 2 4 6 8 10 12 14 16 18 20 22 24
STATISTICS: PROBABILITY
0
0.05
0.1
0.15
0.2
0.25
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
X
Fre
qu
en
cy X
s2 = 4s2 = 8s2 = 12s2 = 16
x = 10The shape of the curve
depends on the variance or standard deviation: the spread of values about the
mean
Σ = 100%
Some dataset are normally distributed – but NOT all.
Height (mm)
Fre
qu
ency
(%
)
02468
1012
0 2 4 6 8 10 12 14 16 18 20 22 24
Σ = 100%
STATISTICS: PROBABILITY
The normal curve has fixed mathematical properties, irrespective of: The scale on which it is drawnThe magnitude or units of its meanThe magnitude or units of its Standard Deviation
…….and these render it susceptible to STATISTICAL ANALYSIS…
Can use the normal distribution to calculate probabilities
STATISTICS: PROBABILITIES
CALCULATING PROBABILITIES
NORMAL DISTRIBUTION
What are probabilities? z-DISTRIBUTION
t-DISTRIBUTION
Probabilities
To calculate the probability of a particular value x being drawn from a normally distributed population of data, you need to know the mean AND the standard deviation of the data
X = value you are consideringμ = population meanσ = population standard deviation
Z = (x – μ)
σ Equation 1
Z-values form the Z-DISTRIBUTION…. Z is based on data that are normally distributed, so the Z distribution is also normally distributed.
STATISTICS: CALCULATING PROBABILITIES
Z = how many standard deviations away from the mean is the value x
If Z is small number then x ≈ meanIf Z is large number then x ≈ mean
Once we know the Z-value we use statistical tables to calculate the associated probability…
Z 0 1 2 3 4 5
0.0 0.5000 0.4960 0.4920 0.4880 0.4840 0.48010.1 0.4602 0.4562 0.4522 0.4483 0.4443 0.44040.2 0.4207 0.4168 0.4129 0.4090 0.4052 0.40130.3 0.3821 0.3783 0.3745 0.3707 0.3669 0.36320.4 0.3466 0.3409 0.3372 0.3336 0.3300 0.3264
0.5 0.3085 0.3050 0.3015 0.2981 0.2946 0.29120.6 0.2743 0.2709 0.2676 0.2643 0.2611 0.25780.7 0.2420 0.2389 0.2358 0.2327 0.2297 0.22660.8 0.2119 0.2090 0.2061 0.2033 0.2005 0.19770.9 0.1841 0.1814 0.1788 0.1762 0.1736 0.1711
1.0 0.1587 0.1562 0.1539 0.1515 0.1492 0.14691.1 0.1357 0.1335 0.1314 0.1292 0.1271 0.12511.2 0.1151 0.1131 0.1112 0.1093 0.1075 0.10561.3 0.0968 0.0951 0.0934 0.0918 0.0901 0.08851.4 0.0808 0.0793 0.0778 0.0764 0.0749 0.0735
0
2
4
6
8
10
12
14
16
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9 2
2.1
2.2
Height (m)
Fre
quen
cy
Z = (x – μ)
σ
Z = (1.95 – 1.55)
0.3Z = (0.4)
0.3
Z = 1.33
?
What is the probability of a randomly drawing a student measuring more than 1.95 m from the population if μ = 1.55 m and σ = 0.3 m?
0.450.550.650.750.850.951.051.151.251.351.451.551.651.751.851.952.052.152.252.352.452.55
Height (m)
Fre
qu
ency
-3.67-3.33-3.00-2.67-2.33-2.00-1.67-1.33-1.00-0.67-0.330.000.330.671.001.331.672.002.332.673.003.33
Z
Fre
qu
ency 0.0918
Z 0 1 2 3 4 5
0.0 0.5000 0.4960 0.4920 0.4880 0.4840 0.48010.1 0.4602 0.4562 0.4522 0.4483 0.4443 0.44040.2 0.4207 0.4168 0.4129 0.4090 0.4052 0.40130.3 0.3821 0.3783 0.3745 0.3707 0.3669 0.36320.4 0.3466 0.3409 0.3372 0.3336 0.3300 0.3264
0.5 0.3085 0.3050 0.3015 0.2981 0.2946 0.29120.6 0.2743 0.2709 0.2676 0.2643 0.2611 0.25780.7 0.2420 0.2389 0.2358 0.2327 0.2297 0.22660.8 0.2119 0.2090 0.2061 0.2033 0.2005 0.19770.9 0.1841 0.1814 0.1788 0.1762 0.1736 0.1711
1.0 0.1587 0.1562 0.1539 0.1515 0.1492 0.14691.1 0.1357 0.1335 0.1314 0.1292 0.1271 0.12511.2 0.1151 0.1131 0.1112 0.1093 0.1075 0.10561.3 0.0968 0.0951 0.0934 0.0918 0.0901 0.08851.4 0.0808 0.0793 0.0778 0.0764 0.0749 0.0735
STATISTICS: CALCULATING PROBABILITIES – an example
Step 1: calculate Z using known information
P(0.0918) of randomly drawing a student measuring 1.95 m from the population [P > 1.95 = 0.0918]
REMEMBER: The higher the P-value the more
probable the event
1st decimal place
2nd decimal place
PROBABILITY
P(A)
p > 1.95
Z = 1.33
Step 2: Look up Z in Z-tables
Now you try:
A population of bone measurements is normally distributed with μ = 60 mm and σ = 10 mm. What is the probability of selecting a bone with a length greater than 66 mm?
Z = (x – μ)
σ
Step 1: calculate Z using known information
Step 2: Look up the P-value in the Z-tables
Z = 0.60
Therefore p = 0.2743
ANSWER:
In Excel:• Enter x, μ and σ into
different cells• Formula: =(x – μ)/ σ
POPULATION DYNAMICS POPULATION DYNAMICS Required background knowledge:
• Data and variability concepts
Data collection
• Measures of central tendency (mean, median, mode, variance, stdev)
• Normal distribution and Standard Error
• Student’s t-test and 95% confidence intervals
• Chi-Square tests
• MS Excel
NOTE:
Standard Deviation = tells you the variation around the mean
Standard Error = tells you how well you’ve estimated the mean
Standard Deviation vs. Standard Error
0
2
4
6
8
10
12
14
16
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9 2 2.
12.
2
Height (m)
Fre
quen
cy
X2
X1
X3
X4
Population normal Population normal = distribution= distribution
0
2
4
6
8
10
12
14
16
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9 2 2.
12.
2
Sample Means
Fre
quen
cy
Sample mean = Sample mean = normal distributionnormal distribution
Standard Error
X2
X1
X3
X4
X6
X5
X7
X8
X10
X9
X11
X12
σ2
nσ2
x =Equation 2
n=7n=7n=1n=1
00
n=1n=122
n=4n=4
As n increases
….So σ decreases
Standard Error
The variance of the mean
σ2
nσ2
x =
Standard Error
Square Root both sides
σ2
nσ
x =
σx =
Equation 2
σ
n√ STANDARD
ERROR
0
2
4
6
8
10
12
14
16
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9 2 2.
12.
2
Height (m)
Fre
quen
cy
Population normal = Population normal = distributiondistribution
0
2
4
6
8
10
12
14
16
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9 2 2.
12.
2
Sample Means
Fre
quen
cy
Sample mean = normal Sample mean = normal distributiondistribution
Standard Error
A normal deviate referring to the normal distribution
of Xi values
Z = (x – μ)
σ
A normal deviate referring to the normal distribution
of means
Z = (x – μ)
σx
What is the probability of obtaining a random sample of nine measurements with a mean greater than 50.0 mm, from a population having a mean of 47 mm and a standard deviation of 12.0 mm?
Standard Error
N = 9, X = 50.0 mm, μ = 47.0 mm, σ = 12.0 mm
σx
= 12.0
√ 9= 4= 12
3
Z = (50.0 – 47.0) = 3 = 0.75
4 4
Z = (x – μ)
σx
Step 1: calculate Z using known information
σx =
Equation 2
σ
n√
Step 2: Look up Z in Z-tables
Z 0 1 2 3 4 5
0.0 0.5000 0.4960 0.4920 0.4880 0.4840 0.48010.1 0.4602 0.4562 0.4522 0.4483 0.4443 0.44040.2 0.4207 0.4168 0.4129 0.4090 0.4052 0.40130.3 0.3821 0.3783 0.3745 0.3707 0.3669 0.36320.4 0.3466 0.3409 0.3372 0.3336 0.3300 0.3264
0.5 0.3085 0.3050 0.3015 0.2981 0.2946 0.29120.6 0.2743 0.2709 0.2676 0.2643 0.2611 0.25780.7 0.2420 0.2389 0.2358 0.2327 0.2297 0.22660.8 0.2119 0.2090 0.2061 0.2033 0.2005 0.19770.9 0.1841 0.1814 0.1788 0.1762 0.1736 0.1711
1.0 0.1587 0.1562 0.1539 0.1515 0.1492 0.14691.1 0.1357 0.1335 0.1314 0.1292 0.1271 0.12511.2 0.1151 0.1131 0.1112 0.1093 0.1075 0.10561.3 0.0968 0.0951 0.0934 0.0918 0.0901 0.08851.4 0.0808 0.0793 0.0778 0.0764 0.0749 0.0735
STATISTICS: z-DISTRIBUTION
Z = 0.75
Step 2: Look up Z in Z-tables
P = 0.2266
So there is a 0.2266 is the probability of obtaining a random sample of nine measurements with a mean greater than 50.0 mm, from a population having a mean of 47 mm and a standard deviation of 12.0 mm.
For probability values, always report
4 decimal places
σx =
σ
n√
Now you try:What is the probability of obtaining a random sample of 5 measurements with a mean greater than 60.0 mm, from a population having a mean of 57 mm and a standard deviation of 7.0 mm?Step 1: calculate Z using known information
Step 2: Look up the P-value in the Z-tables
Z = 0.96
Therefore p = 0.1685
ANSWER:
Z = (x – μ)
σx
• Formula: =(x - μ)/(σ /(SQRT(n)))
In Excel:• Enter x, μ, σ and n
In the last couple of equations for Z we have used the population parameters: μ, σ and
BUT we don’t usually have access to population data and must make do with sample estimators x, s and
σx
sx
IF n is very, very large : we use Z distribution to calculate normal deviates
Z = (x – μ)
σx
STATISTICS: z-DISTRIBUTION
σx
sx =
t = (x – μ)
sx Equation 3
If n is not large, we must uset distribution:
σ sPROXY
