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Application of Statistics
Shakeel Ahmad
Department of Physics
A.M.U., Aligarh
All measurements, how carefully done,
involve errors
These errors may be associated with
technical difficulties
Imperfection of measuring instrument, limitation of
human eye and several other factors that can not
be taken into account, like fluctuations in
temperature, motion of air stream around the
instrument
Accuracy
How close a measured value is to the actual (true)
value.
Precision
How close the measured values are to each other.
Low Accuracy
High Precision
High Accuracy
Low Precision
High Accuracy
High Precision
Accuracy signifies the degree of closeness
between the measured value and true value of
the quantity.
Precision is the limit or resolution to which a
quantity is measured. No matter how carefully
we measure, we can never obtain a result more
precise than the measuring device.
The limit of precision of a measuring device is ±½ the
smallest division of the device.
Significant Figures: Accuracy of measurements
• While measuring a quantity, there is a limit to accuracy
of measurement
• This accuracy depends upon the number of digits upto
which the value of a quantity is known.
• It is, therefore, necessary to use appropriate number of
digits to express the accuracy and reliability.
• No purpose is served by writing additional digits. We
can include only one more digit about which we are
not sure.
• The number of digits in a measurement about which
we are reasonably sure plus the first uncertain digit
are called significant figure.
Example:
Length of an object is written as 2.54 cm  3 significant
figures
2.5 is reliable. 3rd figure, i.e. 4 is uncertain
For a length of 638.6  4 sig. fig. The last digit 6 is
uncertain.
Points to remember
1. Number of significant figures does not vary by varying
the units selected
638.6 cm = 6.386 m
2. It is customary to write the decimal after the first digit
638.6 cm = 6.386 x 102cm
0.0438 kg = 4.38 x 10-2kg
Rules for counting significant figures:
1. All non-zero digits are significant figures. If there is a decimal
point, its position does not matter.  23.73 has 4 sig.fig.
2. All zeros between two non-zero digits are significant figures
and it does not matter where the decimal point occur. 
2090.03 has 6 sig.fig.
3. If there is no decimal point, zero at the end is meaning less
and are not counted.  143000 has only 3 sig.fig.
4. If there is a decimal point, zeros at the end are significant. 
36.00 or 0.3600 has 4 sig. fig.
but
Initial zeros after the decimal points are not significant, i.e.,
zeros on the right of decimal and to the left of non-zero digits
are not significant.  0.0034 has only 2 sig.fig.
5. Zero on the left of decimal point for a number < 0 is never
significant
6. A number with 3 significant figures (say) gives an accuracy of 1
part in 100 to 1000
A number with 6 significant figures (say) gives an accuracy of 1
part in 105 to 106.
Rules of Rounding off the digits
9.876  9.88
Last digit > 5
9.874  9.87
Last digit < 5
If the last digit is 5, e.g.,
9.775  9.78 incremented by one if the last digit is odd.
9.785  9.78 no change in the last digit is even
Rounding off should be such that the last digit should be even
Exercise:
Two cesium clocks, if run for 100 years, free from disturbances
may differ by only about 0.02s. What does this imply for the
accuracy of the clock in measuring a time interval of 1s.
Time interval in 100 years = 100 x 365.25 x 24 x60 x 60 s
Difference in the two clocks = 0.02s
Measured time interval = 3155760000 ± 0.02
= 3155760000.02 or 3155759999.98
Both these values have 12 sig. fig.
The accuracy of the clock is 1 part in 1011 to 1012
=
Alternative
ΔT = 0.02s = 2 x 10-2s
T = 100 years = 3.15576 x 109s
ΔT/T = 6 x 10-12  Accuracy ≈ 10-11 to 10-12
Rules for Arithmetic Operations with Significant Figures:
Results from additions and subtractions are retained to the least
decimal places that is present in the numbers being
added/subtracted
Sum or difference of the two or more numbers has significant
figures only in those places where these are in the least precise
amongst the given numbers
3.123 + 40.5 + 2.0123 = 45.7253
The answer should be only in three significant figures, i.e. 45.7
(least accurate is 40.5 and has only 3 sig. fig.)
Similarly
53.312 – 53.3 = 0.012
But the answer should be 0.0, as the least accurate is 53.3 and has
only one digit after decimal point
Multiplication or Division
of two numbers should have the sig. fig. that is present in the least
precise of the given numbers, e.g.,
4.08 x 16 = 65.28 but the answer is 65,
because the less precise number, 16 has only 2 sig.fig.
Similarly, 6300/ 11.97 = 526.31578 but
The answer should be 530, because the less precise is 6300 which
has 2 sig.fig
Note: 3rd digit is > 5 in 526.31578  it is written as 530
Example:
For a rectangular block, l = 4.234m, b = 1.005m, h = 2.01cm.
Find the area and volume to correct sig. fig.
Area = 4.25517m2 = 4.255m2 ( rounded off to 4 sig. fig.)
Volume = 0.0855289m3 = 0.0855m3 ( rounded off to 3 sig. fig.)
Example:
Two masses, m1 = 20.15g and m2 = 20.17 g
Difference = 0.02g (one sig.fig)  sig, fig. are lost in difference
calculations
Errors in Measurements
It is impossible to find a true value (in practice), while
measuring a physical quantity
Error: Difference in the measured and true value of a
physical quantity.
Degree of Accuracy
Accuracy depends on the instrument you are
measuring with. But as a general rule:
The degree of accuracy is half a unit each side of the
unit of measure.
Examples:
When the instrument measures in 1 unit
then any value between 6½ and 7½ is
measured as 7 units
Thus, the value could be between 6½
and 7½ and is written as
7.0 ± 0.5
The error is ± 0.5
When the instrument measures in 2
units
then any value between 7 and 9 is
measured as 8 units
When the value lies between 7 and 9, it
is written as
8±1
The error is ±1
Although Physics is exact science, physical instruments
do not give the exact values of the quantities
measured.
All measurements in Physics/Science are inaccurate
(generally) in some degree
We often say that the true or actual value of a Physical
quantity can not be found.
We, however, assume that exact value exists and we
concern to estimate the limits between which this
value lies.
The closer these limits, more accurate is the
measurement.
Such errors follow no simple law and arise due to
various causes.
A single observer with the same apparatus will record
different values of the same quantity.
Errors are usually combination of accidental error and
systematic errors
Accidental Errors occur due to the observer and can be
minimized
by repeated observation. They are
disordered.
Systematic errors arise due to the observer or
instrument and are usually more troublesome because
repeated observations do not reveal them (usually) even
when their nature of existence is known. It is difficult to
determine and eliminate such errors
Assessment of possible errors in any measured
quantity is of fundamental importance in science.
Types of Errors
• Accidental
• Systematic
• Random
Accidental errors
estimated by certain
statistical concepts.
Systematic or personal
errors
assumed to be small
and neglected. But
sometimes it becomes
rather a serious issue
and special treatments
are required for
elimination
Systematic errors of
instrument
Systematic or Methodic Errors
Errors which tend to be in one direction, positive or
negative and arise on account of
 Short coming of measuring method
 Imperfection of the theory of physical
phenomenon to which the quantity being
measured is related to
 Lack of accuracy of the formula used for
calculations
e.g. Weighing a body on physical balance
The systematic errors may be due to the fact that
buoyant forces, acting on the body and weights, are
not accounted
Such errors may be reduced by
Changing/improving the measuring method
Applying corrections to the formula used
Instrumental Errors: Occur due to
 imperfection of the design
 inaccurate manufacturing of instrument
e.g., stop watch: running may change due to
change in temperature
or
center of the dial may not coincide with the axis of
rotation of its hands.
It is not possible to eliminate such errors completely.
Random Errors: occur irregularly
Arise due to the verity of the factors which can not be
taken into account.
These are not associated with any systematic cause or
with definite law of action.
e.g., reading of a sensitive beam balance may be
affected by the vibrations of the building or settling
down the dust particles on the pan.
Although such errors can not be completely eliminated,
but can be reduced by repeated observations
R
Random Errors are calculated on
the basis of theory of probability
by applying the Gaussian law of
normal distribution
According to this law, probability
of an error +ΔA in a measurement
of a quantity A is the same as the
probability of -ΔA in the very
measurement
Frequency
-ΔA
+ΔA
Arithmetic mean of a large number of observations is
likely to be closer to the true value
Absolute Errors
Let A1, A2, ………….. An are the measured values of
a quantity A in n attempts
The arithmetic mean, Am = ΣAi/n
Since the true value of the quantity is not known,
Am may be considered as the true value.
The magnitude of the deviation
of any
measurement from the arithmetic mean is called
the absolute error of the measurement
Thus, the absolute errors are
ΔA1 = | Am – A1| ΔA1, ΔA2 are magnitudes
ΔA2 = | Am – A2| and are always taken as +ve
Mean Absolute Error:
Arithmetic mean of all the absolute errors
Δ Am = Σ Δ Ai/n  final absolute error
The final results of the measurement is
A = Am ± Δ Am
Any single measurement of quantity A is likely to
be
Am - ΔAm ≤ A ≤ Am + ΔAm
Relative or % of error = ΔAm/Am
or (ΔAm/Am) x 100 %
Example: Diameter of a ball was measured five times whose
absolute error is Δdinst = ±0.01mm
Observations are: d1 = 5.27mm d2 = 5.30mm
d3 = 5.28mm d4 = 5.33mm
d5 = 5.28mm
Find
Mean diameter dmean = (5.27+5.30+5.28+5.32+5.28)/5
= 5.29mm
Absolute errors Δd1 = dm – d1 = 0.02mm
Δd2 = dm – d2 = 0.01mm
Δd3 = dm – d3 = 0.01mm
Δd4 = dm – d4 = 0.03mm
Δd5 = dm – d5 = 0.02mm
Mean absolute error = (0.02+0.01+0.01+0.03+0.01)/5
Δdmean = 0.02mm
Since Δdmean > Δdinst, result is dmean ± Δdmean = 5.29±0.02
Relative error = Δdmean /dmean = 0.02/5.29= 0.04
% of error = 4%
Example: A box is measured to the nearest 2 cm,
and got 24 cm × 24 cm × 20 cm
Measuring to the nearest 2 cm means the true value could
be up to 1 cm smaller or larger.
The three measurements are:
l = 24 ± 1 cm
b = 24 ± 1 cm
h = 20 ± 1 cm
The smallest possible Volume is: 23cm × 23cm × 19cm = 10051 cm3
The measured Volume is: 24cm × 24cm × 20cm = 11520 cm3
The largest possible Volume is: 25cm × 25cm × 21cm = 13125 cm3
Thus the measured value lies between 10051 and 13125
Absolute Error
• From 10051 to 11520 = 1469
• From 11520 to 13125 = 1605
We pick the bigger one, i.e., Absolute error = 1605cm3
Relative Error =
1605 cm3
11520 cm3
Percentage Error = 13.9%
= 0.139...
• Accuracy: How close is the measured value to the
actual value.
• Precision: How close are the measured values with each
other (quite independent of systematic errors).
(X1, X2, …….., Xn) are of high precision if di values are
small whatever the values of(𝑋 − 𝑋0 ).
The accuracy is high if ei values are small → 𝑋 − 𝑋0 is
small too.
• Accuracy includes precision but precision does not
include accuracy.
If true value of quantity = X0, and its recorded value =X
e = (X – X0) is error in X0
X = X0 + e = X0 (1+f);
f = e/x0 is fractional error
e may be positive or negative but we assume |e| << |X0|
𝑋
𝑋0
1
= 1 + 𝑓 𝑜𝑟
=
≈ (1 − 𝑓) 𝑖𝑓 𝑓 ≪ 1
𝑋0
𝑋
1+𝑓
𝑒
𝑒 𝑋
𝑒
𝑒
= .
=
1+𝑓 ≈
𝑎𝑠 𝑓 ≪ 1
𝑋0 𝑋 𝑋0 𝑋
𝑋
For single measurement: estimation of error is widely
wrong. (Lack of precision of instrument, personal +
accidental error)
if a quantity X0 unit is measured n times as
X1 , X2 ,……………….., Xn
We write Xi = X0 + ei ; Xi is any recorded value and ei is error
associated.
Arithmatic mean
𝑋1 + 𝑋2 + … … … … . . 𝑋𝑛
𝑋=
𝑛
𝑒1 + 𝑒2 + … … … … 𝑒𝑛
= 𝑋0 +
𝑛
e1, e2…….. en : some may be +ve and some –ve
(e1 + e2 + ….. + en)/n may be very small and << e
𝑋 − 𝑋0 ≪ 𝑒
𝑋 is close to X0, represents best value.
It is not possible to find e1, e2 ……….. as X0 is not known. We
therefore, examine scatter or dispersion about 𝑋 and not
about X0.
𝑋𝑟 = 𝑋 + 𝑑𝑟 ;
dr is deviation from 𝑋, often called
residual of Xr .
𝑋𝑟 = 𝑋0 + 𝑒𝑟 = 𝑋 + 𝑑𝑟
𝑒𝑟 − 𝑑𝑟 = 𝑋 − 𝑋0 ,
also 𝑒1 + 𝑒2 + … … + 𝑒𝑛 = 𝑛 𝑋 − 𝑋0
and 𝑑1 + 𝑑2 + … … . + 𝑑𝑛 = 0
By repeated measurements of the same quantity,
accidental errors may be corrected to some
degree, but systematic errors can not be
Combination/Propagation of Errors
Errors in Compound Quantities
Errors in sum/difference
A, B the two quantities with their errors as ΔA and ΔB
respectively
Measured value: (A ± ΔA) and (B ± ΔB)
Let Z = A + B
Z + ΔZ = (A ± ΔA) + (B ± ΔB)
= (A + B) ± (ΔA + ΔB)
ΔZ = ± (ΔA + ΔB)
Maximum error in Z = (ΔA + ΔB)
Similarly: for Z = A – B
Z – ΔZ = (A-B) ± (ΔA + ΔB)
Errors in Product/Division
𝐿𝑒𝑡 𝑍 = 𝐴 ∗ 𝐵;
𝑍 ± ∆𝑍 = 𝐴 ± ∆𝐴 ∗ 𝐵 ± ∆𝐵
𝑍 ± ∆𝑍 = 𝐴𝐵 ± ∆𝐴. 𝐵 ± 𝐴. ∆𝐵 ± ∆𝐴. ∆𝐵
𝐷𝑖𝑣𝑖𝑑𝑒 𝑏𝑦 𝑍:
∆𝑍
∆𝐴 ∆𝐵 ∆𝐴. ∆𝐵
1 ±
=1 ±
+
+
𝑍
𝐴
𝐵
𝐴. 𝐵
∆𝑍
∆𝐴 ∆𝐵
=
+
𝑍
𝐴
𝐵
∆𝐴 ∆𝐵
∆𝑍 = 𝐴. 𝐵
+
𝐴
𝐵
Let Z = A/B;
Z ± ∆Z = A ± ∆A / B ± ∆B
∆𝐴
𝐴
∆𝐴
∆𝐵
𝐴
𝑍 ± ∆𝑍 =
=
1±
1±
∆𝐵
𝐵
𝐴
𝐵
𝐵 1±
𝐵
𝐴 1±
𝐴
∆𝐴
∆𝐵
𝑍 ± ∆𝑍 =
1±
1±
𝐵
𝐴
𝐵
=
𝐴
𝐵
1 ±
∆𝐴
𝐴
±
∆𝐵
𝐵
±
∆𝐴.∆𝐵
𝐴.𝐵
∆𝐴
∆𝐵
𝑍 ± ∆𝑍 = Z ± 𝑍.
± 𝑍.
𝐴
𝐵
A ∆A ∆B
∆Z = ±
+
B A
B
−1
Error in Power
𝐿𝑒𝑡 𝑍 = 𝐴2 ;
𝑍 ± ∆𝑍 = 𝐴 ± ∆𝐴 2 = 𝐴2 ± 2𝐴∆𝐴 ± ∆𝐴
= 𝑍 ± 2𝐴∆𝐴
∆𝑍 = 2𝐴. ∆𝐴  Error in Z
∆𝑍
Relative Error in Z,
𝑍
=
2𝐴.∆𝐴
𝐴2
=
2
∆𝐴
2.
𝐴
Relative error in Z is twice the relative error in A.
In General: Relative error in 𝑍 =
∆𝑍
𝑍
=
∆𝐴
𝑝.
𝐴
+
𝐴𝑃 𝐵𝑞
𝐶𝑟
∆𝐵
𝑞.
𝐵
+
is given by
∆𝐶
𝑟.
𝐶
Q.: % of error in mass and speed are 2% and 3%.
1
How much is the max. error in K.E (𝐸 = 𝑚𝑣 2 )
2
∆𝐸
∆𝑚
∆𝑣
2
3
8
=
+ 2.
=
+ 2.
=
𝐸
𝑚
𝑣
100
100
100
∆𝐸
𝐸
= 8%
𝑎3 𝑏 2
𝑐.𝑑
Q: A physical quantity is given by 𝑃 =
% of errors in a, b, c and d are 1%, 3%, 4% and 2%.
If value of P = 3.763, to what value it should be
rounded off.
∆𝑃
∆𝑎
∆𝑏 1 ∆𝑐 ∆𝑑
= 3.
+ 2.
+
+
𝑃
𝑎
𝑏
2 𝑐
𝑑
1
3
1 4
2
= 3.
+ 2.
+
+
=
100
If P = 3.763,
100
∆𝑃 = 𝑃
2 100
13
×
100
100
13
100
= 13%
= 0.48919
The result should be rounded off to 3.8 as there is
uncertainty even in regard to 2nd sig. fig., i.e. 7 in 3.763
Errors in Compound Quantities …… contd.
If y is some function of x, the error in y is obtained due to
error in X using some mathematical technique
Q = a*b, with f1 and f2 the fractional errors in a, b.
Q = 𝑎0 1 + 𝑓1 ∗ 𝑏0 1 + 𝑓2
= 𝑎0 + 𝑎0 𝑓1 ∗ 𝑏0 + 𝑏0 𝑓2
neglect
𝑄 = 𝑎0 𝑏0 + 𝑎0 𝑏0 𝑓1 + 𝑎0 𝑏0 𝑓2 + 𝑎𝑎00𝑏𝑏0 𝑓
𝑓11𝑓𝑓22
𝑄 ≈ 𝑎0 𝑏0 1 + 𝑓1 +𝑓2
Fractional error in Q is 𝑓1 +𝑓2 , i.e. sum of fractional errors in
a and b
Error in Quotient
𝑎
𝑎0 1 + 𝑓1
𝑎0
𝑄= =
=
1 + 𝑓1 ∗ 1 − 𝑓2 + ⋯
𝑏
𝑏0 1 + 𝑓2
𝑏0
≃
𝑎0
𝑏0
1 + 𝑓1 − 𝑓2
Thus the fractional error in a/b is approximately the
difference in fractional errors in a and b
In general if 𝑄 =
𝑎𝑏𝑐…
,
𝑙𝑚𝑛…
fractional error in Q = (sum of fractional error in a, b, c, ….) (sum of fractional error in l, m, n, ….)
Examples
Ohm’s law:
I = E/R, with fE and fR as the fractional errors in E and R
Fractional error in i = fE – fR.
Acceleration due to gravity
𝑔=
=
4𝜋2𝑙
𝑇2
with l = l0(1+f1) and T = T0(1+f2)
4𝜋 2 𝑙0 1 + 𝑓1
2
=
4𝜋 2 𝑙0 1 + 𝑓1
𝑇0 1 + 𝑓2
𝑇0 2 1 + 2𝑓2
4𝜋 2 𝑙0
=
1 + 𝑓1 − 2𝑓2
2
𝑇0
= 𝑔0 1 + 𝑓1 − 2𝑓2
2
Use of Calculus:
X is a measured quantity with δX as error.
Y = f(X) estimated with δY as error.
𝜕𝑌
lim
𝛿𝑋→0 𝜕𝑋
=
𝑑𝑌
𝑑𝑋
𝒆𝒓𝒓𝒐𝒓 𝒊𝒏 𝒀 =
→
𝜕𝑌
𝜕𝑋
≡
𝑑𝑌
𝑑𝑋
if δX is small.
𝒅𝒀
. 𝝏𝑿
𝒅𝑿
As an example, let Y be the area of a circle of radius x,
Then 𝑌 = 𝜋𝑋 2
𝑑𝑌
𝑑𝑋
= 2π𝑋 →
𝜕𝑌 = 2𝜋𝑋𝜕𝑋
𝜕𝑌
𝜕𝑋
= 2𝜋𝑋
𝜕𝑌 𝑚𝑎𝑦 𝑏𝑒 𝑝𝑜𝑠𝑖𝑡𝑖𝑣𝑒 𝑜𝑟 𝑛𝑒𝑔𝑎𝑡𝑖𝑣𝑒 ,
𝑑𝑒𝑝𝑒𝑛𝑑𝑖𝑛𝑔 𝑜𝑛 𝑡ℎ𝑒 𝑠𝑖𝑔𝑛 𝑜𝑓 𝜕𝑋
𝑖𝑓 𝛿𝑋 𝑖𝑠 𝑠𝑚𝑎𝑙𝑙
Fractional error in the area
𝜕𝑌
2𝜋𝑋𝜕𝑋
𝜕𝑋
=
= 2
2
𝑌
𝜋𝑋
𝑋
𝑖. 𝑒. , twice the fractional error in radius.
In general if Q is function of several measured quantities, X, Y, Z,
…… the error in Q due to errors 𝜕𝑋, 𝜕Y, 𝜕Z is
𝜕𝑄
𝜕𝑄
𝜕𝑄 =
𝜕𝑋 +
𝜕𝑌 + … … … … . .
𝜕𝑋
𝜕𝑌
This is often called as principle of superposition of
errors.
In general if Q =f(X,Y,Z, ….)
𝜕𝑄
𝜕𝑄
𝜕𝑄 =
𝜕𝑋 +
𝜕𝑌 + … … … … . .
𝜕𝑋
𝜕𝑌
→ 𝑝𝑟𝑖𝑛𝑐𝑖𝑝𝑙𝑒 𝑜𝑓 𝑠𝑢𝑝𝑒𝑟𝑝𝑜𝑠𝑖𝑡𝑖𝑜𝑛 𝑜𝑓 𝑒𝑟𝑟𝑜𝑟𝑠
error in Q due to X alone and 𝜕𝑌, 𝜕𝑍, … … . = 0
Let
𝜕𝑋 have value between -e1 to +e1
𝜕Y have value between –e2 to e2
𝜕𝑄
2
=
𝜕𝑄
𝜕𝑋
∗ 𝑒1
2
+
𝜕𝑄
𝜕𝑋
∗ 𝑒2
2
+ …………
𝜕𝑄 is the square root of sum of squares of greatest errors due to
errors in each.
Example: Simple pendulum
g=
4𝜋2 𝑙
𝑇2
𝜕𝑔 =
𝜕𝑔
𝜕𝑙
𝜕𝑙
+
𝜕𝑔
𝜕𝑇
𝜕𝑇
4𝜋2
8𝜋2 𝑙
𝜕𝑔 = 2 𝜕𝑙 − 3 𝜕𝑇
𝑇
𝑇
𝜕𝑔
𝜕𝑙
𝜕𝑇
= −2
𝑡ℎ𝑎𝑡
𝑔
𝑙
𝑇
𝑖𝑠 𝑓1 − 2𝑓2
Example: A quantity Y is expressed in terms of a
measured quantity X by the relation, Y = 4X – (2/X). Find
the percentage of error in Y if there is an error of 1% in X.
𝑑𝑌
𝑑𝑋
= 4 + 2/𝑋 2
2
𝜕𝑌 = 4 + 2 . 𝜕𝑋
𝑋
𝜕𝑌
𝑌
percentage error in Y = ∗ 100
100 ∗ 4 + 2/𝑋 2
100 ∗ 4𝑋 2 + 2
=
𝜕𝑋 =
𝜕𝑋
2
(4𝑋 − 2/𝑋)
𝑋(4 𝑋 − 2)
if
𝜕𝑋
𝑋
4𝑋 2 +2
= 1/100, % error in y is 2
4𝑋 −2
For a free falling object, distance is given by
1
𝑔𝑡 2
2
𝑥=
𝑡ℎ𝑒𝑛
𝑑𝑥
= 𝑔𝑡 𝑜𝑟 𝑑𝑥 = 𝑔𝑡. 𝑑𝑡
𝑑𝑡
i.e. change in time dt causes a change in x by dx.
If
𝑔 = 10𝑚/𝑠2 𝑎𝑛𝑑
𝑥=
1
𝑔𝑡 2
2
=
1
2
𝑡 = 4 ± 0.55𝑠
× 10 × 16 = 80 𝑚.
𝑑𝑥 = 𝑔𝑡. 𝑑𝑡 = 10 × 4 × 0.5 = 20 𝑚.
In general 𝒅𝒚 =
𝒅𝒇
𝝏𝒙 ,
𝒅𝒙
often tedious
Some Statistical Ideas
The experimental scientist does not regard statistics as
an excuse for doing bad experiment.
Frequency distributions:
Numerical data on scientific measurements (and
industrial and social statistics) are often represented
graphically to aid their appreciation.
First step
To arrange these in convenient order if they are large in
numbers. This is often done by grouping them into
classes according to their magnitude
or
according to suitable intervals of a variable on which
they depend.
The number of data in a particular class or group is
usually called the frequency for that class
The following represents the frequency distribution of
student’s marks (in %) in an exam
Class
Freq.
Class
Freq.
0-9
2
50-59
32
10-19
5
60-69
25
20-29
6
70-79
10
30-39
14
80-89
2
40-49
22
90-99
2
Width of class is the difference between
the1st numbers of two consecutive
classes, i.e. 10 here
From the table one can appreciate
the distribution of marks
But Graphically.
Frequency Distribution
Polygon
Bar Graph
Frequency Distribution
Histogram
Histogram:
A series of rectangles are
constructed of width equal to class
width and area equal to frequency
of the corresponding class.
The areas are 2, 5, 6, 14,... units.
Total area of histogram
= 120 units = total no. of students.
The heights of rectangles are
proportional to freq. when
classes are of equal width.
In this case, the mean height of
rectangles = mean frequency
The Mean:
Mean freq. is not of particular significance.
What is often important is mean of the data.
If f1, f2, …. are frequencies in various classes
and x1, x2, …. are mid values of variables then
Mean value of the variable is given by
(f1x1 + f2x2+ ……… + fnxn)/(f1 + f2 + …….. + fn),
This is weighted mean of x1, x2, ……. , xn; the weights
being the freq. of corresponding classes.
It is written as
𝑛
𝑖=1 𝑓𝑖 𝑥𝑖 /
𝑛
𝑖=1 𝑓𝑖
or
𝑓𝑥
𝑓
= 𝑥
To evaluate 𝑥 may be sometimes tedious. The arithmetic can be
minimized as:
Let 𝑥𝑖 = 𝑥 ′ 𝑖 + 𝑚, where m is some constant then
𝑓1 𝑥1 + 𝑓2 𝑥2 + … … . +𝑓𝑛 𝑥𝑛
= 𝑓1 𝑥 ′1 + 𝑚 + 𝑓2 𝑥 ′ 2 + 𝑚 + … … 𝑓𝑛 𝑥 ′ 𝑛 + 𝑚
= 𝑓1 𝑥 ′1 + 𝑓2 𝑥 ′ 2 + … 𝑓𝑛 𝑥 ′ 𝑛 + 𝑚 𝑓1 + 𝑓2 + … + 𝑓𝑛
Therefore
𝑓1 𝑥1 + 𝑓2 𝑥2 + …+ 𝑓𝑛 𝑥𝑛
𝑓1 +𝑓2 + …+ 𝑓𝑛
𝒙 = 𝒙′ + 𝒎
=
𝑓1 𝑥 ′ 1 +𝑓2 𝑥 ′ 2 + …𝑓𝑛 𝑥 ′ 𝑛
𝑓1 +𝑓2 + …+ 𝑓𝑛
∶ 𝒙′ 𝒊𝒔 𝒎𝒆𝒂𝒏 𝒐𝒇 𝒙′𝒊
+𝑚
By choosing m conveniently we can make
evaluation of 𝑥 ′ simpilar than evaluation of 𝑥.
𝑥 ′ will be small if m is close to 𝑥
m is often called the working mean or assumed
mean.
Example
From the Table, below, the mean value can be estimated as
172
10
x=
=3
≅ 3.9
44
11
xi
fi
fixi
𝒙′ 𝒊 = 𝒙𝒊 − 𝟑. 𝟓
𝒇𝒊 𝒙′ 𝒊
0.5
1
0.5
-3
-3
1.5
5
7.5
-2
-10
2.5
7
17.5
-1
-77
3.5
9
31.5
0
0
4.5
10
45.0
1
10
5.5
8
44.0
2
16
6.5
4
26.0
3
12
SUM
44
172.0
18
However, examining the data, we find that the mean lies between
3.5 and 4.5
Taking m = 3.5 values of x’i are tabulated in 4th column
Values of fixi are presented in the last column
=
18
44
Therefore 𝑥 =
𝑥′
This gives
𝑥′
+𝑚 =
18
44
+ 3.5 =
10
3
11
≃ 3.9, as found directly
Thus the amount of arithmetic, and consequently the
likelihood of error, is reduced.
The Median :
If a set of observations are arranged in ascending or
descending order of magnitude the observation in the
middle of the set is called MEDIAN.
If number of observations are odd  (2n+1) say then
median is (n+1)th value.
If no. of observations are even  2n say then median is
the mean of nth and (n+1)th term.
e.g. 10, 12, 13, 7, 20, 18, 9, 15, 11
In order 7, 9, 10, 11, 12, 13, 15, 18, 20 then median = 12.
If last value 20 is not there, then median =
11+12
2
= 11.5
To find the Median, place the
numbers you are given in
order and find the middle
number.
To find the median of
{13, 23, 11, 16, 15, 10, 26}.
We put them in order:
{10, 11, 13, 15, 16, 23, 26}
The middle number is 15, so
the median is 15.
(If there are two middle
numbers, average of the two
is taken.)
MODE:
The number which appears most often in a set of numbers.
in {6, 3, 9, 6, 6, 5, 9, 3} the Mode is 6 (it occurs most often).
Mode corresponds to maximum frequency.
The histogram tends to be more and more close to the
frequency curve as the number of observations increases.
The Mean Deviation
The mean of the distances of each value from their
mean.
Three steps to find the mean deviation
• Find the mean of all values
• Find the distance of each value from that mean
(subtract the mean from each value, ignore minus
signs)
• Then find the mean of those distances
Mean Deviation of 3, 6, 6, 7, 8, 11, 15, 16
Mean = (3+6+6+7+8+11+15+16)/8
= 72/8 = 9
distance of each value from that mean is evaluated as
Mean deviation
= (6+3+3+2+1+2+6+7)/8 = 30/8
= 3.75
It tells how far, on average, all values
are from the middle
Value
3
6
6
7
8
11
15
16
Distance
from 9
6
3
3
2
1
2
6
7
Measures of Dispersion :
An important characteristic of a set of data)
or scatter of the data about some value, e.g.
mean.
For the numbers: 40, 50, 60, 70, 80
mean = 60
Scatter or dispersion = 20 on each side.
They are all within limits 60 ± 20.
Various parameters are used to measure
dispersion.
1. Range
2. Mean deviation
3. Standard deviation
Range of freq. distribution
= max value – min value of variable.
It is a simple measure of dispersion but has
limitations due to simplicity.
Mean Deviation :
If x1, x2 …….. xn are set of data and 𝑥 is mean.
Deviation from the mean is
𝑥1 − 𝑥 , 𝑥2 − 𝑥 … … 𝑥𝑛 − 𝑥 𝒐𝒓 𝑑1 , 𝑑2 , … … , 𝑑𝑛
Some of the di values are positive and some negative.
But
𝑑1 + 𝑑2 + … … + 𝑑𝑛 = 0
𝑖. 𝑒.,
𝑥1 +𝑥2 + … … . + 𝑥𝑛 − 𝑛𝑥 = 0
Mean deviation or mean absolute deviation is given by
𝑑1 + 𝑑2 + … . + 𝑑𝑛
1
=
𝑑𝑖
𝑛
𝑛
Example
For the numbers 7, 5, 8, 10, 12, 6
Mean = 8
Mean deviation =
1
6
1+3+0+2+4+2 =2
However, for large data, we use frequency
𝑓1 𝑑1 + 𝑓2 𝑑2 + … . +𝑓𝑛 𝑑𝑛
=
𝑓1 + 𝑓2 + … … + 𝑓𝑛
𝑓𝑖 𝑥𝑖 − 𝑥
𝑓𝑖
Standard Deviation (σ) :
(The most important measure of deviation)
Defined in terms of the deviations from the mean
𝜎2
2
2
= 𝑑1 + 𝑑2 + … … + 𝑑𝑛
=
1
𝑛
𝑥𝑖 − 𝑥
2
1
𝑛=
𝑛
𝑑𝑖 2
2
If x1, x2, ….., xn have frequencies f1, f2, ……, fn
respectively.
𝝈𝟐 = 𝒇𝟏 𝒅𝟏 𝟐 + 𝒇𝟐 𝒅𝟐 𝟐 + … + 𝒇𝒏 𝒅𝒏 𝟐
𝒇𝟏 + 𝒇𝟐 + … . + 𝒇𝒏
𝜎 2 is variance of the data.
𝝈 is root mean square deviation of the data
measured from the mean.
For the numbers 5, 6, 7, 8, 10, 12
Mean = 8 and
34
σ = 9 + 4 + 1 + 0 + 4 + 16 6 =
= 5.67
6
σ = 2.38
2
Frequency Distributions (Three important Types)
1. Binomial
2. Poisson
3. Normal
Most of the distribution, based on scientific observations or
industrial and social statistics approach closely to on or the
other of these three important distributions.
These distributions can
be derived and expressed
mathematically using the theory of prob.
The Binomial distribution:
On tossing a coin
we have two probabilities Heads or tails
i.e., chances of getting heads or tails is 50%
For ‘n’ number of tosses (n is large)
No. of times we get heads ≈ n/2.
For small n, chances of deviation from 50% are large
Tossing a coin 10 times
For another 10 tosses
we may get heads 3 times
we may get heads 6 times
Question: On tossing ‘n’ times what is the chance of
getting heads ‘m’ times? 0≤ m ≤ n.
For ‘n’ tosses, probabilities of getting heads 0, 1,
2, … , n times are given by successive terms of
1
1 𝑛
binomial expansion of +
.
2
2
In general, if probability of occurring certain
events is p and that it will not happen is q, such
that (p+q = 1) then
the probability that it will happen on 0, 1, 2, …….,
n out of n occasions are given by successive terms
of binomial expansion.
𝒒+𝒑
𝒏
=
𝒒𝒏
𝒏 𝒏 − 𝟏 𝒏−𝟐 𝟐
+
𝒒 𝒑
𝟐!
+ … … + 𝒑𝒏
+ 𝒏𝒒𝒏−𝟏 𝒑
On tossing a coin 10 times, prob. of getting heads on
0, 1, 2, ….., 10 times are given by the successive
terms of the expansion of
1
, 10
10
2
×
𝟏
𝟐𝟗
×
1
2
+
1 10
,
2
which are
𝟏 𝟏𝟎×𝟗 𝟏 𝟏 𝟏𝟎×𝟗×𝟖 𝟏 𝟏
,
,
,……..
𝟖
𝟐
𝟕
𝟑
𝟐
𝟐! 𝟐 𝟐
𝟑!
𝟐 𝟐
1
= 10 1, 10, 45, 120, 210, 252, 210, 120, 45, 10, 1
2
Thus chances of getting heads 3 times in 10
𝟏𝟎×𝟗×𝟖 𝟏 𝟏
𝟏𝟓
tosses =
𝟕 𝟑 =
𝟑!
𝟐
𝟐
𝟏𝟐𝟖
If we toss a coin 3 times, we can get
HHH HHT HTH HTT THH THT TTH TTT
Each outcome is equally likely, and there are 8 of
them. So each has a probability of 1/8
So the probability of “two Heads" in “three tosses” is:
Total No. of
Attempts
23 = 8
No. of outcome
Prob. of each
we want
outcome
3
X
1/8
= 3/8
Thus we have
•
•
•
•
P(3-heads) = P(HHH) = 1/8
P(2-Heads) = P(HHT) + P(HTH) + P(THH) = 1/8 + 1/8 + 1/8 = 3/8
P(1-Head) = P(HTT) + P(THT) + P(TTH) = 1/8 + 1/8 + 1/8 = 3/8
P(Zero Heads) = P(TTT) = 1/8
We can write this in terms of a
Random Variable, X
• P(X = 3) = 1/8
• P(X = 2) = 3/8
• P(X = 1) = 3/8
• P(X = 0) = 1/8
We can now make a formula
No. of tosses, n = 3 and No. of heads we want, k = 2
𝑛!
3!
=
=3
𝑘!(𝑛−𝑘)! 2!(3−2)!
Similarly, chances of 5 heads in 9 tosses
𝑛!
𝑘!(𝑛−𝑘)!
9!
=
=
5!(9−5)!
126
While for 9 tosses total attempts = 29 = 512
No of outcome we want = 126
Prob. of each outcome = 1/512
Prob. of getting 5 heads in 9 tosses = 126/512
i.e., P(x=5) = 126/512 = 0.2460
i.e. 25% chance
In general Prob. Of getting k out of n ways
P(k,n) =
𝑛!
𝑘! 𝑛−𝑘 !
pk (1-p)n-k
Which is general Binomial probability formula
Poisson Distribution
The form of binomial distribution varies considerably
depending upon the values of p and n.
An important practical case is when the probability of
occurring an event, p is very small but n is very large so
that n × p is not insignificant.
Taking np = m, it can be shown that the binomial
expansion of (q + p)n approximates closely to the series
𝑚
𝑚
𝑒 (1 +
1
+
𝑚2
2!
+
𝑚3
3!
+....+
𝑚𝑛
),
𝑛!
where e =2.71828
called as Poisson series and any distribution which
correspond to the successive terms is called Poisson
distribution.
Relative frequency r with which an event happens n
times varies as
n
𝑒𝑚r
0
1
1
2
3
s
m
𝑚2
2!
𝑚3
3!
𝑚𝑠
𝑠!
....
For a true probability distribution sum of prob. = 1
For Poisson distribution, sum of relative freq.
𝑒 −𝑚 (1 +
= 𝑒 −𝑚 ×
=1
𝑚2 𝑚3
𝑚𝑠
+
+
+....+ )
2!
3!
𝑠!
𝑒 𝑚 (approx. if s is very large)
𝑚
1
Also with these approximation,
Mean of the distribution is m
and
standard deviation is 𝐦
𝟏
𝟐
In general, prob. Of occurring of an event n times is
𝑒−𝑚 𝑚𝑛
𝑛!
Example
Arrivals of a particular charged particle noticed by a
detector follow a Poisson distribution with an
average of 4.5 particles in every 15 sec.
Obtain a bar-plot of the distribution, assuming a
maximum of 20 arrivals in 15 sec. Also calculate the
probability of fewer than 3 arrivals in this time
interval.
Prob. of arrival of 0, 1, 2, . . . Particles are given by
𝑒−𝑚 𝑚𝑛
P(n) =
𝑛!
Thus with m = 4.5
P(0) =
𝑒−4.5 4.50
= 0.011
0!
Similarly P(1) = 0.04999 and P(2) = 0.11248
Hence prob. of fewer than 3 arrivals
P(0) + P(1) + P(2) = 0.17358
m=3
m=5
m = 10
Example:
Given are the freq. f of n successors in a set of 500 trials.
Find the mean of distribution and verify that the
distribution is roughly of Poisson type. Show that the
mean ≅ variance.
n
0
1
2
3
4
5
6
7
8
9
SUM
f
24
77
110
112
84
50
24
12
5
2
500
n×f
0
77
220
336
336
250
144
84
40
18
1505
𝒆−𝒎 𝒎𝒏
𝒏!
24.6
74.2
84.3
50.8
25.4
11.0
4
1.4
11.6 112.0
X 500
Mean number of success, m = n×f/f = 1505/500 = 3.01
The terms of Poisson series with m = 3.01
e−3.01
3.012 3.013
1 + 3.01 +
+
+ ……
2!
3!
On multiplying by 500, the successive terms are
24.6, 74.2, 11.6, 112.0, 84.3, 50.8, 25.4, 11.0, 4.0, 1.4
which match with the values of f given in the Table
Using an assumed mean = 3, d = n-3, we may tabulate
Hence
n
d=n-3
fd
fd2
0
-3
-72
216
1
-2
-154
308
2
-1
-110
110
3
0
0
0
4
1
84
84
5
2
100
200
6
3
72
216
7
4
48
192
8
5
25
125
9
6
12
72
sum
5
1523
mean = 3 +
1523
variance =
500
5
500
= 3.01
- (0.01)2 = 3.0459 ≃ 3.05
Normal Distribution
• Discovered in 1733 by de Moivre as an approximation to
the binomial distribution when the number of trails is large
• Also obtained by Laplace and Gauss later
• Importance lies in the Central Limit Theorem, which states
that the sum of a large number of independent random
variables (binomial, Poisson, etc.) will approximate a
normal distribution
The equation
−ℎ2
y = A𝑒
𝑥−𝑚
2
,
where A, h , m are constants
is know as normal error curve
y
A
0.6A
σ
0.2A
2σ
x=m
normal distribution
The normal distribution curve
• has is its maximum value at x = m, i.e., y = A
• is symmetrical about the line x = m
The area under the curve is given by
+∞
ℎ2 (𝑥−𝑚)2
𝐴𝑒
dx
=
(A
𝜋)/h
−∞
If A = h/ 𝜋, the area under the curve is unity and the
equation of the curve is
y=
ℎ −ℎ2
𝑒 (𝑥
𝜋
− 𝑚)2
This is frequency curve, called a normal or
Gaussian distribution.
Any Gaussian distribution is determined by two
parameters h and m
m is the mean of the distribution
h is often called precision constant and is related to
the standard deviation σ such that 2σ2h2 = 1
Taking m = <x> and h2 = 2σ2
y
2
2/2𝜎
1
=
𝑒 −(𝑥−<𝑥>)
𝜎 2𝜋
which gives Gaussian distribution with
mean = <x> and standard deviation = σ
Gaussian distribution for the same mean
and different σ
Example
From a sample of fish in a pond it is found that the
mean length of these fish, m is 30cm while 2 = 4cm
We assume that the length is normal random
variable
If we catch one of these fish the what is the
probability that
• it will be at least 31 cm. long?
• it will be no more than 32 cm. long?
• its length will be between 26 and 29 cm?
Prob. that it will be 31 cm long
P
Prob. that it will be less than 32 cm
Prob. that it length will be between 28 and 30 cm
The Normal or Gaussian Law of errors
2
2/2σ
1
The function, y =
𝑒 −(x−<x>)
𝜎 2𝜋
Defines the normal frequency distribution and is
called the Gaussian law of errors.
The two parameters, <x> and 𝜎 charecterize such
distribution
This law states that a set of measurements involving
accidental errors are distributed about the mean.
Mean is the best estimated value of the quantity,
while 𝜎 estimates the best accuracy
The value of the measured quantity is <x> ± α
α = σ/ 𝒏 is standard error of the mean
n is the number of observations.
Standard error of σ is σ/ 𝟐𝐧
Example:
Values of distances covered by a tourist on a vehicle are listed
( in km/day). Find the mean distance and its standard error
782
798
786
774
771
776
sum
Working
mean =
780
Residual d
2
18
d2
4
324
6
-6
-9
-4
36
36
81
16
7
497
Mean = working mean + <residual>
= 780 + 7/6 = 781
Alternative
Mean = 782 + 798 + 786 + 774 + 771 + 776
<x> = 781.16 = 781
Standard deviation, σ = < 𝑥 2 >−< 𝑥 >2
= 610302 − 781 ∗ 781
=9
Standard error of mean = σ/ 𝒏 = 9/ 𝟔 = 3.68 = 4
Standard error of s.d. = σ/ 𝟐𝒏 = 9/ 𝟏𝟐 = 2.6 = 3
Standard Errors of Compound Quantities
If a number of measured quantities have means m1, m2, m3, . .
. . mn with standard errors α1, α2, α3, . . . .,αn respectively then
the standard error of
Quantity
Error
The sum
m1+ m2
𝜶 𝟏𝟐 + 𝜶 𝟐𝟐
The difference
m1- m2
𝜶 𝟏𝟐 + 𝜶 𝟐𝟐
The multiple
km1
The product
m1m2
𝒎𝟐𝟐 𝜶𝟏𝟐 + 𝒎𝟏𝟐 𝜶𝟐𝟐
m1m2m3
𝟐
𝜶
=
𝒎𝟏 𝒎𝟐 𝒎𝟑
k𝜶𝟏
k is some constant
+
Power
𝒎𝟏
𝒑
α = 𝒑𝒎𝒑−𝟏 𝜶𝟏
𝜶𝟏 𝟐
𝒎𝟏
𝜶𝟐 𝟐
𝜶𝟑 𝟐
+
𝒎𝟏
𝒎𝟏
Weighted mean
A mean where some values contribute more than others.
When we do a simple mean (or
average), we give equal weight
to each number.
mean = (1 + 2 + 3 + 4)/4
= 2.5
Each of the four numbers has a
weight of ¼
Mean = ¼×1 + ¼×2 + ¼×3 +
¼×4
= 0.25 + 0.5 + 0.75 +1.0
= 2.5
If weight of 3 is changed to 0.7, the other three
numbers have still equal weights of 0.1 each so that
total weight is 1
Mean = 0.1 × 1 + 0.1 × 2
+ 0.7 × 3+ 0.1 × 4
= 2.8
This weighted mean is now a
little higher ("pulled" there by
the weight of 3).
When some values get more
weight than others
the central point (the mean) can
change:
Least-Squares Fitting
Fitting requires a parametric model that relates the response
data to the predictor data with one or more coefficients. The
result of the fitting process is an estimate of the model
coefficients.
Coefficients are obtained by minimizing the summed squares
of residuals.
Residuals are defined as
𝑟𝑖 = 𝑦𝑜𝑏𝑠 - 𝑦𝑐𝑎𝑙
The least-squares method minimizes the summed square of
residuals. The residual is identified as the error associated
with the data.
Example:
𝑦 = 𝑎 + 𝑏𝑥
𝑎=
b=
y x2 −
n x2 −
x xy
x 2
n xy− x y
n x2 − x 2
intercept
slope
Example:
x
y
ycal
d
d2
1.0
2.4
2.3
0.1
0.01
2.0
3.9
4.1
0.2
0.04
3.0
6.1
5.9
0.2
0.04
4.0
8.3
7.7
0.6
0.36
5.0
9.5
9.5
0.0
0.00
6.0
11.4
11.3
0.1
0.01
Σ x = 21.0 Σ y = 41.6 Σx2 = 91.0
This gives 𝒚 = 𝟎. 𝟓 + 𝟏. 𝟖𝒙
Σ d2 = 0.46
Σxy = 177.6
α2 = Σd2/(n-2) = 0.12
Errors in coefficients
𝜶𝟏 = ±
𝜶𝟐 𝒙𝟐
n x2 − x 2
𝜶𝟐 = ±
𝒏𝜶𝟐
x2 − x 2
n
= 0.63
Fitting
• To find a functional form that describes data
within errors
• Why fit data?
• Extract physical parameters from data
• Test validity of data or model
• Interpolate/extrapolate data
Goodness of fit
• Data are expected to fluctuate by ~ error
• Chi-square per Degree of Freedom (DoF) should
be ~ 1
• DoF: Number of data points - number of fitted
parameters
• Chi-square allows to test if model/fitted function
is compatible with data
How to fit data?
Vary parameters of function until you find a global
maximum of goodness-of-fit criterion
Goodness-of-fit
• chi-square (most common)
• likelihood
• (Kolmogorov-Smirnov test)
observed value
Chi square
𝜒2=
calculated value
𝑥𝑜𝑏𝑠 2 −𝑥𝑐𝑎𝑙 2
𝜎2
estimated uncertainty
Best Fit: Global minimum of Chi-square
X
Y± Error
X2
X.Y
Ycal
d2
χ2
1
5.1 ± 0.6
1
5.1
4.83
0.072
0.202
2
10.2 ± 0.9
4
20.4
9.96
0.057
0.071
3
14.7 ± 1.4
9
44.1
15.09
0.152
0.077
4
19.5 ± 1.8
16
78.0
20.22
0.518
0.160
5
25.9 ± 2.2
25
129.5
25.35
0.302
0.062
6
30.7 ± 3.3
36
184.2
30.48
0.048
0.004
ΣX=21 , ΣY=106.1 ,
Σd2=1.149 ,
ΣX2=91 ,
Σ χ2=0.576
ΣX.Y=461.3
a
2
Y
.
X
    X . X .Y
b

2
n. X   X . X
2
n. X .Y   X . Y
n. X
d


2
  X . X
2
n2
= 0.287
= - 0.3
= 5.13
1  
 2 . X 2
n. X   X . X
2
n. 2
2  
n. X 2   X . X

D.F
= ± 0.118
Ycal = -0.3 ±0.460 + (5.13±0.118)X
Which gives :
And
= ± 0.460
2
0.576

= 0.144
4
Chi-square distribution
Even if the error estimate is correct and the model is
correct
• Chi-square will fluctuate from chi2/DoF < 1 to
chi2/DoF > 1
• The shape of the chi2-distribution depends only on
the number of degrees of freedom
• Probability that data and model agree can be
calculated
Chi-square probability
Percentage of all measurements that have a worse
chi-square than expected
Books
1. Ideas about Errors
by
J. Topping
2. A Practical Guide to Data analysis for
Physical Science Students by Louis Lyons
Few figures taken from the net
Thank you