Statistical Design

1. • DATA: Observations( Shaft Diameter in 1 shift)
• POPULATION: Group of all identical components (Shafts)
• SAMPLE: Some portion of Population
Statistical Techniques
Collection/Processing/Analysis/Interpretation of DATA
Statistics: Drawing conclusion from DATA
SPPU Pune’ SRES’ Sanjivani College of Engineering Kopargaon

2. Papers Sold Frequency
18 2
19 0
20 4
21 0
22 2
23 1
24 0
25 1
Frequency Distributions
Numbers of newspapers sold during last 10 days:
Raw Data: 22, 20, 18, 23, 20, 25, 22, 20, 18, 20
Papers Sold
Class Interval
Class Frequency
15-19 2
20-24 7
25-29 1
Grouped Frequency Distributions
FD: Representation of Raw Data in systematic and useful way
15-19: Lower and upper class limits
19-15 = 4 Class width
SPPU Pune’ SRES’ Sanjivani College of Engineering Kopargaon

3. Record of high temp for 50 cities(0F)
Frequency Distribution: Methods
Histogram:
• Graphical representation of the distribution of numerical data.
• To covey the data to viewers in pictorial form
Class frequency vs Class boundaries
SPPU Pune’ SRES’ Sanjivani College of Engineering Kopargaon

4. Frequency Polygon
Class frequency vs Class mid points

5. Central value
Measures of central tendency :
• Arithmetic Mean
• Median
• Mode
Parameters defining Frequency Distribution
There is a always a central value where the frequency of
observation is maximum while the remaining observations are
spread symmetrically about central value.
SPPU Pune’ SRES’ Sanjivani College of Engineering Kopargaon
Max observations
Min observations Min observations

6. Median – Middle value that separates the higher half from
the lower half of the data set.
Q1 A
Q2 A
Mode – Most frequent value in the data set.
Q3 A

7. µ =
𝑥1+ 𝑥2 + 𝑥3 + 𝑥4 + 𝑥5 + 𝑥6 + 𝑥 𝑁
𝑁
=
Ʃ 𝑥 𝑖
𝑁
Mean - Sum of all measurements /
Number of observations in the data set.
µ =
𝑓1 𝑥1 + 𝑓2 𝑥2 + 𝑓3 𝑥3 + 𝑓4 𝑥4 .. + 𝑓 𝑁 𝑥 𝑁
𝑓1+𝑓2+𝑓3…𝑓 𝑁
=
Ʃ𝑓 𝑖 𝑥 𝑖
𝑁
When x1 ,x2, x3 occur at frequencies f1, f2, f3…
Observed data x1, x2, x3 N: Total Observation
SPPU Pune’ SRES’ Sanjivani College of Engineering Kopargaon

8. The heights of KIDS (at the shoulders) are:
600mm, 470mm, 170mm, 430mm and 300mm.
Mean, the Variance, and the Standard Deviation.
Mean
170mm
300mm
430mm
470mm
600mm
µ =
𝑥1+ 𝑥2 + 𝑥3 + 𝑥4 + 𝑥5 + 𝑥6 ……. + 𝑥 𝑁
𝑁
=
Ʃ 𝑋 𝑖
𝑁

9. Difference of each KIDs height from the Mean
Square root of Variance
σ = 𝟐𝟏𝟕𝟎𝟒 = 147.32... = 147
Standard Deviation (σ)

10. Standard deviation σ
• SD is a measure of how Each Value in a data set
Deviates from the Mean.
• It measures how concentrated the data are around
the mean.
• Smaller standard deviation : More Concentrated
data
SPPU Pune’ SRES’ Sanjivani College of Engineering Kopargaon

11. µ =
𝑥1+ 𝑥2 + 𝑥3 + 𝑥4 + 𝑥5 + 𝑥6 . + 𝑥 𝑁
𝑁
=
Ʃ 𝑥 𝑖
𝑁
Mean - Sum of all measurements /
Number of observations in the data set.
Observed data x1, x2, x3 N: Total Observation
Variance and SD
V =
(𝑥1−µ)2+(𝑥2−µ)2+(𝑥3−µ)2… +⋯(𝑥𝑁−µ)2
𝑁
 =
(𝑥1−µ)2+(𝑥2−µ)2+(𝑥3−µ)2… + …(𝑥𝑁−µ)2
𝑁
=
Ʃ(𝑥i−µ)2
𝑁
SPPU Pune’ SRES’ Sanjivani College of Engineering Kopargaon

12. µ =
𝑓1 𝑥1 + 𝑓2 𝑥2 + 𝑓3 𝑥3 + 𝑓4 𝑥4 .. + 𝑓 𝑁 𝑥 𝑁
𝑓1+𝑓2+𝑓3…𝑓 𝑁
=
Ʃ𝑓 𝑖 𝑥 𝑖
𝑁
When x1 ,x2, x3 occur at frequencies f1, f2, f3…
V =
f1(𝑥1−µ)2+f2(𝑥2−µ)2+f3(𝑥3−µ)2 … +⋯fn(𝑥𝑁−µ)2
𝑁
 =
f1(𝑥1−µ)2+f2(𝑥2−µ)2+f3(𝑥3−µ)2 … + …fn(𝑥𝑁−µ)2
𝑁
=
Ʃ 𝑓𝑖(𝑥i−µ)2
𝑁
 =
Ʃ 𝑓𝑖(𝑥i)2−
[Ʃ (𝑓𝑖𝑥i)]2
𝑁
𝑁
 =
Ʃ 𝑓𝑖(𝑥i)2−
Ʃ [(𝑓𝑖𝑥i)]2
𝑁
𝑁 − 1
Population Sample

13. Class interval
Mpa Frequency
261-280 2
281-300 12
301-320 50
321-340 32
341-360 4
xi fi fi*xi fi*xi^2
270 2 540 145800
290 12 3480 1009200
310 50 15500 4805000
330 32 10560 3484800
350 4 1400 490000
Total 100 31480 9934800
µ =
Ʃ𝑓 𝑖 𝑥 𝑖
𝑁
=
31480
100
=314.8 Mpa
 =
Ʃ 𝑓𝑖(𝑥i)2−
Ʃ [(𝑓𝑖𝑥i)]2
𝑁
𝑁 − 1 15.86 𝑀𝑝𝑎
𝑉𝑎𝑟𝑖𝑎𝑛𝑐𝑒 = 251.48 𝑀𝑝𝑎
Q1
100 test specimen of GCI are tested on UTM to determine
Sult. Followings are the results:

14. Normal Distribution curve
Things close to ND
• Heights of people
• Size of things produced by machines
• Errors in measurements
• Blood pressure
• Marks in a test
Data tends to be around a central value.
Remaining observations are spread symmetrically about it.

16. 1) Random ND Curve
Area under curve is = N
SPPU Pune’ SRES’ Sanjivani College of Engineering Kopargaon
Random Variable: X
X vs F(x)
Types of ND curve 2) Standard ND Curve
(Natural)
1) Random ND Curve

17. 1) Random ND Curve
Area under curve is = 1
SPPU Pune’ SRES’ Sanjivani College of Engineering Kopargaon
Standard Variable: Z
Z vs F(Z)
Types of ND curve 2) Standard ND Curve
(Natural)
2) Standard ND Curve
(Natural)
Zi =
𝑥𝑖−𝞵
σ
F(z) =
1
2Π
𝑒−𝑧2/2
.

18. X= 𝞵 +Z σ
Z=1 Z=2
Z=3
X= 𝞵 + σ X= 𝞵 +2 σ X= 𝞵 +3 σ
Z =
𝑥−𝞵
σ
X= 𝞵 +Z σ
1) Random ND Curve 2) Standard ND Curve

19. Design and Natural Tolerance
Natural Tolerance = ± 3 σ
Limit within which all allowable items fall
Shaft dia 30 ± 0.01
UL, X1=30.01
LL, X2=29.99
𝞵 =30
Z1 =
𝑥1−𝞵
σ =−1.25
Z2 =
𝑥2−𝞵
σ =1.25
Design Tolerance
Specification Limit set by designer from the consideration of
matching of the two components.
1
2
1) DT < ± 3 σ
Rejection is Inevitable
2) DT = ± 3 σ
No Rejection
3) DT > ± 3 σ
Should be about
± 4 σ No Rejection

20. Z1 =
𝑥1−𝞵
σ
=+1.25
Area under Normal curve from 0 to Z
A1= 0.3944

21. Area under Normal curve from 0 to Z

22. 100 test specimen of GCI are tested on UTM to determine Sult.
Followings are the results:
Q2 200 Bearings bushes.
Internal diameter is normally distributed with the mean of 30.010 mm and
SD of 0.008 mm. UL:30.02 & LL:30.00 mm. Calculate % of rejected bushes.
Z1 =
𝑥1−𝞵
σ
=+1.25
N=200
X1=30.02
X2=30.00
µ=30.010
σ =0.008
Z2 =
𝑥2−𝞵
σ
=−1.25
Standard variable
A1=0.3944 A2=0.3944
Rejected bushes % =R∗100
=21.12%
A=A1+A2= 0.7888
SPPU Pune’ SRES’ Sanjivani College of Engineering Kopargaon
R=1-A =0.2112

23. σ =𝟎. 𝟎𝟏𝟓 𝐦𝐦 for all machines
Mean diameter of shafts fabricated on 03 machines is fond to be:
µ1 = 24.99 mm µ2= 25mm µ3= 25.01mm
Tolerance specifies by the designer for the diameter of shaft is:
25.000± 0.025 mm :
X1 = 25.00 − 0.025 X2 = 25.00 + 0.025
Machine A Machine B Machine C
µ 24.99 25 25.01
Z1 =
𝑥1−𝞵
𝜎
-1 -1.67 -2.33
Z2 =
𝑥𝟐−𝞵
σ
+2.33 +1.67 +1
𝐴1 = 𝐴𝑟𝑒𝑎 0 𝑡𝑜 𝑍1 0.3413 0.4525 0.4901
𝐴2 = 𝐴𝑟𝑒𝑎 0 𝑡𝑜 𝑍2 0.4901 0.4525 0.3413
A=A1+A2 0.8314 0.9050 0.8314
R=(1-A)*100 16.86% 9.5% 16.86%
Q3

24. Tension test. 120 specimen.
UTS is normally distributed with mean of 300Mpa. SD:25 Mpa.
1)How many specimen will have UTS less than 275 Mpa.
2)How many specimen have UTS between 275 and 350 Mpa
Z 1=
𝑥1−𝞵
σ
Z 1=
275−300
25
Z 1= −1
A1 =0.3413
for Z=1
R=(0.5-A1)100
=15.87%
R=0.1587*120
=19 Specimen
Z 2=
3505−300
25
=+2
A2 =0.4772
for Z=2
A =A1+A2=0.8185 =81.85%
A =0.8185 *120=98 specimen
Q4

25. Population Combinations
Bearing:
Inner diameter
Shaft:
Outer diameter
Population 1 Population 2
D1, D2, D3 d1, d2.Random Variable
µD =
𝐷1+ 𝐷2 +𝐷3.
3
µd =
𝑑1 +𝑑2.
2
Mean[µ]
D =
(D1−µD)2+(D2−µD)2+(D3−µD)2.
3
d =
(d1−µd)2+(d2−µd)2.
2

Combining 2 or More populations
e.g. Bearing-Shaft Assembly (Clearance)

26. C1 =D1 −d1 C2 =D1 −d2 C3 =D2 −d1 C4 =D2 −d2 C5 =D3−d1 C6 =D3 −d2
Z1 =
𝑥M−𝞵c
σc
Z2 =
𝑥m−𝞵c
σc
X1 X2 X3 X4 X5 X6
c = D2 + d2
For clearance : Assembly (Satandard variable for clearance )
Clearance C= D-d
Standard deviation for Clearance
XM= Max clearance
Xm= Min clearance

27. Clearance fit between Journal and Bush of bearing: 40H6-e7.
Dimensions of two components are normally distributed
Design tolerance = Natural tolerance.
From stability point of view Max and Min clearance =0.08 and
0.06 mm. Determine the probability of % of rejected assembly
0.008= 3 B
B =0.002667mm
3 J =0.0125
J =0.004167m
Q5
40H6-e7.
40H6= 40.016 & 40 mm
µB= (40.016+40)/2 =40.008
T= (40.016 ̶ 40)/2= 0.008
=µB ± T
= 40.008 ±0.008 mm
40e7 = 39.950 & 39.925
µJ= (39.950+39.925)/2 =39.375
T = (39.950 ̶ 39.925)/2 =0.0125
=µJ ± T
=39.9375 ± 0.0125mm
Population
BUSH 40H6
Population
JOURNAL 40e7
Design tolerance = Natural tolerance.
40
H6 e7
es ei es ei
+16 0 -50 -75

28. Population of Clearance C
Probability of REJECTION
µC =µB − µJ = 0.0705 𝑚𝑚
C= B2 + 𝐽2 =0.004947 mm
Z1 =
X1−𝞵C
σC
Z = 0 to 1.92 A1=0.4726
R=(1-0.9556)*100=4.44%
Z2 =
X2−𝞵C
σC
0.08−𝞵C
σC
0.06−𝞵C
σC
+1.92
−2.12 Z = 0 to 2.12 A2=0.4830
A=0.9556
Max and Min clearance =0.08 and 0.06 mm

29. Transition fit between Recess and the spigot of rigid coupling: 60H6-j5.
Dimensions of two components are normally distributed
Specified (Design) tolerance = natural tolerance.
Determine the probability of Interference fit between 2 parts
60H6 = 60.019 & 60 mm
Specified (Design) tolerance = Natural tolerance.
0.0095 = 3 R
R =0.00317mm
µR=60.0095
60.0095 ±0.0095 mm
Q6
Population
Recess 60 H6
Population
Spigot 60 j5
60j5 = 60.006 & 59.993
0.0065 = 3 s
s =0.00217mm
µS=59.9995
59.9995 ± 0.0065mm

30. Population of Interference: I
Probability of interference fit
µI =µS − µR = −0.01 𝑚𝑚
I =  𝑆2 +  𝑅2 = 0.00384 mm
Z =
𝐼−𝞵I
σ𝐼
Z =
0−𝞵I
σ𝐼
Z = +2.6
Z = 0 to +2.6 A=0.4953
R=(0.5-0.4953)*100=0.47%
No interference point I=0

31. Assembly of 3 components A, B and C .The dimensions of 3
components are normal distributed. Natural tolerance=Design
tolerance for 3 components. Determine the % of assemblies
where interference may occur.
Population A
µA=40.00
Design Tolerance = Natural tolerance.
0.009 = 3 A A =0.03mm
Population B
µA=40.00 0.009 = 3 B B=0.03mm
Population A & B are addition type (X)
µX =µA + µB =µX =100
X = A2 + B2 0.0424 𝑚𝑚

32. Population X & C (Interference): I
µI =µX − µC = −0.09
I = X2 + C2 = 0.052 𝑚𝑚
Z =
𝐼−𝞵I
σ𝐼
Z = 0 to +1.73 A= 0.4582
R=(0.5−0.4582)*100=4.18 %
% of assemblies with rejection
Z =
0−𝞵I
σ𝐼
Z = +1.73
No interference point I=0
µC= 100.09
T=0.09

33. Overall as well as individual dimensions are normally
distributed,. Natural tolerances are equal to design tolerances.
Tolerance of B ?
µX =µA + µB
µA = 10 µX = 40µB = 30
Design Tolerance = Natural tolerance.
0.9=3 X X =0.3
0.6=3 A A =0.2
X = 𝐴2 + 𝐵2
B =0.2236
Design Tolerance =3 B Tolerance of B =0.6708
µB = 30 𝑫𝒊𝒎𝒆𝒏𝒔𝒊𝒐𝒏𝒔 𝒐𝒇 𝑩 𝟑𝟎 ±0.6788

34. Reliability
Reliability, describes the ability of a system / component to function under
stated conditions for a specified period of time.
FoS does not address reliability.
Design based on Reliability and not on FoS
Probabilistic approach to design

35. 1st Population of strength (Syt)
D= S𝑦𝑡2 + Sw2
µSyt
Syt
2nd Population of Stress (Sw)
µSw
µD =µSyt
− µSw
Normal curve for population of MoS
In terms of Z
Z =
m−𝞵D
σD Z =
0−𝞵D
σD
Probability of failure: ___%
If Stress=Strength m=0
Margin of safety
Sw
3rd Population of Margin of safety MoS (m)
Reliability: ____%

36. Strength of material is ND with mean of 230 Mpa and SD 30 Mpa.
Stress induced in components is ND with mean of 150 Mpa and SD 15 Mpa.
Determine reliability in design.
Population of strength
m= S2 + t2
µS = 230 S =30
Population of Stress µt = 150
Population of Margin of safety
µm =µS − µt =80
Reliability MoS
Z =
m−𝞵m
σm
Z = −2.39
Z =
0−𝞵m
σm
A=0.4916
Re=0.4916+0.5=0.9916 99.16%
m=33.54
t =15
Probability of failure: 0.84%Reliability: 99.16 %
S=t
m=0