Experiment 1
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Experiment No. 1(i) DATA ANALYSIS:
No of roller
Diameter ( x i ) mm.
No of roller
Diameter ( x i ) mm.
01 02 03 04 05 06 07 08 09 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43
19.10 19.13 18.91 19.11 19.14 19.10 19.05 19.13 19.10 18.95 19.13 19.05 19.17 19.19 19.17 19.16 19.07 19.05 19.04 19.13 19.03 19.12 19.08 19.08 18.97 19.01 19.10 19.01 19.05 19.03 19.20 18.96 19.06 19.00 19.05 19.09 19.07 19.07 18.94 19.13 19.14 18.99 19.16
44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86
19.05 19.03 19.14 19.17 19.06 19.07 19.08 19.11 19.05 19.04 19.05 19.02 19.11 19.10 19.08 19.09 19.03 19.25 19.10 19.05 19.06 19.03 19.17 19.09 19.11 19.06 19.11 18.96 19.14 18.88 19.19 19.02 19.00 19.00 19.06 19.12 19.04 19.14 19.08 19.03 19.10 18.96 19.08 1
No of roller Diameter ( x i ) mm. 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129
19.00 19.12 19.08 19.02 19.24 19.12 19.02 19.02 19.10 19.12 19.01 19.15 19.10 19.12 19.12 1918 19.09 19.06 19.19 19.20 19.05 19.07 19.16 19.07 19.11 19.10 19.09 19.08 18.97 19.10 18.91 19.09 19.03 19.11 19.06 19.11 19.08 19.05 19.10 19.16 19.12 19.07 19.00
No of roller
Diameter ( x i ) mm.
No of roller
Diameter ( x i ) mm.
130 131 132 133 134 135 136
19.10 19.04 19.01 19.08 19.10 19.00 19.07
137 138 139 140 141 142 143
19.09 19.07 19.18 19.08 19.13 19.11 19.00
CALCULATION: We get, ∑ = 2861.46 ∑
= = 19.0764 N = 150 max = 19.25 Xi min = 18.88 =
∑(
)
. = 0.065725
For interval 01 X1=18.880 and x2= 18.917 z1= (18.880-19.0764)/0.065725= - 2.99 z2 = ((18.917-19.0764)/0.065725)= - 2.43 From table we get, Area under the interval from normal curve, ∆ 1= - 2.99------------0.4986 ∆ 2= - 2.43-------------0.4925 E1=∆ x150=0.0061 x 150 =0.915 For interval 02 X1= 18.917and x2= 18.954 z1= (18.917-19.0764)/0.065725= - 2.43 z2 = ((18.954-19.0764)/0.065725)= - 1.86 From table we get, Area under the interval from normal curve, ∆ 1= - 2.43------------0.4925 ∆ 2= - 1.86-------------0.4686 E2=∆ x150=0.0239 x 150 =3.585 For interval 03 X1= 18.954and x2= 18.991 z1= (18.954and -19.0764)/0.065725= - 1.86 z2 = ((18.991-19.0764)/0.065725)= - 1.23 From table we get, Area under the interval from normal curve, ∆ 1= - 2.43------------0.4686 ∆ 2= - 1. 23-------------0.3907 E3=∆ x150=0.0779 x 150 =11.685
2
No of roller Diameter ( x i ) mm. 144 145 146 147 148 149 150
19.11 19.09 19.03 19.13 19.03 19.05 18.88
For interval 04 X1= 18.991and x2= 19.028 z1= (18.991-19.0764)/0.065725= - 1.23 z2 = ((19.028-19.0764)/0.065725)= - 0.74 From table we get, Area under the interval from normal curve, ∆ 1= - 1. 23-------------0.3907 ∆ 2= - 0.74 -------------0.2704 E4=∆ x150=0.1266 x 150 =18.99 For interval 05 X1= 19.028and x2= 19.065 z1= (19.028-19.0764)/0.065725= - 0.74 z2 = ((19.065-19.0764)/0.065725)= - 0.01 From table we get, Area under the interval from normal curve, ∆ 1= - 0.74 -------------0.2704 ∆ 2= - 0.01-------------0.0040 E5=∆ x150=0.2664 x 150 =39.96 For interval 06 X1= 19.065 and x2= 19.102 z1= ((19.065-19.0764)/0.065725)= - 0.01 z2 = ((19.102-19.0764)/0.065725)= 0.39 From table we get, Area under the interval from normal curve, ∆ 1= - 0.01-------------0.0040 ∆ 2= 0.39 -------------0.1517 E6=∆ x150=0.1477 x 150 =22.155 For interval 07 X1= 19.102and x2= 19.139 z1= ((19.102-19.0764)/0.065725)= 0.39 z2 = ((19.139-19.0764)/0.065725)= 0.95 From table we get, Area under the interval from normal curve, ∆ 1= 0.39 -------------0.1517 ∆ 2= 0.95 -------------0.3289 E7=∆ x150=0.1477 x 150 =26.58 For interval 08 X1= 19.139 and x2= 19.176 z1= ((19.139-19.0764)/0.065725)= 0.95 z2 = ((19.176-19.0764)/0.065725)= 1.52 From table we get, Area under the interval from normal curve, ∆ 1= 0.95 -------------0.3289 ∆ 2= 1.52 -------------0.4357 E9=∆ x150=0.1068 x 150 =16.02 For interval 09 X1= 19.176 and x2= 19.213 z1= ((19.176-19.0764)/0.065725)= 1.52 z2 = ((19.213-19.0764)/0.065725)= 2.08 From table we get, Area under the interval from normal curve, ∆ 1= 1.52 -------------0.4357 3
∆ 2= 2.08 -------------0.4812 E9=∆ 2x150=0.0455 x 150 =6.825 For interval 10 X1= 19.213 and x2= 19.250 z1= ((19.213-19.0764)/0.065725)= 2.08 z2 = ((19.250-19.0764)/0.065725)= 2.64 From table we get, Area under the interval from normal curve, ∆ 1= 2.08 -------------0.4812 ∆ 2= 2.64 -------------0.4959 E10=∆ 2x150=0.0147 x 150 =2.205 Now, − = 3.085 − = −1.585 − = −5.685 − = −0.99 − = −7.96 − = 17.845 − = −1.58 − = −2.02 − = 0.175 − = −0.205 ℵ
= ∑
(
)
=
. .
= 2.9112
Again, No of degree or degree of freedom = No of interval or cell − No of quantities needed from the sample to calculate expected frequency. ⇒ Degree of freedom = 10 − 3 ∴ν = 7 But, α = 0.05 From supplied chart we get for ν = 7 and α = 0.05 , χ 2crit = 14.067 So, observed frequency distribution does fit with normal distribution.
4
Interval
Range
Observed frequency ( oi )
Expected frequency ( ei )
Oi-ei
01 02 03 04 05 06 07 08 09 10
18.880-18.917 18.917-18.954 18.954-18.991 18.991-19.028 19.028-19.065 19.065-19.102 19.102-19.139 19.139-19.176 19.176-19.213 19.213-19.250
4 2 6 18 32 40 25 14 7 2
0.915 3.585 11.685 18.99 39.96 22.155 26.58 16.02 6.825 2.205
3.085 -1.585 -5.685 -0.99 -7.96 17.845 -1.58 -2.02 0.175 -0.205
RESULT AND GRAPH: Histogram of Observed frequency: 45 40
Observed Frequency
35 30 25 20
40 32
15
25 10
18 14
5 4 0
7
6 2
2
Diameter Xi (mm)
5
Histogram of Expected frequency: 45
39.96
Expected Frequency
40 35 30
26.58 22.155
25
18.99
20 15
16.02
11.685
10 5
6.825 0.915
3.585
2.205
0
Diameter Xi (mm)
Comparison of Observed and Expected Frequency: 45 40
Observed Frequency Expected Frequency
35 Frequency
30 25 20 15 10 5 0
Diameter Xi (mm)
Mean = 19.0764 Standard deviation = 0.065725 = 2.9112 χ 2crit = 14.067 and ℵ
Experiment:1(ii) DATA AND CALCULATION:
Vernier constant: Bevel protractor Vernier constant 1° = 60’ 24 vb divisions are meeting with 23 MS divisions = 57.5′ MS divisions ∴ 1 VS division are meeting with ∴ Difference, Vernier constant = 60 − 57.5 = 2.5 2 = 5′ ∠ABC :
Main scale reading = 25° Vernier scale reading = 20’
Main scale reading = 28° Vernier scale reading = 25’
Main scale reading = 80° Vernier scale reading = 5’
Main scale reading = 86° Vernier scale reading = 10’
Main scale reading = 90° Vernier scale reading = 10’ ∴∠EFA = 76o + 9 × 5 ' = 76o 45 '
Main scale reading = 77° Vernier scale reading = 15’
7
RESULT: ∠ABC ∠BCD ∠CDE
∠ DEF ∠EFA ∠FAB
205°15’ 151°35’ 80°25’
8
86°10’ 90°10’ 102°45’
No of roller
Diameter ( x i ) mm.
No of roller
Diameter ( x i ) mm.
01 02 03 04 05 06 07 08 09 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43
19.10 19.13 18.91 19.11 19.14 19.10 19.05 19.13 19.10 18.95 19.13 19.05 19.17 19.19 19.17 19.16 19.07 19.05 19.04 19.13 19.03 19.12 19.08 19.08 18.97 19.01 19.10 19.01 19.05 19.03 19.20 18.96 19.06 19.00 19.05 19.09 19.07 19.07 18.94 19.13 19.14 18.99 19.16
44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77 78 79 80 81 82 83 84 85 86
19.05 19.03 19.14 19.17 19.06 19.07 19.08 19.11 19.05 19.04 19.05 19.02 19.11 19.10 19.08 19.09 19.03 19.25 19.10 19.05 19.06 19.03 19.17 19.09 19.11 19.06 19.11 18.96 19.14 18.88 19.19 19.02 19.00 19.00 19.06 19.12 19.04 19.14 19.08 19.03 19.10 18.96 19.08 1
No of roller Diameter ( x i ) mm. 87 88 89 90 91 92 93 94 95 96 97 98 99 100 101 102 103 104 105 106 107 108 109 110 111 112 113 114 115 116 117 118 119 120 121 122 123 124 125 126 127 128 129
19.00 19.12 19.08 19.02 19.24 19.12 19.02 19.02 19.10 19.12 19.01 19.15 19.10 19.12 19.12 1918 19.09 19.06 19.19 19.20 19.05 19.07 19.16 19.07 19.11 19.10 19.09 19.08 18.97 19.10 18.91 19.09 19.03 19.11 19.06 19.11 19.08 19.05 19.10 19.16 19.12 19.07 19.00
No of roller
Diameter ( x i ) mm.
No of roller
Diameter ( x i ) mm.
130 131 132 133 134 135 136
19.10 19.04 19.01 19.08 19.10 19.00 19.07
137 138 139 140 141 142 143
19.09 19.07 19.18 19.08 19.13 19.11 19.00
CALCULATION: We get, ∑ = 2861.46 ∑
= = 19.0764 N = 150 max = 19.25 Xi min = 18.88 =
∑(
)
. = 0.065725
For interval 01 X1=18.880 and x2= 18.917 z1= (18.880-19.0764)/0.065725= - 2.99 z2 = ((18.917-19.0764)/0.065725)= - 2.43 From table we get, Area under the interval from normal curve, ∆ 1= - 2.99------------0.4986 ∆ 2= - 2.43-------------0.4925 E1=∆ x150=0.0061 x 150 =0.915 For interval 02 X1= 18.917and x2= 18.954 z1= (18.917-19.0764)/0.065725= - 2.43 z2 = ((18.954-19.0764)/0.065725)= - 1.86 From table we get, Area under the interval from normal curve, ∆ 1= - 2.43------------0.4925 ∆ 2= - 1.86-------------0.4686 E2=∆ x150=0.0239 x 150 =3.585 For interval 03 X1= 18.954and x2= 18.991 z1= (18.954and -19.0764)/0.065725= - 1.86 z2 = ((18.991-19.0764)/0.065725)= - 1.23 From table we get, Area under the interval from normal curve, ∆ 1= - 2.43------------0.4686 ∆ 2= - 1. 23-------------0.3907 E3=∆ x150=0.0779 x 150 =11.685
2
No of roller Diameter ( x i ) mm. 144 145 146 147 148 149 150
19.11 19.09 19.03 19.13 19.03 19.05 18.88
For interval 04 X1= 18.991and x2= 19.028 z1= (18.991-19.0764)/0.065725= - 1.23 z2 = ((19.028-19.0764)/0.065725)= - 0.74 From table we get, Area under the interval from normal curve, ∆ 1= - 1. 23-------------0.3907 ∆ 2= - 0.74 -------------0.2704 E4=∆ x150=0.1266 x 150 =18.99 For interval 05 X1= 19.028and x2= 19.065 z1= (19.028-19.0764)/0.065725= - 0.74 z2 = ((19.065-19.0764)/0.065725)= - 0.01 From table we get, Area under the interval from normal curve, ∆ 1= - 0.74 -------------0.2704 ∆ 2= - 0.01-------------0.0040 E5=∆ x150=0.2664 x 150 =39.96 For interval 06 X1= 19.065 and x2= 19.102 z1= ((19.065-19.0764)/0.065725)= - 0.01 z2 = ((19.102-19.0764)/0.065725)= 0.39 From table we get, Area under the interval from normal curve, ∆ 1= - 0.01-------------0.0040 ∆ 2= 0.39 -------------0.1517 E6=∆ x150=0.1477 x 150 =22.155 For interval 07 X1= 19.102and x2= 19.139 z1= ((19.102-19.0764)/0.065725)= 0.39 z2 = ((19.139-19.0764)/0.065725)= 0.95 From table we get, Area under the interval from normal curve, ∆ 1= 0.39 -------------0.1517 ∆ 2= 0.95 -------------0.3289 E7=∆ x150=0.1477 x 150 =26.58 For interval 08 X1= 19.139 and x2= 19.176 z1= ((19.139-19.0764)/0.065725)= 0.95 z2 = ((19.176-19.0764)/0.065725)= 1.52 From table we get, Area under the interval from normal curve, ∆ 1= 0.95 -------------0.3289 ∆ 2= 1.52 -------------0.4357 E9=∆ x150=0.1068 x 150 =16.02 For interval 09 X1= 19.176 and x2= 19.213 z1= ((19.176-19.0764)/0.065725)= 1.52 z2 = ((19.213-19.0764)/0.065725)= 2.08 From table we get, Area under the interval from normal curve, ∆ 1= 1.52 -------------0.4357 3
∆ 2= 2.08 -------------0.4812 E9=∆ 2x150=0.0455 x 150 =6.825 For interval 10 X1= 19.213 and x2= 19.250 z1= ((19.213-19.0764)/0.065725)= 2.08 z2 = ((19.250-19.0764)/0.065725)= 2.64 From table we get, Area under the interval from normal curve, ∆ 1= 2.08 -------------0.4812 ∆ 2= 2.64 -------------0.4959 E10=∆ 2x150=0.0147 x 150 =2.205 Now, − = 3.085 − = −1.585 − = −5.685 − = −0.99 − = −7.96 − = 17.845 − = −1.58 − = −2.02 − = 0.175 − = −0.205 ℵ
= ∑
(
)
=
. .
= 2.9112
Again, No of degree or degree of freedom = No of interval or cell − No of quantities needed from the sample to calculate expected frequency. ⇒ Degree of freedom = 10 − 3 ∴ν = 7 But, α = 0.05 From supplied chart we get for ν = 7 and α = 0.05 , χ 2crit = 14.067 So, observed frequency distribution does fit with normal distribution.
4
Interval
Range
Observed frequency ( oi )
Expected frequency ( ei )
Oi-ei
01 02 03 04 05 06 07 08 09 10
18.880-18.917 18.917-18.954 18.954-18.991 18.991-19.028 19.028-19.065 19.065-19.102 19.102-19.139 19.139-19.176 19.176-19.213 19.213-19.250
4 2 6 18 32 40 25 14 7 2
0.915 3.585 11.685 18.99 39.96 22.155 26.58 16.02 6.825 2.205
3.085 -1.585 -5.685 -0.99 -7.96 17.845 -1.58 -2.02 0.175 -0.205
RESULT AND GRAPH: Histogram of Observed frequency: 45 40
Observed Frequency
35 30 25 20
40 32
15
25 10
18 14
5 4 0
7
6 2
2
Diameter Xi (mm)
5
Histogram of Expected frequency: 45
39.96
Expected Frequency
40 35 30
26.58 22.155
25
18.99
20 15
16.02
11.685
10 5
6.825 0.915
3.585
2.205
0
Diameter Xi (mm)
Comparison of Observed and Expected Frequency: 45 40
Observed Frequency Expected Frequency
35 Frequency
30 25 20 15 10 5 0
Diameter Xi (mm)
Mean = 19.0764 Standard deviation = 0.065725 = 2.9112 χ 2crit = 14.067 and ℵ
Experiment:1(ii) DATA AND CALCULATION:
Vernier constant: Bevel protractor Vernier constant 1° = 60’ 24 vb divisions are meeting with 23 MS divisions = 57.5′ MS divisions ∴ 1 VS division are meeting with ∴ Difference, Vernier constant = 60 − 57.5 = 2.5 2 = 5′ ∠ABC :
Main scale reading = 25° Vernier scale reading = 20’
Main scale reading = 28° Vernier scale reading = 25’
Main scale reading = 80° Vernier scale reading = 5’
Main scale reading = 86° Vernier scale reading = 10’
Main scale reading = 90° Vernier scale reading = 10’ ∴∠EFA = 76o + 9 × 5 ' = 76o 45 '
Main scale reading = 77° Vernier scale reading = 15’
7
RESULT: ∠ABC ∠BCD ∠CDE
∠ DEF ∠EFA ∠FAB
205°15’ 151°35’ 80°25’
8
86°10’ 90°10’ 102°45’
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