In the previous section, we saw the different types of errors that can occur while taking measurements. In this section, we will see how the 'quantity of errors' can be calculated. In other words, we will see the method to calculate 'how much error occurred', when the experiment was conducted
1. Suppose that, the length of a cylinder is being measured
• Let n trials be done
♦ In the first trial, length obtained is a1 cm
♦ In the second trial, length obtained is a2 cm
♦ In the third trial, length obtained is a3 cm
♦ . . . .
♦ . . . .
♦ In the nth trial, length obtained is an cm
2. Each of the above measurements may be different from one another
• We cannot say which among the n values, is the true length
3. In such a situation, the ‘mean of the n values’ will be very close to the true value
• But how can we prove that 'mean is indeed very close to the true value'?
4. For proving it, let us recall the 'basics about mean (average)' that we learned in our previous classes:
• Consider a group of seven students. It is decided to make uniforms for them
• Their heights are: 142 cm, 141 cm, 138 cm , 145 cm, 139 cm, 143 cm and 144 cm
• How much clothes should be bought?
Solution:
(i) Sum of the heights works out to 992 cm
(ii) If we divide this sum by 7, we will get the average height
• So average height = 992⁄7 = 142 cm
(iii) Assume that all 7 students have a height of 142 m
• Let 2 m of cloth be required for a student of 142 cm height
(iv) Then for the 7 students, (2 × 7) = 14 m of cloth will be required
• This is an easy method for calculating the total quantity of cloth, especially if the number of students in the group or class is large
(v) But doubts arise:
• The student of 138 cm height do not need 2 m of cloth
♦ Some cloth will go as waste
♦ Same is the case with students of heights less than 142 cm
• Similarly, the student of 145 cm will need more than 2 m
♦ His uniform will be incomplete
♦ Same is the case with students of heights more than 142 cm
(vi) But in reality, such a wastage or insufficiency do not occur. The reason can be explained using the schematic diagram in fig.2.25 below:
• In fig.a, all students stand in ascending order of their heights
• The fourth student have the average height of 142 cm. He is shown in yellow color in fig.b
• The average height is marked by a cyan line in fig.c
• The white rectangles represents the wastage caused by the first 3 students
♦ They are below the cyan line
• The magenta rectangles represent the insufficiency suffered by the last 3 students
♦ They are above the cyan line
• If the mean is calculated accurately, the wastage will be very nearly equal to the insufficiency
• That is., the total heights of the white rectangles will be equal to the total heights of the magenta rectangles
• In short, the negatives will be cancelled by the positives. This is the advantage of using the mean value
5. When we calculate the mean of the measured values, a similar cancellation occurs. Let us see how:
(i) We have n number of readings
(ii) Half of those readings will be greater than the true value
♦ The errors in this case will be all positive
♦ Because, Error = Measured value - True value
♦ The errors of such over estimated readings are represented by the magenta rectangles in fig.2.25 above
(iii) The other half will be lesser than the true value
♦ The errors in this case will be all negative
♦ Because, Error = Measured value - True value
♦ The errors of such under estimated readings are represented by the white rectangles in fig.2.25 above
(iv) How can we be sure that exactly half readings are over estimates and the other half are under estimates?
• The answer is that, it is an assumption
• It is a reasonable assumption. Because, 'likely hood of over estimating' is same as the 'likely hood of under estimating'
• Nobody would make all readings greater than the true value
• Also, nobody would make all readings lesser than the true value
• Every body wants the readings to be as accurate as possible
• The 'possibility for making positive errors' is same as the 'possibility for making negative errors'
6. So we can confirm that, the mean value is very close to the true value
• We know the formula for calculating the mean value: $\mathbf\small{a_{mean}=\frac{a_1+a_2+a_3+\;.\;.\;.\;+\;a_n}{n}}$
• In short form, this formula is written as: $\mathbf\small{a_{mean}=\frac{\sum\limits_{i=1}^{i=n}{a_i} }{n}}$
7. So we have obtained the mean value. The mean value will be very close to the true value
• So we can now find 'how much error was made in each reading'
• Let 𝚫a1 be the error made in the first reading
• We have: Error = Measured value - True value
♦ So 𝚫a1 = a1 - amean
♦ 𝚫a2 = a2 - amean
♦ 𝚫a3 = a3 - amean
♦ . . . .
♦ . . . .
♦ 𝚫an = an - amean.
8. The above 𝚫a values will be +ve in some cases and -ve in the rest of the cases
• For our present discussion, we do not need the sign
• We need only 'how much error' occurred in each reading
• So we take absolute values:
♦ |𝚫a1| = |(a1 - amean)|
♦ |𝚫a2| = |(a2 - amean)|
♦ |𝚫a3| = |(a3 - amean)|
♦ . . . .
♦ . . . .
♦ |𝚫an| = |(an - amean)|
9. The mean of the above absolute error values will give us the final absolute error or the mean absolute error
• It is denoted as 𝚫amean
• This 𝚫amean can be calculated using the same method: $\mathbf\small{\Delta a_{mean}=\frac{|\Delta a_1|+|\Delta a_2|+|\Delta a_3|+\;.\;.\;.\;+\;|\Delta a_n|}{n}}$
• In short form, this formula is written as: $\mathbf\small{\Delta a_{mean}=\frac{\sum\limits_{i=1}^{i=n}{|\Delta a_i|} }{n}}$
10. So now we have two quantities: amean and 𝚫amean.
■ We must clearly understand the difference between the two
• amean is the value which is very close to the true value
• 𝚫amean is an indication of 'how much error' occurred when the experiment was conducted
11. So the true value will be within the range: $\mathbf\small{a_{\rm mean}\pm \Delta a_{\rm mean}}$
• That is., (amean - 𝚫amean) ≤ true value ≤ (amean + 𝚫amean)
• Relative error is a ratio. It can be obtained using the relation:
Relative error = $\mathbf\small{\frac{\Delta a_{mean}}{a_{mean}}}$
2. This ratio helps us to compare two quantities:
(i) The error (ii) amean (which is close to the true value)
3. If the error is very small (when compared to the true value), we will get a small ratio
• In that case, the error can be considered as negligible
4. If the error is large (when compared to the true value), we will get a large ratio
• In that case, the error cannot be considered as negligible
An example is given below. It is written in 3 steps:
(i) Consider the experiment to find the distance between earth and a star
(ii) If the error made is a few kilometers, it is negligible. We will get a small relative error
(iii) If the error made is thousands of kilometers, it is not negligible. We will get a large relative error
5. Once we calculate the relative error, we can easily convert it into percentage format
• All we need to do is: multiply by 100
• The result thus obtained is called the percentage error. It is denoted as $\mathbf\small{\delta a}$
• So we get: $\mathbf\small{\delta a=\left(\frac{\Delta a_{mean}}{a_{mean}}\right)\times 100 \text{%}}$
Solved example 2.19
We measure the period of oscillation of a simple pendulum. In successive measurements, the readings turn out to be 2.63 s, 2.56 s, 2.42 s, 2.71 s and 2.80 s. Calculate the absolute errors, relative error or percentage error
Solution:
The calculations are done in the table below:
1. We get: amean = 2.624 s
• This is written in column 3
• This result is got by addition of various numbers
• So the result must not have more decimal places than the least in the given data
• Thus, the result cannot have more than 2 decimal places
• Rounding off, we get: amean = 2.62 s
(After addition, we divide the sum by (n = 5). But n is a factor. It has infinite number of significant numbers)
2. The absolute errors are calculated in the fourth column
• We get: Mean absolute error 𝚫amean = 0.107 s
• This is also obtained by addition. Addition of numbers having two decimal places
• So rounding off, we get: 𝚫amean = 0.11 s
3. Now we can write three statements about that pendulum:
(i) The true value of the time period of that pendulum is very close to 2.62 s
(ii) The mean absolute error is 0.11 s
(iii) There is a probability that, the actual time period lies within the range:
(2.62 - 0.11) to (2.62 + 0.11)
• That is., 2.51 ≤ actual time period ≤ 2.73 s
4. Calculation of relative error
• We have: Relative error = $\mathbf\small{\frac{\Delta a_{mean}}{a_{mean}}}$
• Substituting the values, we get:
Relative error = $\mathbf\small{\frac{0.11}{2.62}=0.04}$
• Percentage error = Relative error ×100 = 0.04 × 100 = 4%
Solved example 2.20
Diameter of a wire as measured by a screw gauge was found to be 2.620, 2.625, 2.630, 2.628 and 2.626 cm. Calculate
(i) Mean value of diameter
(ii) absolute error in each measurement
(iii) Mean absolute error
(iv) relative error
(v) Percentage error
(vi) Express the result in terms of percentage error
Solution:
The calculations are done in the table below:
1. We get: amean = 2.626 cm
• This is written in column 3
• This is the answer for part (i)
2. The absolute error of each measurement is written in column 4
• These are the answers for part (ii)
3. The mean absolute error is written in column 5
• This is the answer for part (iii)
4. Calculation of relative error
• We have: Relative error = $\mathbf\small{\frac{\Delta a_{mean}}{a_{mean}}}$
• Substituting the values, we get:
Relative error = $\mathbf\small{\frac{0.003}{2.626}=0.001}$
• This is the answer for part (iv)
5. Percentage error = Relative error ×100 = 0.001 × 100 = 0.1%
• This is the answer for part (v)
6. Diameter of the wire can be written as:
d = 2.626 cm ± 0.1%
• This is the answer for part (vi)
1. Suppose that, the length of a cylinder is being measured
• Let n trials be done
♦ In the first trial, length obtained is a1 cm
♦ In the second trial, length obtained is a2 cm
♦ In the third trial, length obtained is a3 cm
♦ . . . .
♦ . . . .
♦ In the nth trial, length obtained is an cm
2. Each of the above measurements may be different from one another
• We cannot say which among the n values, is the true length
3. In such a situation, the ‘mean of the n values’ will be very close to the true value
• But how can we prove that 'mean is indeed very close to the true value'?
4. For proving it, let us recall the 'basics about mean (average)' that we learned in our previous classes:
• Consider a group of seven students. It is decided to make uniforms for them
• Their heights are: 142 cm, 141 cm, 138 cm , 145 cm, 139 cm, 143 cm and 144 cm
• How much clothes should be bought?
Solution:
(i) Sum of the heights works out to 992 cm
(ii) If we divide this sum by 7, we will get the average height
• So average height = 992⁄7 = 142 cm
(iii) Assume that all 7 students have a height of 142 m
• Let 2 m of cloth be required for a student of 142 cm height
(iv) Then for the 7 students, (2 × 7) = 14 m of cloth will be required
• This is an easy method for calculating the total quantity of cloth, especially if the number of students in the group or class is large
(v) But doubts arise:
• The student of 138 cm height do not need 2 m of cloth
♦ Some cloth will go as waste
♦ Same is the case with students of heights less than 142 cm
• Similarly, the student of 145 cm will need more than 2 m
♦ His uniform will be incomplete
♦ Same is the case with students of heights more than 142 cm
(vi) But in reality, such a wastage or insufficiency do not occur. The reason can be explained using the schematic diagram in fig.2.25 below:
 |
| Fig.2.25 |
• The fourth student have the average height of 142 cm. He is shown in yellow color in fig.b
• The average height is marked by a cyan line in fig.c
• The white rectangles represents the wastage caused by the first 3 students
♦ They are below the cyan line
• The magenta rectangles represent the insufficiency suffered by the last 3 students
♦ They are above the cyan line
• If the mean is calculated accurately, the wastage will be very nearly equal to the insufficiency
• That is., the total heights of the white rectangles will be equal to the total heights of the magenta rectangles
• In short, the negatives will be cancelled by the positives. This is the advantage of using the mean value
5. When we calculate the mean of the measured values, a similar cancellation occurs. Let us see how:
(i) We have n number of readings
(ii) Half of those readings will be greater than the true value
♦ The errors in this case will be all positive
♦ Because, Error = Measured value - True value
♦ The errors of such over estimated readings are represented by the magenta rectangles in fig.2.25 above
(iii) The other half will be lesser than the true value
♦ The errors in this case will be all negative
♦ Because, Error = Measured value - True value
♦ The errors of such under estimated readings are represented by the white rectangles in fig.2.25 above
(iv) How can we be sure that exactly half readings are over estimates and the other half are under estimates?
• The answer is that, it is an assumption
• It is a reasonable assumption. Because, 'likely hood of over estimating' is same as the 'likely hood of under estimating'
• Nobody would make all readings greater than the true value
• Also, nobody would make all readings lesser than the true value
• Every body wants the readings to be as accurate as possible
• The 'possibility for making positive errors' is same as the 'possibility for making negative errors'
6. So we can confirm that, the mean value is very close to the true value
• We know the formula for calculating the mean value: $\mathbf\small{a_{mean}=\frac{a_1+a_2+a_3+\;.\;.\;.\;+\;a_n}{n}}$
• In short form, this formula is written as: $\mathbf\small{a_{mean}=\frac{\sum\limits_{i=1}^{i=n}{a_i} }{n}}$
7. So we have obtained the mean value. The mean value will be very close to the true value
• So we can now find 'how much error was made in each reading'
• Let 𝚫a1 be the error made in the first reading
• We have: Error = Measured value - True value
♦ So 𝚫a1 = a1 - amean
♦ 𝚫a2 = a2 - amean
♦ 𝚫a3 = a3 - amean
♦ . . . .
♦ . . . .
♦ 𝚫an = an - amean.
8. The above 𝚫a values will be +ve in some cases and -ve in the rest of the cases
• For our present discussion, we do not need the sign
• We need only 'how much error' occurred in each reading
• So we take absolute values:
♦ |𝚫a1| = |(a1 - amean)|
♦ |𝚫a2| = |(a2 - amean)|
♦ |𝚫a3| = |(a3 - amean)|
♦ . . . .
♦ . . . .
♦ |𝚫an| = |(an - amean)|
9. The mean of the above absolute error values will give us the final absolute error or the mean absolute error
• It is denoted as 𝚫amean
• This 𝚫amean can be calculated using the same method: $\mathbf\small{\Delta a_{mean}=\frac{|\Delta a_1|+|\Delta a_2|+|\Delta a_3|+\;.\;.\;.\;+\;|\Delta a_n|}{n}}$
• In short form, this formula is written as: $\mathbf\small{\Delta a_{mean}=\frac{\sum\limits_{i=1}^{i=n}{|\Delta a_i|} }{n}}$
10. So now we have two quantities: amean and 𝚫amean.
■ We must clearly understand the difference between the two
• amean is the value which is very close to the true value
• 𝚫amean is an indication of 'how much error' occurred when the experiment was conducted
11. So the true value will be within the range: $\mathbf\small{a_{\rm mean}\pm \Delta a_{\rm mean}}$
• That is., (amean - 𝚫amean) ≤ true value ≤ (amean + 𝚫amean)
Relative error or Percentage error
1. Instead of mean absolute error, we often use the relative error or percentage error• Relative error is a ratio. It can be obtained using the relation:
Relative error = $\mathbf\small{\frac{\Delta a_{mean}}{a_{mean}}}$
2. This ratio helps us to compare two quantities:
(i) The error (ii) amean (which is close to the true value)
3. If the error is very small (when compared to the true value), we will get a small ratio
• In that case, the error can be considered as negligible
4. If the error is large (when compared to the true value), we will get a large ratio
• In that case, the error cannot be considered as negligible
An example is given below. It is written in 3 steps:
(i) Consider the experiment to find the distance between earth and a star
(ii) If the error made is a few kilometers, it is negligible. We will get a small relative error
(iii) If the error made is thousands of kilometers, it is not negligible. We will get a large relative error
5. Once we calculate the relative error, we can easily convert it into percentage format
• All we need to do is: multiply by 100
• The result thus obtained is called the percentage error. It is denoted as $\mathbf\small{\delta a}$
• So we get: $\mathbf\small{\delta a=\left(\frac{\Delta a_{mean}}{a_{mean}}\right)\times 100 \text{%}}$
Now we will see two solved examples
Solved example 2.19
We measure the period of oscillation of a simple pendulum. In successive measurements, the readings turn out to be 2.63 s, 2.56 s, 2.42 s, 2.71 s and 2.80 s. Calculate the absolute errors, relative error or percentage error
Solution:
The calculations are done in the table below:
1. We get: amean = 2.624 s
• This is written in column 3
• This result is got by addition of various numbers
• So the result must not have more decimal places than the least in the given data
• Thus, the result cannot have more than 2 decimal places
• Rounding off, we get: amean = 2.62 s
(After addition, we divide the sum by (n = 5). But n is a factor. It has infinite number of significant numbers)
2. The absolute errors are calculated in the fourth column
• We get: Mean absolute error 𝚫amean = 0.107 s
• This is also obtained by addition. Addition of numbers having two decimal places
• So rounding off, we get: 𝚫amean = 0.11 s
3. Now we can write three statements about that pendulum:
(i) The true value of the time period of that pendulum is very close to 2.62 s
(ii) The mean absolute error is 0.11 s
(iii) There is a probability that, the actual time period lies within the range:
(2.62 - 0.11) to (2.62 + 0.11)
• That is., 2.51 ≤ actual time period ≤ 2.73 s
4. Calculation of relative error
• We have: Relative error = $\mathbf\small{\frac{\Delta a_{mean}}{a_{mean}}}$
• Substituting the values, we get:
Relative error = $\mathbf\small{\frac{0.11}{2.62}=0.04}$
• Percentage error = Relative error ×100 = 0.04 × 100 = 4%
Solved example 2.20
Diameter of a wire as measured by a screw gauge was found to be 2.620, 2.625, 2.630, 2.628 and 2.626 cm. Calculate
(i) Mean value of diameter
(ii) absolute error in each measurement
(iii) Mean absolute error
(iv) relative error
(v) Percentage error
(vi) Express the result in terms of percentage error
Solution:
The calculations are done in the table below:
1. We get: amean = 2.626 cm
• This is written in column 3
• This is the answer for part (i)
2. The absolute error of each measurement is written in column 4
• These are the answers for part (ii)
3. The mean absolute error is written in column 5
• This is the answer for part (iii)
4. Calculation of relative error
• We have: Relative error = $\mathbf\small{\frac{\Delta a_{mean}}{a_{mean}}}$
• Substituting the values, we get:
Relative error = $\mathbf\small{\frac{0.003}{2.626}=0.001}$
• This is the answer for part (iv)
5. Percentage error = Relative error ×100 = 0.001 × 100 = 0.1%
• This is the answer for part (v)
6. Diameter of the wire can be written as:
d = 2.626 cm ± 0.1%
• This is the answer for part (vi)
In the next section, we will see combination of errors
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