
In an experiment to measure focal length of a concave mirror, the values of focal length is successive observations are $17.3\;{\text{cm}}$,$17.8\;{\text{cm}}$, $18.3\;{\text{cm}}$, $18.2\;{\text{cm}}$, $17.9\;{\text{cm}}$ and $18.0\;{\text{cm}}$. Calculate mean absolute error and percentage error. Express the result in a proper way.
Answer
226.8k+ views
Hint: Firstly, calculate the mean of the observed focal lengths of the concave mirror. Secondly, find the percentage error in each observation by subtracting it from mean value obtained. Finally, calculate absolute error and percentage error of all the mean values.
Complete step by solution:
We know that the mean of a number terms is given by the formula written below.
\[m = \dfrac{{{\text{sum}}\;{\text{of}}\;{\text{terms}}}}{{{\text{number}}\;{\text{of}}\;{\text{terms}}}}\]
Now, calculate the mean of all the observations of focal length,
\[{f_{{\text{mean}}}} = \dfrac{{17.3 + 17.8 + 18.3 + 18.2 + 17.9 + 18.0}}{6}\]
Simplifying above expression,
\[{f_{{\text{mean}}}} = 17.9\;{\text{cm}}\]
Now, we know that the absolute errors, $\Delta x$ can be obtained from by subtracting mean value obtained from each of the observations, i.e. $\Delta x = {x_{{\text{actual}}}} - {x_{{\text{mean}}}}$.
Let the errors obtained in each observation are ${e _1},\;{e_2},\;{e_3},\;...{e_n}$, then use the formula $\Delta x = {x_{{\text{actual}}}} - {x_{{\text{mean}}}}$ to find the error.
\[{e_1} = 17.3 - 17.9\]
\[ = - 0.6\]
\[{e_2} = 17.8 - 17.9\]
$ = - 0.1$
\[{e_3} = 18.3 - 17.9\]
$ = 0.4$
\[{e_4} = 18.2 - 17.9\]
$ = 0.3$
${e_5} = 17.9 - 17.9$
$ = 0.0$
\[{e_6} = 18 - 17.9\]
\[ = 0.1\]
We know that the formula for finding error is given by,
\[\Delta {f_{{\text{mean}}}} = \dfrac{{\left| {{e_1}} \right| + \left| {{e_2}} \right| + \left| {{e_3}} \right| + ... + \left| {{e_n}} \right|}}{{{\text{number}}\;{\text{of}}\;{\text{terms}}}}\]
Now, substitute the errors values obtained in the above formula.
\[\Delta {f_{{\text{mean}}}} = \dfrac{{\left| { - 0.6} \right| + \left| { - 0.1} \right| + \left| { - 0.4} \right| + \left| {0.3} \right| + \left| 0 \right| + \left| {0.1} \right|}}{6}\]
\[ = 0.25\]
Since the absolute error is obtained, we just need to find the percentage error.
As the formula for percentage error is given by,
${\text{percentage}}\;{\text{error}} = \dfrac{{{\text{error}}}}{{{\text{mean}}\;{\text{value}}}} \times 100$
Now, substitute the values obtained into above formula,
${\text{percentage}}\;{\text{error}} = \dfrac{{0.25}}{{17.9}} \times 100$
$ = 1.39\% $
Therefore, the absolute error is $0.25$ and the error percentage is $1.39\;\% $.
Note:In these types of questions, find the mean value of all the given terms. Once the mean value is obtained, you can subtract it from each observation to find the absolute value.
The absolute value can be obtained by subtracting the actual value, ${x_a}$ , from measured value ${x_0}$, i.e., $\Delta x = {x_0} - {x_a}$. You can directly add the modulus into the formula of absolute value to get the result.
Complete step by solution:
We know that the mean of a number terms is given by the formula written below.
\[m = \dfrac{{{\text{sum}}\;{\text{of}}\;{\text{terms}}}}{{{\text{number}}\;{\text{of}}\;{\text{terms}}}}\]
Now, calculate the mean of all the observations of focal length,
\[{f_{{\text{mean}}}} = \dfrac{{17.3 + 17.8 + 18.3 + 18.2 + 17.9 + 18.0}}{6}\]
Simplifying above expression,
\[{f_{{\text{mean}}}} = 17.9\;{\text{cm}}\]
Now, we know that the absolute errors, $\Delta x$ can be obtained from by subtracting mean value obtained from each of the observations, i.e. $\Delta x = {x_{{\text{actual}}}} - {x_{{\text{mean}}}}$.
Let the errors obtained in each observation are ${e _1},\;{e_2},\;{e_3},\;...{e_n}$, then use the formula $\Delta x = {x_{{\text{actual}}}} - {x_{{\text{mean}}}}$ to find the error.
\[{e_1} = 17.3 - 17.9\]
\[ = - 0.6\]
\[{e_2} = 17.8 - 17.9\]
$ = - 0.1$
\[{e_3} = 18.3 - 17.9\]
$ = 0.4$
\[{e_4} = 18.2 - 17.9\]
$ = 0.3$
${e_5} = 17.9 - 17.9$
$ = 0.0$
\[{e_6} = 18 - 17.9\]
\[ = 0.1\]
We know that the formula for finding error is given by,
\[\Delta {f_{{\text{mean}}}} = \dfrac{{\left| {{e_1}} \right| + \left| {{e_2}} \right| + \left| {{e_3}} \right| + ... + \left| {{e_n}} \right|}}{{{\text{number}}\;{\text{of}}\;{\text{terms}}}}\]
Now, substitute the errors values obtained in the above formula.
\[\Delta {f_{{\text{mean}}}} = \dfrac{{\left| { - 0.6} \right| + \left| { - 0.1} \right| + \left| { - 0.4} \right| + \left| {0.3} \right| + \left| 0 \right| + \left| {0.1} \right|}}{6}\]
\[ = 0.25\]
Since the absolute error is obtained, we just need to find the percentage error.
As the formula for percentage error is given by,
${\text{percentage}}\;{\text{error}} = \dfrac{{{\text{error}}}}{{{\text{mean}}\;{\text{value}}}} \times 100$
Now, substitute the values obtained into above formula,
${\text{percentage}}\;{\text{error}} = \dfrac{{0.25}}{{17.9}} \times 100$
$ = 1.39\% $
Therefore, the absolute error is $0.25$ and the error percentage is $1.39\;\% $.
Note:In these types of questions, find the mean value of all the given terms. Once the mean value is obtained, you can subtract it from each observation to find the absolute value.
The absolute value can be obtained by subtracting the actual value, ${x_a}$ , from measured value ${x_0}$, i.e., $\Delta x = {x_0} - {x_a}$. You can directly add the modulus into the formula of absolute value to get the result.
Recently Updated Pages
The relation between efficiency eta of a heat engine class 11 physics JEE_Main

A block of mass m is connected to another block of class 11 physics JEE_Main

A body of mass 2Kg is driven by an engine delivering class 11 physics JEE_Main

One end of string of length l is connected to a particle class 11 physics JEE_Main

An ideal liquid is oscillating in a V tube as shown class 11 physics JEE_Main

Two persons of masses 55kg and 65kg respectively are class 11 physics JEE_Main

Trending doubts
JEE Main 2026: City Intimation Slip and Exam Dates Released, Application Form Closed, Syllabus & Eligibility

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

How to Convert a Galvanometer into an Ammeter or Voltmeter

Ideal and Non-Ideal Solutions Explained for Class 12 Chemistry

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Laws of Motion Class 11 Physics Chapter 4 CBSE Notes - 2025-26

JEE Advanced 2026 - Exam Date (Released), Syllabus, Registration, Eligibility, Preparation, and More

Mechanical Properties of Fluids Class 11 Physics Chapter 9 CBSE Notes - 2025-26

Thermodynamics Class 11 Physics Chapter 11 CBSE Notes - 2025-26

Units And Measurements Class 11 Physics Chapter 1 CBSE Notes - 2025-26

