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A substance was known to contain 49.06 -0.02% of a constituent X. The result obtained by two analysis using tge same substance and the same general technique were Analysis 1 - 49.01, 49.21, 49.08 Analysis 2 - 49.40, 49.44, 49.42 Calculate the relative mean eroor and relative mean deviation in the two sets of results and comment on the accuracy and precision of them?
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A substance was known to contain 49.06 -0.02% of a constituent X. The ...
**Relative Mean Error:**
The relative mean error (RME) is a measure of the accuracy of a set of results. It indicates how close the average value of the results is to the true value. The RME is calculated using the formula:
RME = (Σ|X_i - X_true| / N) / X_true * 100
where Σ|X_i - X_true| is the sum of the absolute differences between each individual result (X_i) and the true value (X_true), N is the number of results, and X_true is the true value.
**Analysis 1:**
The average of the results from Analysis 1 is (49.01 + 49.21 + 49.08) / 3 = 49.10.
To calculate the RME, we need to determine the true value of the constituent X. The substance is known to contain 49.06 - 0.02% of X. Therefore, the true value can be calculated as:
X_true = 49.06 - (0.02/100) * 49.06 = 49.05
Now we can calculate the RME using the formula:
RME = (|49.01 - 49.05| + |49.21 - 49.05| + |49.08 - 49.05|) / 3 / 49.05 * 100 = 0.195%
**Analysis 2:**
The average of the results from Analysis 2 is (49.40 + 49.44 + 49.42) / 3 = 49.42.
Using the same true value of X as calculated before (49.05), we can calculate the RME:
RME = (|49.40 - 49.05| + |49.44 - 49.05| + |49.42 - 49.05|) / 3 / 49.05 * 100 = 0.711%
**Relative Mean Deviation:**
The relative mean deviation (RMD) is a measure of the precision of a set of results. It indicates how close the individual results are to the average value. The RMD is calculated using the formula:
RMD = (Σ|X_i - X_mean| / N) / X_mean * 100
where Σ|X_i - X_mean| is the sum of the absolute differences between each individual result (X_i) and the mean value (X_mean), N is the number of results, and X_mean is the mean value.
**Analysis 1:**
Using the average of the results from Analysis 1 (49.10), we can calculate the RMD:
RMD = (|49.01 - 49.10| + |49.21 - 49.10| + |49.08 - 49.10|) / 3 / 49.10 * 100 = 0.059%
**Analysis 2:**
Using the average of the results from Analysis 2 (49.42), we can calculate the RMD:
RMD = (|49.40 - 49.42| + |49.44 - 49.42| + |49.42 - 49.42|) / 3 / 49.42 * 100 = 0.016%
The relative mean error (RME) is a measure of the accuracy of a set of results. It indicates how close the average value of the results is to the true value. The RME is calculated using the formula:
RME = (Σ|X_i - X_true| / N) / X_true * 100
where Σ|X_i - X_true| is the sum of the absolute differences between each individual result (X_i) and the true value (X_true), N is the number of results, and X_true is the true value.
**Analysis 1:**
The average of the results from Analysis 1 is (49.01 + 49.21 + 49.08) / 3 = 49.10.
To calculate the RME, we need to determine the true value of the constituent X. The substance is known to contain 49.06 - 0.02% of X. Therefore, the true value can be calculated as:
X_true = 49.06 - (0.02/100) * 49.06 = 49.05
Now we can calculate the RME using the formula:
RME = (|49.01 - 49.05| + |49.21 - 49.05| + |49.08 - 49.05|) / 3 / 49.05 * 100 = 0.195%
**Analysis 2:**
The average of the results from Analysis 2 is (49.40 + 49.44 + 49.42) / 3 = 49.42.
Using the same true value of X as calculated before (49.05), we can calculate the RME:
RME = (|49.40 - 49.05| + |49.44 - 49.05| + |49.42 - 49.05|) / 3 / 49.05 * 100 = 0.711%
**Relative Mean Deviation:**
The relative mean deviation (RMD) is a measure of the precision of a set of results. It indicates how close the individual results are to the average value. The RMD is calculated using the formula:
RMD = (Σ|X_i - X_mean| / N) / X_mean * 100
where Σ|X_i - X_mean| is the sum of the absolute differences between each individual result (X_i) and the mean value (X_mean), N is the number of results, and X_mean is the mean value.
**Analysis 1:**
Using the average of the results from Analysis 1 (49.10), we can calculate the RMD:
RMD = (|49.01 - 49.10| + |49.21 - 49.10| + |49.08 - 49.10|) / 3 / 49.10 * 100 = 0.059%
**Analysis 2:**
Using the average of the results from Analysis 2 (49.42), we can calculate the RMD:
RMD = (|49.40 - 49.42| + |49.44 - 49.42| + |49.42 - 49.42|) / 3 / 49.42 * 100 = 0.016%
Community Answer
A substance was known to contain 49.06 -0.02% of a constituent X. The ...
Observers1)49.01;49.21; 49.08.Mean =49.10%.
Relative mean error = (49.10- 49.06) / 49.06= 0.08% .
Relative mean deviation = 1/3 (0.09+0.11 +0.02)×100/49.10= 0.15%.
Observer2) 49.40;49.44 ;49.42. Mean =49.42%.
Relative mean error = (49.42 - 49.06)/49.06=0.73%.
Relative mean deviation = 1/3(0.02+0.02+0.00)×100/49.42=0.03%.
The analysis of observer 1 were therefore accurate and precise; those of observer 2 were unusually precise, but less accurate than those of observer 1. Some small source of constant error appears to be present in the results of 2.
Relative mean error = (49.10- 49.06) / 49.06= 0.08% .
Relative mean deviation = 1/3 (0.09+0.11 +0.02)×100/49.10= 0.15%.
Observer2) 49.40;49.44 ;49.42. Mean =49.42%.
Relative mean error = (49.42 - 49.06)/49.06=0.73%.
Relative mean deviation = 1/3(0.02+0.02+0.00)×100/49.42=0.03%.
The analysis of observer 1 were therefore accurate and precise; those of observer 2 were unusually precise, but less accurate than those of observer 1. Some small source of constant error appears to be present in the results of 2.
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A substance was known to contain 49.06 -0.02% of a constituent X. The result obtained by two analysis using tge same substance and the same general technique were Analysis 1 - 49.01, 49.21, 49.08 Analysis 2 - 49.40, 49.44, 49.42 Calculate the relative mean eroor and relative mean deviation in the two sets of results and comment on the accuracy and precision of them?
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A substance was known to contain 49.06 -0.02% of a constituent X. The result obtained by two analysis using tge same substance and the same general technique were Analysis 1 - 49.01, 49.21, 49.08 Analysis 2 - 49.40, 49.44, 49.42 Calculate the relative mean eroor and relative mean deviation in the two sets of results and comment on the accuracy and precision of them? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about A substance was known to contain 49.06 -0.02% of a constituent X. The result obtained by two analysis using tge same substance and the same general technique were Analysis 1 - 49.01, 49.21, 49.08 Analysis 2 - 49.40, 49.44, 49.42 Calculate the relative mean eroor and relative mean deviation in the two sets of results and comment on the accuracy and precision of them? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A substance was known to contain 49.06 -0.02% of a constituent X. The result obtained by two analysis using tge same substance and the same general technique were Analysis 1 - 49.01, 49.21, 49.08 Analysis 2 - 49.40, 49.44, 49.42 Calculate the relative mean eroor and relative mean deviation in the two sets of results and comment on the accuracy and precision of them?.
A substance was known to contain 49.06 -0.02% of a constituent X. The result obtained by two analysis using tge same substance and the same general technique were Analysis 1 - 49.01, 49.21, 49.08 Analysis 2 - 49.40, 49.44, 49.42 Calculate the relative mean eroor and relative mean deviation in the two sets of results and comment on the accuracy and precision of them? for Chemistry 2024 is part of Chemistry preparation. The Question and answers have been prepared according to the Chemistry exam syllabus. Information about A substance was known to contain 49.06 -0.02% of a constituent X. The result obtained by two analysis using tge same substance and the same general technique were Analysis 1 - 49.01, 49.21, 49.08 Analysis 2 - 49.40, 49.44, 49.42 Calculate the relative mean eroor and relative mean deviation in the two sets of results and comment on the accuracy and precision of them? covers all topics & solutions for Chemistry 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A substance was known to contain 49.06 -0.02% of a constituent X. The result obtained by two analysis using tge same substance and the same general technique were Analysis 1 - 49.01, 49.21, 49.08 Analysis 2 - 49.40, 49.44, 49.42 Calculate the relative mean eroor and relative mean deviation in the two sets of results and comment on the accuracy and precision of them?.
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A substance was known to contain 49.06 -0.02% of a constituent X. The result obtained by two analysis using tge same substance and the same general technique were Analysis 1 - 49.01, 49.21, 49.08 Analysis 2 - 49.40, 49.44, 49.42 Calculate the relative mean eroor and relative mean deviation in the two sets of results and comment on the accuracy and precision of them?, a detailed solution for A substance was known to contain 49.06 -0.02% of a constituent X. The result obtained by two analysis using tge same substance and the same general technique were Analysis 1 - 49.01, 49.21, 49.08 Analysis 2 - 49.40, 49.44, 49.42 Calculate the relative mean eroor and relative mean deviation in the two sets of results and comment on the accuracy and precision of them? has been provided alongside types of A substance was known to contain 49.06 -0.02% of a constituent X. The result obtained by two analysis using tge same substance and the same general technique were Analysis 1 - 49.01, 49.21, 49.08 Analysis 2 - 49.40, 49.44, 49.42 Calculate the relative mean eroor and relative mean deviation in the two sets of results and comment on the accuracy and precision of them? theory, EduRev gives you an
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