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If the arithmetic mean of a set of n measurements was found to be m, a

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If the arithmetic mean of a set of n measurements was found to be m, a  [#permalink]

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New post 29 Jun 2017, 13:44
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If the arithmetic mean of a set of n measurements was found to be m, and then it was discovered that there was an error of +k% in each measurement. What is the percentage error in the mean?

(1) m = 6
(2) n = 7

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Re: If the arithmetic mean of a set of n measurements was found to be m, a  [#permalink]

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New post 29 Jun 2017, 16:02
Let us say that there are n measurements and each represented by\(a_1,a_2,a_3,a_4,a_5.......a_n\)

Mean of the above measurements is given by =\(\frac{(a_1+a_2+a_3+a_4+a_5+.......+a_n)}{n}\) = m -------Equation 1

now if each original value had an error of +K%, this means

\((1+ \frac{k}{100})\)Original value = new value

Therefore, Original value = new value * \([\frac{100}{(100+k)}]\)

Correct Mean = \([a_1*[\frac{100}{(100+k)}]+a_2*[\frac{100}{(100+k)}]+a_3*[\frac{100}{(100+k)}]+a_4*[\frac{100}{(100+k)}]+.......+a_n*[\frac{100}{(100+k)}]]/n\)

Correct Mean =\([\frac{100}{(100+k)}]\frac{[a1+a2+a3+a4+....+an]}{n}\) ------------------Equation 2

Replacing the value from Equation 1 in Equation 2

Correct Mean =\(\frac{100m}{(100+k)}\)

Percentage error in the Mean = (Given Mean - Old Mean)*100 / Given Mean = [(m - \(\frac{100m}{(100+k)}\))*100]/m = [1 - \(\frac{100}{(100+k)}\)]*100 = \(\frac{(100+k-100)}{(100+k)}\) * 100 = \(\frac{100k}{100+k}\)


So we need the of k to find the Percentage error in the Mean

Statement 1: m = 6
This statement does not given the value of k, Hence is insufficient.

Statement 2: n = 7
This statement does not given the value of k, Hence is insufficient.

Combining both statement is also not sufficient. Answer is E
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Re: If the arithmetic mean of a set of n measurements was found to be m, a  [#permalink]

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New post 29 Jun 2017, 18:55
duahsolo wrote:
If the arithmetic mean of a set of n measurements was found to be m, and then it was discovered that there was an error of +k% in each measurement. What is the percentage error in the mean?

(1) m = 6
(2) n = 7



avg. /mean(M) = Sum(s) / No. of samples(s)

Actual mean without error= M = s/ n

Mean with error(Let M') = s(1+k%) / n

%age error = M' - M /M

= s(1+k%)/n - s/n / s/n

= k/100

so for knowing the percentage error in the mean we must know k...

(1) no info of k...insuff

(2) no info of k...insuff

combined..again
no info of k...insuff

Ans E
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Re: If the arithmetic mean of a set of n measurements was found to be m, a  [#permalink]

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New post 26 Jun 2018, 06:24
+1 for option E. The error for mean will be k% itself. Both options individually and together are not sufficient. Hence option E it is.
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Re: If the arithmetic mean of a set of n measurements was found to be m, a &nbs [#permalink] 26 Jun 2018, 06:24
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