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

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Manager
Joined: 02 Jun 2015
Posts: 191
Location: Ghana
If the arithmetic mean of a set of n measurements was found to be m, a  [#permalink]

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29 Jun 2017, 12:44
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Difficulty:

35% (medium)

Question Stats:

67% (01:29) correct 33% (01:22) wrong based on 75 sessions

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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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Posts: 331
Re: If the arithmetic mean of a set of n measurements was found to be m, a  [#permalink]

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29 Jun 2017, 15:02
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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29 Jun 2017, 17:55
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
Senior Manager
Joined: 08 Jun 2015
Posts: 435
Location: India
GMAT 1: 640 Q48 V29
GMAT 2: 700 Q48 V38
GPA: 3.33
Re: If the arithmetic mean of a set of n measurements was found to be m, a  [#permalink]

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26 Jun 2018, 05:24
+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, 05:24
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# If the arithmetic mean of a set of n measurements was found to be m, a

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