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A set consists of six terms : x, 10, 20, 30, 40 and 50. What should be
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27 Jul 2023, 06:29
27 Jul 2023, 06:29
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A set consists of six terms : x, 10, 20, 30, 40 and 50. What should be the value of x so that standard deviation of the six terms is minimum ?
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A set consists of six terms : x, 10, 20, 30, 40 and 50. What should be
[#permalink]
17 Sep 2023, 05:11
17 Sep 2023, 05:11
Expert Reply
OE
Standard deviation signifies, deviation of the terms from its mean value
That means, more deviated are the terms from its mean, more is SD.
To keep standard deviation minimum, we add a term which is closest to the
mean and a number which is closest to mean is mean itself.
Consider the set of terms 10, 20, 30, 40, 50.
Here mean is 30. (As the terms are in AP, mean and median is same).
Now, to have minimum standard deviation after adding a new term to the above set,
the new term should be equal to the mean.
i.e. if we add 30 to (10, 20, 30, 40, 50), then we can see that 30 is not at all deviated
from the mean.
As standard deviation signifies ‘deviation from mean’, we did not let the term deviate
from the mean.
Ans. (30)
Standard deviation signifies, deviation of the terms from its mean value
That means, more deviated are the terms from its mean, more is SD.
To keep standard deviation minimum, we add a term which is closest to the
mean and a number which is closest to mean is mean itself.
Consider the set of terms 10, 20, 30, 40, 50.
Here mean is 30. (As the terms are in AP, mean and median is same).
Now, to have minimum standard deviation after adding a new term to the above set,
the new term should be equal to the mean.
i.e. if we add 30 to (10, 20, 30, 40, 50), then we can see that 30 is not at all deviated
from the mean.
As standard deviation signifies ‘deviation from mean’, we did not let the term deviate
from the mean.
Ans. (30)
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Joined: 11 Sep 2023
Posts: 43
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Re: A set consists of six terms : x, 10, 20, 30, 40 and 50. What should be
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10 Oct 2023, 07:03
10 Oct 2023, 07:03
Carcass wrote:
OE
Sum of all terms = Average × number of terms
Sum of weights for 10 students = 43 × 10 = 430
It is given that two students, having average weight 45kg leaves the class.
So, Sum of weights of two students = Average weight × number of students
= 45×2
= 90
Now, Sum of weights of remaining 8 students = Sum of weight of 10 students – Sum
of weight of 2 students
= 430 – 90
= 340
Average weight of 8 students =340/8=42.5
Sum of all terms = Average × number of terms
Sum of weights for 10 students = 43 × 10 = 430
It is given that two students, having average weight 45kg leaves the class.
So, Sum of weights of two students = Average weight × number of students
= 45×2
= 90
Now, Sum of weights of remaining 8 students = Sum of weight of 10 students – Sum
of weight of 2 students
= 430 – 90
= 340
Average weight of 8 students =340/8=42.5
wrong question I guess
