riteshgupta wrote:
Can any one answer the below with a bit of detail, so that S.D concept is cleared????
During an experiment, some water was removed from each of 6 water tanks. If the standard deviation of the volumes of water in the tanks at the beginning of the experiment was 10 gallons, what was the standard deviation of the volumes of water in the tanks at the end of the experiment?
(1) For each tank, 30 percent of the volume of water that was in the tank at the beginning of the experiment was removed during the experiment.
(2) The average (arithmetic mean) volume of water in the tanks at the end of the experiment was 63 gallons.
We have
Standard Deviation = \(\sigma =\sqrt{{\frac{1}{6}}*[(x_1 - M)^2 + (x_2 - M)^2 + ... + (x_6 - M)^2} = 10\)
Here \(M\) is the mean.
The subscript 6 is for the 6 water tanks
Now,
Evaluating statement 1 onlyFor each tank 30% of the water was removed.
The mean becomes \((1- \frac{30}{100})M = 0.7M\)
Also each of \(x_1, x_2, ..., x_6\) becomes \(0.7x_1, 0.7x_2, ..., 0.7x_6\) respectively.
Hence,
Standard Deviation = \(\sigma =0.7* \sqrt{{\frac{1}{6}}*[(x_1 - M)^2 + (x_2 - M)^2 + ... + (x_6 - M)^2} = 0.7*10 = 7\)
Hence this statement alone is sufficient.
Choices B, C, E are eliminated.
Evaluating statement 2 onlyThe mean after the reduction, \(M = 63\) gallons.
We have no idea of the initial value of \(M\) or \(x_1, x_2, ..., x_6\) or how these values have changed.
Hence, this statement alone is insufficient.
Choice D is ruled out.
Answer is A.
Regards,
Shouvik.