Multiple choice

In TVS-2 batch every student was weighed on two weighing machines A and B Machine A and machine B both have some technical errors Machine A gives the reading equal to four times the original weight of a student and machine B gives the reading equal to 5 kg more than the actual weight If variances calculated by readings of machines A and B were $\displaystyle V_{A}: : : and: : : V_{B} $ respectively and mean deviations are $\displaystyle M_{A}: : : and: : : M_{B} $ respectively then

  1. $\displaystyle M_{A}=4M_{B}$
  2. $\displaystyle \frac{V_{A}}{V_{B}}=16\left ( \frac{M_{A}}{M_{B}} \right )$
  3. $\displaystyle \frac{V_{A}}{V_{B}}= \frac{M_{A}}{M_{B}}$
  4. $\displaystyle V_{A}=4V_{B}$
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D Correct answer

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