Central Tendency & DispersionPYQ Jan 21Question 3131 of 473
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It is given that the mean (X\displaystyle X) is 10 and standard deviation (s.d.) is 3.2. If the observations are increased by 4, then the new mean and standard deviations are:

Options

AX=10\displaystyle X=10, s.d. =7.2\displaystyle =7.2
BX=14\displaystyle X=14, s.d. =3.2\displaystyle =3.2
CX=14\displaystyle X=14, s.d. =7.2\displaystyle =7.2
DX=10\displaystyle X=10, s.d. =3.2\displaystyle =3.2
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Correct Answer

✅ Option b — X=14\displaystyle X=14, s.d. =3.2\displaystyle =3.2

All Options:

  • AX=10\displaystyle X=10, s.d. =7.2\displaystyle =7.2
  • BX=14\displaystyle X=14, s.d. =3.2\displaystyle =3.2
  • CX=14\displaystyle X=14, s.d. =7.2\displaystyle =7.2
  • DX=10\displaystyle X=10, s.d. =3.2\displaystyle =3.2

Detailed Solution & Explanation

**Given:** Original Mean Xˉ=10\displaystyle \bar{X} = 10, Original SD = 3.2 Each observation is increased by 4. **Effect on Mean:** When a constant c\displaystyle c is added to every observation, the mean also increases by c\displaystyle c: Xˉnew=Xˉ+c=10+4=14\bar{X}_{new} = \bar{X} + c = 10 + 4 = 14 **Effect on Standard Deviation:** Standard Deviation measures the spread of data around the mean. Adding a constant to all observations shifts the entire dataset uniformly — it does NOT change the spread. Therefore: SDnew=SDold=3.2SD_{new} = SD_{old} = 3.2 **Verification:** If xi\displaystyle x_i are the original observations and xi′=xi+4\displaystyle x_i' = x_i + 4: - New mean: xˉ′=xˉ+4=14\displaystyle \bar{x}' = \bar{x} + 4 = 14 - Deviation: xi′−xˉ′=(xi+4)−(xˉ+4)=xi−xˉ\displaystyle x_i' - \bar{x}' = (x_i + 4) - (\bar{x} + 4) = x_i - \bar{x} (unchanged) - Hence SD unchanged. **New Mean = 14, New SD = 3.2** Hence, **Option B** is the correct answer.

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