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Statistics question

2019 · 10 Apr · Shift 2 · Q32
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Statistics question

2019 · 10 Apr · Shift 2 · Q32

JEE MainMathematicsStatisticsMCQ+4 / −1
If both the mean and the standard deviation of 50 observations x1, x2,..., x50 are equal to 16, then the mean of (x1 – 4)2 , (x2 – 4)2 ,....., (x50 – 4)2 is :
  1. A
    400
  2. B
    480
  3. C
    380
  4. D
    525
View written solutionFree

Correct answer: A

  1. Let the mean of the 50 observations be xˉ=16\bar{x}=16xˉ=16.

  2. The standard deviation is also given as 161616. Using the relation σ2=x2‾−(xˉ)2,\sigma^2 = \overline{x^2} - (\bar{x})^2,σ2=x2−(xˉ)2, where x2‾\overline{x^2}x2 is the mean of the squares of the observations.

Since σ=16\sigma=16σ=16, σ2=162=256.\sigma^2 = 16^2 = 256.σ2=162=256. So, x2‾−162=256\overline{x^2} - 16^2 = 256x2−162=256 x2‾−256=256\overline{x^2} - 256 = 256x2−256=256 x2‾=512.\overline{x^2} = 512.x2=512.

  1. We need the mean of (x1−4)2,(x2−4)2,…,(x50−4)2.(x_1-4)^2,(x_2-4)^2,\ldots,(x_{50}-4)^2.(x1​−4)2,(x2​−4)2,…,(x50​−4)2.

Its mean is (x−4)2‾=x2−8x+16‾=x2‾−8xˉ+16.\overline{(x-4)^2} = \overline{x^2 - 8x + 16} = \overline{x^2} - 8\bar{x} + 16.(x−4)2​=x2−8x+16​=x2−8xˉ+16. Substitute the known values: (x−4)2‾=512−8(16)+16\overline{(x-4)^2} = 512 - 8(16) + 16(x−4)2​=512−8(16)+16 =512−128+16= 512 - 128 + 16=512−128+16 =400.= 400.=400.

  1. Therefore, the required mean is 400.\boxed{400}.400​.

  2. Option check:

  • A: 400400400 ✅
  • B: 480480480 ❌
  • C: 380380380 ❌
  • D: 525525525 ❌
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