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Statistics question

2018 · Shift 0 · Q35
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Statistics question

2018 · Shift 0 · Q35

JEE MainMathematicsStatisticsMCQ+4 / −1
If ∑i=19(xi−5)=9\sum\limits_{i = 1}^9 {\left( {{x_i} - 5} \right)} = 9i=1∑9​(xi​−5)=9 and ∑i=19(xi−5)2=45\sum\limits_{i = 1}^9 {{{\left( {{x_i} - 5} \right)}^2}} = 45i=1∑9​(xi​−5)2=45, then the standard deviation of the 9 items x1,x2,.......,x9{x_1},{x_2},.......,{x_9}x1​,x2​,.......,x9​ is
  1. A
    3
  2. B
    9
  3. C
    4
  4. D
    2
View written solutionFree

Correct answer: D

  1. Let the mean of the 9 observations be xˉ\bar{x}xˉ.

Given ∑i=19(xi−5)=9\sum_{i=1}^9 (x_i-5)=9∑i=19​(xi​−5)=9 So, ∑i=19xi−9⋅5=9\sum_{i=1}^9 x_i - 9\cdot 5 = 9∑i=19​xi​−9⋅5=9 ∑i=19xi=54\sum_{i=1}^9 x_i = 54∑i=19​xi​=54 Hence, xˉ=549=6\bar{x}=\frac{54}{9}=6xˉ=954​=6

  1. We are also given ∑i=19(xi−5)2=45\sum_{i=1}^9 (x_i-5)^2=45∑i=19​(xi​−5)2=45

We need standard deviation about the mean xˉ=6\bar{x}=6xˉ=6: σ=19∑i=19(xi−xˉ)2\sigma=\sqrt{\frac{1}{9}\sum_{i=1}^9 (x_i-\bar{x})^2}σ=91​∑i=19​(xi​−xˉ)2​ So first compute ∑i=19(xi−6)2\sum_{i=1}^9 (x_i-6)^2∑i=19​(xi​−6)2

  1. Use the identity xi−5=(xi−6)+1x_i-5=(x_i-6)+1xi​−5=(xi​−6)+1 Therefore, (xi−5)2=((xi−6)+1)2=(xi−6)2+2(xi−6)+1(x_i-5)^2=((x_i-6)+1)^2=(x_i-6)^2+2(x_i-6)+1(xi​−5)2=((xi​−6)+1)2=(xi​−6)2+2(xi​−6)+1 Summing over i=1i=1i=1 to 999, ∑i=19(xi−5)2=∑i=19(xi−6)2+2∑i=19(xi−6)+9\sum_{i=1}^9 (x_i-5)^2=\sum_{i=1}^9 (x_i-6)^2+2\sum_{i=1}^9 (x_i-6)+9∑i=19​(xi​−5)2=∑i=19​(xi​−6)2+2∑i=19​(xi​−6)+9

Now, ∑i=19(xi−6)=∑i=19xi−9⋅6=54−54=0\sum_{i=1}^9 (x_i-6)=\sum_{i=1}^9 x_i-9\cdot 6=54-54=0∑i=19​(xi​−6)=∑i=19​xi​−9⋅6=54−54=0 So, 45=∑i=19(xi−6)2+0+945=\sum_{i=1}^9 (x_i-6)^2+0+945=∑i=19​(xi​−6)2+0+9 ∑i=19(xi−6)2=36\sum_{i=1}^9 (x_i-6)^2=36∑i=19​(xi​−6)2=36

  1. Hence variance is σ2=369=4\sigma^2=\frac{36}{9}=4σ2=936​=4 So standard deviation is σ=4=2\sigma=\sqrt{4}=2σ=4​=2

  2. Checking options:

  • A: 333 ❌
  • B: 999 ❌
  • C: 444 ❌
  • D: 222 ✅

Therefore, the correct answer is D.

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