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Statistics question

2019 · 9 Jan · Shift 2 · Q34
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Statistics question

2019 · 9 Jan · Shift 2 · Q34

JEE MainMathematicsStatisticsMCQ+4 / −1
A data consists of n observations : x1, x2, . . . . . . ., xn. If ∑i=1n(xi+1)2=9n\sum\limits_{i = 1}^n {{{\left( {{x_i} + 1} \right)}^2}} = 9ni=1∑n​(xi​+1)2=9n and ∑i=1n(xi−1)2=5n,\sum\limits_{i = 1}^n {{{\left( {{x_i} - 1} \right)}^2}} = 5n,i=1∑n​(xi​−1)2=5n, then the standard deviation of this data is :
  1. A
    2
  2. B
    5\sqrt 55​
  3. C
    5
  4. D
    7\sqrt 77​
View written solutionFree

Correct answer: B

  1. Let ∑i=1nxi=S,∑i=1nxi2=Q.\sum_{i=1}^n x_i = S, \qquad \sum_{i=1}^n x_i^2 = Q.∑i=1n​xi​=S,∑i=1n​xi2​=Q.

We are given: ∑i=1n(xi+1)2=9n\sum_{i=1}^n (x_i+1)^2 = 9n∑i=1n​(xi​+1)2=9n and ∑i=1n(xi−1)2=5n.\sum_{i=1}^n (x_i-1)^2 = 5n.∑i=1n​(xi​−1)2=5n.

  1. Expand both sums: ∑i=1n(xi+1)2=∑i=1n(xi2+2xi+1)=Q+2S+n=9n\sum_{i=1}^n (x_i+1)^2 = \sum_{i=1}^n (x_i^2+2x_i+1)=Q+2S+n=9n∑i=1n​(xi​+1)2=∑i=1n​(xi2​+2xi​+1)=Q+2S+n=9n So, Q+2S=8n...(1)Q+2S=8n \quad ...(1)Q+2S=8n...(1)

Similarly, ∑i=1n(xi−1)2=∑i=1n(xi2−2xi+1)=Q−2S+n=5n\sum_{i=1}^n (x_i-1)^2 = \sum_{i=1}^n (x_i^2-2x_i+1)=Q-2S+n=5n∑i=1n​(xi​−1)2=∑i=1n​(xi2​−2xi​+1)=Q−2S+n=5n So, Q−2S=4n...(2)Q-2S=4n \quad ...(2)Q−2S=4n...(2)

  1. Solve equations (1) and (2): Adding, 2Q=12n  ⟹  Q=6n.2Q=12n \implies Q=6n.2Q=12n⟹Q=6n. Subtracting, 4S=4n  ⟹  S=n.4S=4n \implies S=n.4S=4n⟹S=n.

Hence, xˉ=Sn=1.\bar{x}=\frac{S}{n}=1.xˉ=nS​=1.

  1. Variance of the data is σ2=1n∑i=1nxi2−(1n∑i=1nxi)2\sigma^2=\frac{1}{n}\sum_{i=1}^n x_i^2-\left(\frac{1}{n}\sum_{i=1}^n x_i\right)^2σ2=n1​∑i=1n​xi2​−(n1​∑i=1n​xi​)2 =Qn−(Sn)2=6nn−12=6−1=5.=\frac{Q}{n}-\left(\frac{S}{n}\right)^2=\frac{6n}{n}-1^2=6-1=5.=nQ​−(nS​)2=n6n​−12=6−1=5.

Therefore, the standard deviation is σ=5.\sigma=\sqrt{5}.σ=5​.

  1. Checking options:
  • A: 222 ✗
  • B: 5\sqrt{5}5​ ✓
  • C: 555 ✗
  • D: 7\sqrt{7}7​ ✗

So the correct answer is Option B.

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