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Statistics question

2019 · 12 Apr · Shift 1 · Q38
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Statistics question

2019 · 12 Apr · Shift 1 · Q38

JEE MainMathematicsStatisticsMCQ+4 / −1
If the data x1, x2,......., x10 is such that the mean of first four of these is 11, the mean of the remaining six is 16 and the sum of squares of all of these is 2,000 ; then the standard deviation of this data is :
  1. A
    2\sqrt 22​
  2. B
    2
  3. C
    2 2\sqrt 22​
  4. D
    4
View written solutionFree

Correct answer: B

  1. Find the total sum of the data

The mean of the first 4 observations is 111111, so their sum is 4×11=44.4 \times 11 = 44.4×11=44.

The mean of the remaining 6 observations is 161616, so their sum is 6×16=96.6 \times 16 = 96.6×16=96.

Hence, the total sum of all 10 observations is 44+96=140.44 + 96 = 140.44+96=140.

Therefore, the mean of all 10 observations is xˉ=14010=14.\bar{x} = \frac{140}{10} = 14.xˉ=10140​=14.

  1. Use the standard deviation formula

Given that ∑xi2=2000.\sum x_i^2 = 2000.∑xi2​=2000.

For n=10n=10n=10, the variance is σ2=∑xi2n−xˉ2.\sigma^2 = \frac{\sum x_i^2}{n} - \bar{x}^2.σ2=n∑xi2​​−xˉ2.

Substituting the values: σ2=200010−142=200−196=4.\sigma^2 = \frac{2000}{10} - 14^2 = 200 - 196 = 4.σ2=102000​−142=200−196=4.

So the standard deviation is σ=4=2.\sigma = \sqrt{4} = 2.σ=4​=2.

  1. Match with the options

The correct option is:

  • B: 222
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