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Statistics question

2018 · 16 Apr · Shift 1 · Q41
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Statistics question

2018 · 16 Apr · Shift 1 · Q41

JEE MainMathematicsStatisticsMCQ+4 / −1
The mean and the standard deviation(s.d.) of five observations are9 and 0, respectively. If one of the observations is changed such that the mean of the new set of five observations becomes 10, then their s.d. is :
  1. A
    0
  2. B
    1
  3. C
    2
  4. D
    4
View written solutionFree

Correct answer: C

  1. Use the given mean and standard deviation of the original 5 observations

Since the standard deviation is 000, all five observations must be equal to the mean.

So the original observations are: 9,9,9,9,99,9,9,9,99,9,9,9,9

  1. One observation is changed and new mean becomes 101010

Originally, the sum of the five observations is: 5×9=455 \times 9 = 455×9=45

After changing one observation, the new mean is 101010, so the new sum is: 5×10=505 \times 10 = 505×10=50

Thus, the changed observation must increase the total by 555.

So one of the 999's becomes: 9+5=149+5=149+5=14

Hence the new observations are: 14,9,9,9,914,9,9,9,914,9,9,9,9

  1. Compute the new standard deviation

Mean of new set =10=10=10.

Deviations from mean: 14−10=4, 9−10=−1, 9−10=−1, 9−10=−1, 9−10=−114-10=4,\, 9-10=-1,\, 9-10=-1,\, 9-10=-1,\, 9-10=-114−10=4,9−10=−1,9−10=−1,9−10=−1,9−10=−1

Squares of deviations: 42=16, (−1)2=1,1,1,14^2=16,\, (-1)^2=1,1,1,142=16,(−1)2=1,1,1,1

Sum of squared deviations: 16+1+1+1+1=2016+1+1+1+1=2016+1+1+1+1=20

Variance: 205=4\frac{20}{5}=4520​=4

Standard deviation: 4=2\sqrt{4}=24​=2

  1. Match with the options

The new standard deviation is: 222

So the correct option is C.

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