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Statistics question

2003 · Shift 0 · Q75
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Statistics question

2003 · Shift 0 · Q75

JEE MainMathematicsStatisticsMCQ+4 / −1
The median of a set of 9 distinct observations is 20.5. If each of the largest 4 observations of the set is increased by 2, then the median of the new set :
  1. A
    is increased by 2
  2. B
    is decreased by 2
  3. C
    is two times the original median
  4. D
    remains the same as that of the original set
View written solutionFree

Correct answer: D

  1. Arrange the 9 distinct observations in ascending order

    Let the observations be x1<x2<x3<x4<x5<x6<x7<x8<x9.x_1 < x_2 < x_3 < x_4 < x_5 < x_6 < x_7 < x_8 < x_9.x1​<x2​<x3​<x4​<x5​<x6​<x7​<x8​<x9​.

  2. Use the definition of median for 9 observations

    Since there are 9 observations (an odd number), the median is the middle term, i.e. the 5th observation.

    Hence, x5=20.5.x_5 = 20.5.x5​=20.5.

  3. Modify the largest 4 observations

    The largest 4 observations are x6,x7,x8,x9.x_6, x_7, x_8, x_9.x6​,x7​,x8​,x9​.

    Each of these is increased by 2, so the new set becomes x1,x2,x3,x4,x5,x6+2,x7+2,x8+2,x9+2.x_1, x_2, x_3, x_4, x_5, x_6+2, x_7+2, x_8+2, x_9+2.x1​,x2​,x3​,x4​,x5​,x6​+2,x7​+2,x8​+2,x9​+2.

  4. Check the median of the new set

    The first five observations remain unchanged: x1<x2<x3<x4<x5.x_1 < x_2 < x_3 < x_4 < x_5.x1​<x2​<x3​<x4​<x5​.

    Also, since only the last four terms were increased, the 5th term is still x5x_5x5​.

    Therefore, the median of the new set is still x5=20.5.x_5 = 20.5.x5​=20.5.

  5. Conclusion

    The median does not change.

    Therefore, the correct option is: D: remains the same as that of the original set\boxed{\text{D: remains the same as that of the original set}}D: remains the same as that of the original set​

  6. Comparison with stored correct answer

    Stored correct answer: D

    Derived answer: D

    These match.

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