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An observation was recorded wrongly: find the correct mean, variance or standard deviation

This type of question, asked 13 times: 2021 Paper 18, 26 Jun 2022, 28 Jun 2022, 27 Jul 2022, 25 Jan 2023, 8 Apr 2023, 13 Apr 2023, 15 Apr 2023, 27 Jan 2024, 6 Apr 2024, 24 Jan 2025, 7 Apr 2025, 24 Jan 2026 S1. All 13 questions below, oldest first.

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  1. Question 1

    2021 Paper 18

    62
    The mean and standard deviation of 20 observations were calculated as 10 and 2.5 respectively. It was found that by mistake one data value was taken as 25 instead of 35. If α\alpha and β\sqrt{\beta} are the mean and standard deviation respectively for correct data, then (α,β)(\alpha,\beta) is :

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  2. Question 2

    26 Jun 2022

    167
    The mean and standard deviation of 50 observations are 15 and 2 respectively. It was found that one incorrect observation was taken such that the sum of correct and incorrect observations is 70. If the correct mean is 16, then the correct variance is equal to:

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  3. Question 3

    28 Jun 2022

    83
    The mean and standard deviation of 15 observations are found to be 8 and 3 respectively. On rechecking it was found that, in the observations, 20 was misread as 5 . Then, the correct variance is equal to

  4. Question 4

    27 Jul 2022

    82
    The mean and variance of 10 observations were calculated as 15 and 15 respectively by a student who took by mistake 25 instead of 15 for one observation. Then, the correct standard deviation is

  5. Question 5

    25 Jan 2023

    66
    The mean and variance of the marks obtained by the students in a test are 10 and 4 respectively. Later, the marks of one of the students is increased from 8 to 12 . If the new mean of the marks is 10.2, then their new variance is equal to:

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  6. Question 6

    8 Apr 2023

    151
    Let the mean and variance of 12 observations be 92\frac{9}{2} and 4 respectively. Later on, it was observed that two observations were considered as 9 and 10 instead of 7 and 14 respectively. If the correct variance is mn\frac{m}{n}, where mm and nn are co-prime, then m+nm + n is equal to

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  7. Question 7

    13 Apr 2023

    171
    The mean and standard deviation of the marks of 10 students were found to be 50 and 12 respectively. Later, it was observed that two marks 20 and 25 were wrongly read as 45 and 50 respectively. Then the correct variance is

  8. Question 8

    15 Apr 2023

    63
    The mean and standard deviation of 10 observations are 20 and 8 respectively. Later on, it was observed that one observation was recorded as 50 instead of 40. Then the correct variance is:

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  9. Question 9

    27 Jan 2024

    171
    The mean and standard deviation of 15 observations were found to be 12 and 3 respectively. On rechecking it was found that an observation was read as 10 in place of 12 . If μ\mu and σ2\sigma^{2} denote the mean and variance of the correct observations respectively, then 15(μ+μ2+σ2)15\left( \mu +\mu^{2}+\sigma^{2} \right) is equal to .............

  10. Question 10

    6 Apr 2024

    64
    The mean and standard deviation of 20 observations are found to be 10 and 2, respectively. On rechecking, it was found that an observation by mistake was taken 8 instead of 12. The correct standard deviation is

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  11. Question 11

    24 Jan 2025

    64
    For a statistical data x1,x2,…,x10x_{1},x_{2},\ldots,x_{10} of 10 values, a student obtained the mean as 5.5 and ∑i=110xi2=371\sum_{i = 1}^{10} x_{i}^{2}= 371. He later found that he had noted two values in the data incorrectly as 4 and 5 , instead of the correct values 6 and 8 , respectively. The variance of the corrected data is

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  12. Question 12

    7 Apr 2025

    62
    The mean and standard deviation of 100 observations are 40 and 5.1, respectively. By mistake one observation is taken as 50 instead of 40. If the correct mean and the correct standard deviation are μ\mu and σ\sigma respectively, then 10(μ+σ)10(\mu+\sigma) is equal to.

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  13. Question 13

    24 Jan 2026 S1

    68
    The mean and variance of a data of 10 observations are 10 and 2, respectively. If an observation α\alpha in this data is replaced by β\beta, then the mean and variance become 10.1 and 1.99, respectively. Then α+β\alpha + \beta equals.

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