Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Statistics question

2020 · 8 Jan · Shift 2 · Q24
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Mathematics
  4. /Statistics
  5. /2020 · 8 Jan · Shift 2 · Q24

Statistics question

2020 · 8 Jan · Shift 2 · Q24

JEE MainMathematicsStatisticsMCQ+4 / −1
The mean and variance of 20 observations are found to be 10 and 4, respectively. On rechecking, it was found that an observation 9 was incorrect and the correct observation was 11. Then the correct variance is
  1. A
    3.98
  2. B
    3.99
  3. C
    4.01
  4. D
    4.02
View written solutionFree

Correct answer: B

  1. Given data

    • Number of observations: n=20n = 20n=20
    • गलत (incorrectly computed) mean: xˉ=10\bar{x} = 10xˉ=10
    • गलत variance: σ2=4\sigma^2 = 4σ2=4

  2. Use variance formula σ2=∑xi2n−xˉ2\sigma^2 = \frac{\sum x_i^2}{n} - \bar{x}^2σ2=n∑xi2​​−xˉ2 So, 4=∑xi220−1024 = \frac{\sum x_i^2}{20} - 10^24=20∑xi2​​−102 4=∑xi220−1004 = \frac{\sum x_i^2}{20} - 1004=20∑xi2​​−100 ∑xi220=104\frac{\sum x_i^2}{20} = 10420∑xi2​​=104 ∑xi2=2080\sum x_i^2 = 2080∑xi2​=2080

  3. Find the incorrect total sum ∑xi=nxˉ=20×10=200\sum x_i = n\bar{x} = 20 \times 10 = 200∑xi​=nxˉ=20×10=200

  4. Correct the wrong observation One value 999 was used instead of the correct value 111111.

    Therefore corrected sum is ∑xi′=200−9+11=202\sum x_i' = 200 - 9 + 11 = 202∑xi′​=200−9+11=202

    Hence corrected mean is xˉ′=20220=10.1\bar{x}' = \frac{202}{20} = 10.1xˉ′=20202​=10.1

  5. Correct the sum of squares ∑xi′2=2080−92+112\sum x_i'^2 = 2080 - 9^2 + 11^2∑xi′2​=2080−92+112 =2080−81+121= 2080 - 81 + 121=2080−81+121 =2120= 2120=2120

  6. Compute corrected variance σ′2=∑xi′220−(xˉ′)2\sigma'^2 = \frac{\sum x_i'^2}{20} - (\bar{x}')^2σ′2=20∑xi′2​​−(xˉ′)2 =212020−(10.1)2= \frac{2120}{20} - (10.1)^2=202120​−(10.1)2 =106−102.01= 106 - 102.01=106−102.01 =3.99= 3.99=3.99

  7. Match with options 3.993.993.99 corresponds to Option B.

Final Answer: B. 3.993.993.99

PreviousNext

More from Statistics

  • Let the observations xi (1 ≤ i ≤ 10) satisfy the equations, i=1∑10​(x1​−5)= 10 and i=1∑10​(x1​−5)2= 40. If μ and λ are the mean…2020 · MCQ
  • The mean and variance of seven observations are 8 and 16, respectively. If 5 of the observations are 2, 4, 10, 12, 14, then the product of the remaining two observations is :2019 · MCQ
  • A student scores the following marks in five tests : 45, 54, 41, 57, 43. His score is not known for the sixth test. If the mean score is 48 in the six tests, then the standard deviation of the marks in six tests is2019 · MCQ
  • If the standard deviation of the numbers –1, 0, 1, k is 5​ where k > 0, then k is equal to2019 · MCQ
  • The mean and the median of the following ten numbers in increasing order 10, 22, 26, 29, 34, x, 42, 67, 70, y are 42 and 35 respectively, then xy​ is equal to2019 · MCQ
  • 5 students of a class have an average height 150 cm and variance 18 cm2. A new student, whose height is 156 cm, joined them. The variance (in cm2) of the height of these six students is :2019 · MCQ
  • A data consists of n observations : x1, x2, . . . . . . ., xn. If i=1∑n​(xi​+1)2=9n and i=1∑n​(xi​−1)2=5n, then the standard deviation of this…2019 · MCQ
  • If for some x ∈ R, the frequency distribution of the marks obtained by 20 students in a test is : then the mean of the marks is Includes table2019 · MCQ