Math Problem Statement

The mean and sample standard deviation of the dataset consisting of 9 observations is 14 and 10 respectively. Later it is noted that one observation 11 is wrongly noted as 6. What is the sample variance of the original dataset? (Correct to 2 decimal place accuracy)

Solution

To find the sample variance of the original dataset, we need to correct the wrongly noted observation and then recalculate the variance. Let's follow the steps:

  1. Mean and standard deviation of the incorrect dataset:

    • Mean (xˉ\bar{x}) = 14
    • Standard deviation (s) = 10
  2. Calculating the sum of the incorrect dataset: Since the mean is the sum of all observations divided by the number of observations (n=9n = 9): xi=xˉ×n=14×9=126\sum x_i = \bar{x} \times n = 14 \times 9 = 126

  3. Identifying the incorrect observation: The incorrect observation is 6, and the correct observation is 11.

  4. Sum of the correct dataset: Adjust the sum by replacing 6 with 11: xicorrected=1266+11=131\sum x_i^{\text{corrected}} = 126 - 6 + 11 = 131

  5. Mean of the correct dataset: xˉcorrected=xicorrectedn=131914.56\bar{x}_{\text{corrected}} = \frac{\sum x_i^{\text{corrected}}}{n} = \frac{131}{9} \approx 14.56

  6. Calculating the sum of squares of deviations for the incorrect dataset: The formula for the sample variance is: s2=1n1(xixˉ)2s^2 = \frac{1}{n-1} \sum (x_i - \bar{x})^2 Therefore: (xixˉ)2=s2×(n1)=102×8=800\sum (x_i - \bar{x})^2 = s^2 \times (n-1) = 10^2 \times 8 = 800

  7. Correcting the sum of squares of deviations: We need to adjust this sum for the corrected observation: Incorrect deviation sum=(614)2=64\text{Incorrect deviation sum} = (6 - 14)^2 = 64 Correct deviation sum=(1114.56)212.74\text{Correct deviation sum} = (11 - 14.56)^2 \approx 12.74 (xixˉcorrected)2=80064+12.74=748.74\sum (x_i - \bar{x}_{\text{corrected}})^2 = 800 - 64 + 12.74 = 748.74

  8. Calculating the sample variance for the corrected dataset: scorrected2=(xixˉcorrected)2n1=748.74893.59s^2_{\text{corrected}} = \frac{\sum (x_i - \bar{x}_{\text{corrected}})^2}{n-1} = \frac{748.74}{8} \approx 93.59

Therefore, the sample variance of the original dataset is approximately 93.5993.59.

Would you like any details or have any questions about this process? Here are some related questions you might find useful:

  1. How do you calculate the sample variance?
  2. What is the difference between sample variance and population variance?
  3. How does correcting an observation affect the dataset's mean and variance?
  4. What is the importance of the mean in a dataset?
  5. How do you calculate the standard deviation from the variance?

Tip: Always double-check your data for any incorrect entries as even a small error can significantly affect statistical calculations.

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Math Problem Analysis

Mathematical Concepts

Statistics
Descriptive Statistics
Variance

Formulas

Sample Variance formula

Theorems

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Suitable Grade Level

Undergraduate