The mean and sample standard deviation of the dataset consisting of 7 observations is 19 and 8 respectively. Later it is noted that one observation 17 is wrongly noted as 14. What is the sample variance of the original dataset? (Conect to 2 decimal place accuracy)

1

We are given the following information:

- Mean of the dataset, \( \bar{x} = 19 \)
- Sample standard deviation, \( s = 8 \)
- Number of observations, \( n = 7 \)
- One observation was wrongly recorded as 14 instead of 17.

### Step 1: Calculate the sum of the original dataset
We know that the mean is given by:

\[
\bar{x} = \frac{\sum x_i}{n}
\]

So, the sum of the original dataset \( \sum x_i \) is:

\[
\sum x_i = \bar{x} \times n = 19 \times 7 = 133
\]

### Step 2: Correct the sum of the dataset
Since one observation was wrongly recorded as 14 instead of 17, the correct sum of the dataset will be:

\[
\text{Correct sum} = 133 + (17 - 14) = 133 + 3 = 136
\]

### Step 3: Calculate the correct mean
The correct mean is:

\[
\bar{x}_{\text{correct}} = \frac{\text{Correct sum}}{n} = \frac{136}{7} = 19.43
\]

### Step 4: Calculate the original sum of squared deviations from the mean
We know the sample standard deviation \( s \) is related to the sum of squared deviations by the formula:

\[
s^2 = \frac{\sum (x_i - \bar{x})^2}{n - 1}
\]

Thus, the sum of squared deviations from the mean is:

\[
\sum (x_i - \bar{x})^2 = s^2 \times (n - 1) = 8^2 \times (7 - 1) = 64 \times 6 = 384
\]

### Step 5: Adjust the sum of squared deviations for the correction
Now, we need to adjust this sum of squared deviations for the corrected observation. The adjustment for the squared deviation of 14 being replaced by 17 can be computed as follows:

- The original deviation for 14 was \( (14 - 19)^2 = (-5)^2 = 25 \).
- The correct deviation for 17 is \( (17 - 19.43)^2 = (-2.43)^2 \approx 5.90 \).

Thus, the adjustment to the sum of squared deviations is:

\[
\text{Adjustment} = 5.90 - 25 = -19.10
\]

Therefore, the corrected sum of squared deviations is:

\[
\text{Corrected sum of squared deviations} = 384 - 19.10 = 364.90
\]

### Step 6: Calculate the corrected sample variance
The corrected sample variance is:

\[
s_{\text{correct}}^2 = \frac{\text{Corrected sum of squared deviations}}{n - 1} = \frac{364.90}{6} \approx 60.82
\]

Thus, the corrected sample variance is approximately **60.82**.

---

Would you like to go over any details or ask questions?

Here are 5 related questions to expand on this concept:
1. How would the process change if the wrong value was larger than the correct one?
2. What is the effect on the standard deviation when a value closer to the mean is corrected?
3. How does variance change if multiple values in the dataset are wrong?
4. What is the difference between sample variance and population variance?
5. How can you use the sample variance to estimate the population variance?

**Tip**: Always double-check your data before calculating statistics to avoid errors that could skew your results!

The mean and sample standard deviation of the dataset consisting of 7 observations is 19 and 8 respectively. Later it is noted that one observation 17 is wrongly noted as 14. What is the sample variance of the original dataset?

Learn how to correct a dataset and find the sample variance when one value is wrongly noted. In this problem, the mean is 19 and the standard deviation is 8, with a correction from 14 to 17. A step-by-step solution covers the adjustment of the sum of squared deviations and calculation of corrected variance. Suitable for high school statistics students.

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