Question

The mean and the standard deviation of 20 items is found to be 10 and 2
respectively. At the time of checking it was found that one items with value 8 was
incorrect. Calculate the mean and standard deviation if the wrong item is
omitted

Answer 1

Answer to Question #85578 - Math - Statistics and Probability

Question: The mean and the standard deviation of 20 items is found to be 10 and 2 respectively. At the time of checking it was found that one item with value 8 was incorrect. Calculate the mean and standard deviation if the wrong item is omitted.

Solution: Given that the mean is 10, standard deviation is 2, and there are 20 items. If the values are $x_{1}, x_{2}, \ldots, x_{20}$, then we have,

$$mean = \bar{x} = \frac{1}{20} \sum_{i=1}^{20} x_i = 10$$ $$\text{and standard deviation} = \sqrt{\frac{1}{20} \sum_{i=1}^{20} (x_i - \bar{x})^2} = 2$$

From equation (1), we have, $\sum_{i=1}^{20} x_i = 200$

and from equation (2),

$$\begin{aligned} &\frac{1}{20} \sum_{i=1}^{20} (x_i - \bar{x})^2 = 4 \\ &\Rightarrow \frac{1}{20} \sum_{i=1}^{20} x_i^2 - \bar{x}^2 = 4 \\ &\Rightarrow \frac{1}{20} \sum_{i=1}^{20} x_i^2 = 4 + 10^2 = 104 \\ &\Rightarrow \sum_{i=1}^{20} x_i^2 = 104 \times 20 = 2080 \end{aligned}$$

Let $x_k$ denote the wrong entry for some $k \in \{1, 2, \ldots, 20\}$. So, $x_k = 8$.

If we omit $x_k$, then there are 19 entries. So the new mean is,

$$mean = \frac{1}{19} \left( \sum_{i=1}^{20} x_i \right) - x_k = \frac{200 - 8}{19} = \frac{192}{19} \approx 10.105$$

Let's call this new mean $\bar{x}' = \frac{192}{19}$

$$\begin{aligned} Variance &= \frac{1}{19} \left( \left( \sum_{i=1}^{20} x_i^2 \right) - x_k^2 \right) - \bar{x}'^2 \\ &= \frac{1}{19} (2080 - 8^2) - \left( \frac{192}{19} \right)^2 \\ &\approx 3.9889 \end{aligned}$$

Therefore, standard deviation $\approx \sqrt{3.9889} \approx 1.997$

Answer:

- Mean = 10.105

- Standard deviation = 1.997

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