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 on Question #54873 – 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

1) Let's first find the correct mean. By the definition of the mean we have:

$$\bar{x} = \frac{\sum x_i}{n},$$

where, $\bar{x}$ is the mean, $\sum x_i$ is the sum of the items, $n$ is the number of the items.

Substituting the mean and the number of the items into the formula we can calculate the sum of the items:

$$\sum x_i = \bar{x}n = 10 \cdot 20 = 200.$$

Because one item with value 8 was incorrect and it is omitted we must subtract this item from the sum of the items:

$$\sum (x_i)_C = 200 - 8 = 192.$$

Thus, we have the number of items $n = 19$ and correct sum of the items $\sum x_i = 192$ and can calculate the correct mean:

$$\bar{x}_C = \frac{\sum (x_i)_C}{n} = \frac{192}{19} = 10.1.$$

2) Let's find the correct standard deviation. From the condition of the question we know that the standard deviation $\sigma = 2$. Then the variance $\sigma^2$ will be:

$$\sigma^2 = \frac{\sum x_i^2}{n} - \left(\frac{\sum x_i}{n}\right)^2 = 2^2 = 4,$$ $$\frac{\sum x_i^2}{20} - 10^2 = 4,$$ $$\sum x_i^2 = (4 + 100) \cdot 20 = 2080.$$

Because wrong item 8 is omitted we get:

$$\sum \left(x _ {i} ^ {2}\right) _ {C} = 2 0 8 0 - 8 ^ {2} = 2 0 8 0 - 6 4 = 2 0 1 6.$$

Therefore, we can calculate the correct standard deviation:

$$\sigma_ {C} = \sqrt {\frac {\sum (x _ {i} ^ {2}) _ {C}}{n} - (\frac {\sum (x _ {i}) _ {C}}{n}) ^ {2}} = \sqrt {\frac {2 0 1 6}{1 9} - (\frac {1 9 2}{1 9}) ^ {2}} = \sqrt {\frac {2 0 1 6 \cdot 1 9 - 1 9 2 ^ {2}}{1 9 ^ {2}}} = \frac {\sqrt {1 4 4 0}}{1 9} = 1. 9 9.$$

Answer:

1) $\bar{x}_C = 10.1$

2) $\sigma_{C} = 1.99$

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