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**Hint**: For finding the arithmetic mean of ‘n’ positive numbers, we use the formula $ \dfrac{{{a}_{1}}+{{a}_{2}}+{{a}_{3}}+....+{{a}_{n}}}{n} $ where $ {{a}_{1}},{{a}_{2}}....{{a}_{n}} $ are ‘n’ positive numbers. If we have grouped data and we know the frequency of the data, then the arithmetic mean is given by the formula $ \dfrac{\sum\limits_{i=1}^{n}{{{f}_{i}}{{x}_{i}}}}{\sum\limits_{i=1}^{n}{{{f}_{i}}}} $ where $ {{f}_{i}} $ is the frequency of the data element $ {{x}_{i}} $ . We will use the later formula and find out the arithmetic mean of the given data.

**:**

__Complete step-by-step answer__Now, from the given data we can see that it is a grouped data and hence the formula $ \dfrac{\sum\limits_{i=1}^{n}{{{f}_{i}}{{x}_{i}}}}{\sum\limits_{i=1}^{n}{{{f}_{i}}}} $ will be used.

Thus, the arithmetic mean of the given data will be calculated as:

$ \Rightarrow \dfrac{3\left( 220 \right)+1\left( 230 \right)+2\left( 240 \right)+6\left( 250 \right)+5\left( 260 \right)+4\left( 270 \right)+3\left( 280 \right)}{3+1+2+6+5+4+3} $

$ \begin{align}

& \Rightarrow \dfrac{660+230+480+1500++1300+1080+840}{24} \\

& \Rightarrow \dfrac{6090}{24} \\

& \Rightarrow 253.75 \\

\end{align} $

Therefore, the arithmetic mean of the given data is 253.75

**So, the correct answer is “Option B”.**

**Note**: This question can also be done by the formula $ \dfrac{{{a}_{1}}+{{a}_{2}}+{{a}_{3}}+....+{{a}_{n}}}{n} $ by the following way:

Here, we can consider the wages as distinct numbers for all the employees. For example, we have been given that Rs. 220 is the wage for 3 employees so we can take 220 as $ {{a}_{1}},{{a}_{2}} $ and $ {{a}_{3}} $ and so on. The formula we used above has actually been derived from this very formula and thus the answer will be the same in both cases. But we use the formula $ \dfrac{\sum\limits_{i=1}^{n}{{{f}_{i}}{{x}_{i}}}}{\sum\limits_{i=1}^{n}{{{f}_{i}}}} $ because this makes calculation easy and thus removes the scope of committing mistakes. In grouped data, using the formula $ \dfrac{{{a}_{1}}+{{a}_{2}}+{{a}_{3}}+....+{{a}_{n}}}{n} $ will result in a very long and tedious calculation and increase the chances for mistakes and wouldn’t really work for data with high frequencies. Thus, for grouped data we should always use the formula $ \dfrac{\sum\limits_{i=1}^{n}{{{f}_{i}}{{x}_{i}}}}{\sum\limits_{i=1}^{n}{{{f}_{i}}}} $ .

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