Answer

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**Hint:**Here in this question we have to calculate the mode and for which we have the values given. For this, we will use the mode formula given by $Mode = L + \left[ {\left( {{f_m} - {f_{m - 1}}} \right)/\left\{ {\left( {{f_m} - {f_{m - 1}}} \right) + \left( {{f_m} - {f_{m + 1}}} \right)} \right\}} \right] \times W$ . And then substituting the values in the formula and solving it, we will get the answer.

**Formula used:**

Mode formula is given by,

$Mode = L + \left[ {\left( {{f_m} - {f_{m - 1}}} \right)/\left\{ {\left( {{f_m} - {f_{m - 1}}} \right) + \left( {{f_m} - {f_{m + 1}}} \right)} \right\}} \right] \times W$

Here,

$L$ , will be the lower class boundary of the modal group.

${f_{m - 1}}$ , will be the frequency of the group before the modal group.

${f_m}$ , will be the frequency of the modal group.

${f_{m + 1}}$ , will be the frequency of the group after the modal group and

$W$ , will be the width of the group.

**Complete step-by-step answer:**So first of all we will see the values given to us and then we will proceed further for solving it.

So, we have

$L = 30$

${f_{m - 1}} = 30$

${f_m} = 6$

${f_{m + 1}} = 5$

$W = 10$

So now we will substitute the values in the formula, we will get

$ \Rightarrow 30 + \left[ {\left( {6 - 3} \right)/\left\{ {\left( {6 - 3} \right) + \left( {6 - 5} \right)} \right\}} \right] \times 10$

And on solving the above equation, we will get

$ \Rightarrow 30 + \left( {\dfrac{3}{{4}}} \right) \times 10$

And on solving it, we will get

$ \Rightarrow 37.5$

Therefore, the mode salary of the following observations will be equal to $37.5$

**Hence, the option $\left( a \right)$ is correct.**

**Note:**As we know that the mode is the value that occurs or we can say happens the highest number of times. So we can say that the mean, median, and mode will be the three types and we can say the kinds of averages. So if there is no number in the list then they are not repeated then there will not be any mode in the list.

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