Answer

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**Hint:**Here, in this question, we will solve it by using the step deviation method. So the mid values will be calculated by taking out the average of the class interval. And the number of students will become frequent. So by taking the $\sum {{f_i}{u_1}} $ and by using the formula, we will get the mean.

**Formula used:**

Formula for solving the mean is given by,

$ \Rightarrow Mean(\bar x) = A + h\left[ {\dfrac{1}{N}\sum {{f_i}{u_1}} } \right]$

Here, $A$ , will be the mid-term of the mid values

$N$ , will be the total number of frequencies.

$\sum {{f_i}{u_1}} $ , will be the sum

**Complete step-by-step answer:**So with the values given we will find out the required blacks. Therefore,

${d_i} = {x_i} - A$ and similarly ${u_i} = \dfrac{{{d_i}}}{h}$ . So with the table, we had calculated all those values. And the table will be represented as

Class Interval | Mid value $x_i$ | Frequency $f_i$ | $d_i$ | $u_i$ | $u_i$$f_i$ |

0-10 | 5 | 4 | -20 | -2 | -8 |

10-20 | 15 | 6 | -10 | -1 | -6 |

20-30 | 25=A | 10 | 0 | 0 | 0 |

30-40 | 35 | 20 | 10 | 1 | 10 |

40-50 | 45 | 10 | 20 | 2 | 10 |

N=50,

Total=26

So here after calculating all those values, we have

$h = 10,A = 25,N = 50,\sum {{f_i}{u_1}} = 26$ . So on substituting this all values in the formula and the formula is given by,

$ \Rightarrow Mean(\bar x) = A + h\left[ {\dfrac{1}{N}\sum {{f_i}{u_1}} } \right]$

So on substituting the values, we get the equation as

$ \Rightarrow Mean(\bar x) = 25 + 10\left[ {\dfrac{1}{{50}}\sum {26} } \right]$

Son solving the braces first we will get number as

$ \Rightarrow 30.2$

Hence, the mean will be equal to $30.2$ .

**Note:**So the basic step of deviation use is to find the mean as in this we get easily available. We can say that mean, median and mode are the three essential parts through which we can see the different perspectives of the same data. Median, which is being used for finding the middle number. Whereas mode is being used when any number is used on a frequent basis. So these are the basic ideas about the mean, median and mode.

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