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**Hint:**First assume the missing frequency of class 49-52. Then take the mid values of each class as ${x_i}$ and frequency ${f_i}$. The mean value is equivalent to the fraction between the addition of a product of mid-value with frequency and the total frequency.

**Complete Step by Step Solution:**

Given the mean for the given frequency distribution is 47.2.

Let the missing frequency be $x$.

The frequency distribution table for the given data is as follows:

Class | Frequency (${f_i}$) | Mid-value (${x_i}$) | ${f_i}{x_i}$ |

40-43 | 31 | 41.5 | 1286.5 |

43-46 | 58 | 44.5 | 2581 |

46-49 | 60 | 47.5 | 2850 |

49-52 | $x$ | 50.5 | \[50.5x\] |

52-55 | 27 | 53.5 | 1444.5 |

Total | $\sum {{f_i}} = 176 + x$ | $\sum {{f_i}{x_i}} = 8162 + 50.5x$ |

We know that the general formula to find the mean value is,

Mean $ = \dfrac{{\sum {{x_i}{f_i}} }}{{\sum {{x_i}} }}$

Now, we will substitute the value for the sum of the product of frequency and midpoint and the value for the sum of total frequency.

$ \Rightarrow 47.2 = \dfrac{{8162 + 50.5x}}{{176 + x}}$

Cross-multiply the terms,

$ \Rightarrow 8307.2 + 47.2x = 8162 + 50.5x$

Move variable part on one side and constant part on another side,

$ \Rightarrow 50.5x - 47.2x = 8307.2 - 8162$

Subtract the like terms,

$ \Rightarrow 3.3x = 145.2$

Divide both sides by 3.3,

$\therefore x = 44$

**Hence the missing frequency is 44.**

**Note:**In such types of problems, the class will not be taken only mid-point should be taken because the interval cannot be multiplied to the frequency. If we don’t remember the formula, we can multiply each midpoint with frequency and add all of them then divide it with the sum of frequency.

In the mean formula, while computing $\sum {fx} $, don’t take the sum of $f$ and $x$ separately and then multiply them. It will be difficult. Students should carefully make the frequency distribution table; there are high chances of making mistakes while copying and computing data.

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