Please give me the answer in English plz

The formula for the mean ($\bar{x}$) using an assumed mean ($A$) is:

Substitute the values from the table:

The mean length is $\boxed{84.5 \text{ cm}}$.

Total frequency $N = \sum f = 200$.
The median position is $\frac{N}{2} = \frac{200}{2} = 100^{th}$ value.
From the cumulative frequency column, the $100^{th}$ value falls in the class 80-89 (cf = 127).
So, the median class is 80-89.

Lower boundary of median class ($L$) = 79.5
Frequency of median class ($f_m$) = 52
Cumulative frequency of the class before the median class ($cf_b$) = 75
Class width ($c$) = 10
The formula for the median is:

Substitute the values:

The median is $\boxed{84.31 \text{ cm}}$.

The formula for standard deviation ($\sigma$) using an assumed mean is:

Substitute the values from the table:

The standard deviation is $\boxed{14.80 \text{ cm}}$.

From the total frequency:

From the mean formula $\bar{x} = \frac{\sum fx}{\sum f}$:

From Equation 1, express $a$ in terms of $b$: $a = 26 - b$.
Substitute this into Equation 2:

Now substitute $b=14$ back into $a = 26 - b$:

The values are $\boxed{a = 12}$ and $\boxed{b = 14}$.

The frequencies are 7, $a=12$, 22, 36, $b=14$, 9.
The highest frequency is 36, which corresponds to the class 42-44.
So, the modal class is 42-44.

Lower boundary of modal class ($L_m$) = 41.5
Frequency of modal class ($f_m$) = 36
Frequency of the class before the modal class ($f_1$) = 22
Frequency of the class after the modal class ($f_2$) = 14
Class width ($c$) = 3
The formula for the mode is:

Substitute the values:

The mode is $\boxed{42.67 \text{ cm}}$.