Calculate the mean weight of livestock using the assumed mean method and solve a combination problem for committee selection.

Here's the solution for question 5:

Step 1: Calculate the midpoint ($x$) for each weight class.

• 20-24: $\frac{20+24}{2} = 22$
• 25-29: $\frac{25+29}{2} = 27$
• 30-34: $\frac{30+34}{2} = 32$
• 35-39: $\frac{35+39}{2} = 37$
• 40-44: $\frac{40+44}{2} = 42$
• 45-49: $\frac{45+49}{2} = 47$

Step 2: Calculate the deviation ($d$) for each midpoint from the assumed mean ($A=36$).
$d = x - A$

• $22 - 36 = -14$
• $27 - 36 = -9$
• $32 - 36 = -4$
• $37 - 36 = 1$
• $42 - 36 = 6$
• $47 - 36 = 11$

Step 3: Calculate the product of frequency ($f$) and deviation ($d$) for each class ($fd$).

• $5 \times (-14) = -70$
• $15 \times (-9) = -135$
• $12 \times (-4) = -48$
• $9 \times (1) = 9$
• $6 \times (6) = 36$
• $3 \times (11) = 33$

Step 4: Calculate the sum of frequencies ($\sum f$) and the sum of $fd$ ($\sum fd$).
$\sum f = 5 + 15 + 12 + 9 + 6 + 3 = 50$
$\sum fd = -70 - 135 - 48 + 9 + 36 + 33 = -175$

Step 5: Use the formula for the mean weight using the assumed mean method.

$$\text{Mean} (\bar{x}) = A + \frac{\sum fd}{\sum f}$$ $$\bar{x} = 36 + \frac{-175}{50}$$ $$\bar{x} = 36 - 3.5$$ $$\bar{x} = 32.5$$

The mean weight of the Livestock is $\boxed{32.5}$.

Here's the solution for question 6:

This is a combination problem, as the order of persons in a committee does not matter. The formula for combinations is $C(n, k) = \binom{n}{k} = \frac{n!}{k!(n-k)!}$.
Total number of persons ($n$) = 9.
Committee size ($k$) = 4.