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Calculate the mean weight of livestock using the assumed mean method from the given frequency distribution.

Mathematics

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Answers

3 parts
a

126\text{126}

b

21\text{21}

c

70\text{70}

Step-by-step solution

7 steps

Here are the solutions to the problems.

:
To find the mean weight using the assumed mean method, we first determine the midpoint (xx) for each weight class, then calculate the deviation (dd) from the assumed mean (A=36A=36), and finally apply the formula xˉ=A+∑fd∑f\bar{x} = A + \frac{\sum fd}{\sum f}.

Step 1: Calculate the midpoints (xx), deviations (d=x−36d = x - 36), and fdfd values for each class.

Weight RangeMidpoint (xx)No. of Livestock (ff)d=x−36d = x - 36fdfd
20 - 24225-14-70
25 - 292715-9-135
30 - 343212-4-48
35 - 3937919
40 - 44426636
45 - 494731133

Step 2: Sum the frequencies (∑f\sum f) and the fdfd values (∑fd\sum fd).

∑f=5+15+12+9+6+3=50\sum f = 5 + 15 + 12 + 9 + 6 + 3 = 50 ∑fd=−70−135−48+9+36+33=−175\sum fd = -70 - 135 - 48 + 9 + 36 + 33 = -175

Step 3: Apply the assumed mean formula.

xˉ=A+∑fd∑f\bar{x} = A + \frac{\sum fd}{\sum f} xˉ=36+−17550\bar{x} = 36 + \frac{-175}{50} xˉ=36−3.5\bar{x} = 36 - 3.5 xˉ=32.5\bar{x} = 32.5

The mean weight of the livestock is 32.5\boxed{\text{32.5}}.

:
We need to form committees of 4 persons from a group of 9 persons. This is a combination problem, as the order of selection does not matter. The formula for combinations is C(n,k)=(nk)=n!k!(n−k)!C(n, k) = \binom{n}{k} = \frac{n!}{k!(n-k)!}.

Part a· 1 step

there are no restrictions?

  1. Calculate the number of ways to choose 4 persons from 9 without any restrictions.

    C(9,4)=9!4!(9−4)!=9!4!5!C(9, 4) = \frac{9!}{4!(9-4)!} = \frac{9!}{4!5!} C(9,4)=9×8×7×64×3×2×1C(9, 4) = \frac{9 \times 8 \times 7 \times 6}{4 \times 3 \times 2 \times 1} C(9,4)=9×2×7=126C(9, 4) = 9 \times 2 \times 7 = 126

    There are 126\boxed{\text{126}} possible committees.

Part b

21\text{21}

both Kwame and Rita must be on the committee?

  1. If Kwame and Rita must be on the committee, 2 spots are already filled.

    We need to choose 4−2=24 - 2 = 2 more persons.

  2. The remaining pool of people to choose from is 9−2=79 - 2 = 7 persons (excluding Kwame and Rita).

  3. Calculate the number of ways to choose 2 persons from these 7.

    C(7,2)=7!2!(7−2)!=7!2!5!C(7, 2) = \frac{7!}{2!(7-2)!} = \frac{7!}{2!5!} C(7,2)=7×62×1C(7, 2) = \frac{7 \times 6}{2 \times 1} C(7,2)=21C(7, 2) = 21

    There are 21\boxed{\text{21}} possible committees.

Part c

70\text{70}

either Kwame or Rita, but not both, must be on the committee?

This means we consider two mutually exclusive cases:
Case 1: Kwame is on the committee, and Rita is not.
Case 2: Rita is on the committee, and Kwame is not.

  1. Calculate Case 1 (Kwame is on, Rita is not).

    If Kwame is on, 1 spot is filled. We need to choose 4−1=34 - 1 = 3 more persons.
    If Rita is not on, she is excluded from the selection pool.
    The remaining pool of people to choose from is 9−1(Kwame)−1(Rita)=79 - 1 (\text{Kwame}) - 1 (\text{Rita}) = 7 persons.
    Number of ways for Case 1: Choose 3 persons from these 7.

    C(7,3)=7!3!(7−3)!=7!3!4!C(7, 3) = \frac{7!}{3!(7-3)!} = \frac{7!}{3!4!} C(7,3)=7×6×53×2×1C(7, 3) = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} C(7,3)=35C(7, 3) = 35

  2. Calculate Case 2 (Rita is on, Kwame is not).

    This is symmetrical to Case 1.
    If Rita is on, 1 spot is filled. We need to choose 4−1=34 - 1 = 3 more persons.
    If Kwame is not on, he is excluded from the selection pool.
    The remaining pool of people to choose from is 9−1(Rita)−1(Kwame)=79 - 1 (\text{Rita}) - 1 (\text{Kwame}) = 7 persons.
    Number of ways for Case 2: Choose 3 persons from these 7.

    C(7,3)=35C(7, 3) = 35

  3. Add the possibilities from both cases.

    Total ways = 35+35=7035 + 35 = 70.
    There are 70\boxed{\text{70}} possible committees.

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