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puteri [66]
2 years ago
8

Exercise 5.6.6: Selecting a committee of senators. About A country has two political parties, the Demonstrators and the Repudiat

ors. Suppose that the national senate consists of 100 members, 44 of which are Demonstrators and 56 of which are Rupudiators. (a) How many ways are there to select a committee of 10 senate members with the same number of Demonstrators and Repudiators? Suppose that each party must select a speaker and a vice speaker. How many ways are there for the two speakers and two vice speakers to be selected? Feedback?
Mathematics
1 answer:
krok68 [10]2 years ago
3 0

Answer:

There are 4,148,350,734,528 ways

Step-by-step explanation:

We have

  • 44 senators which are Demonstrators.
  • 56 senators which are Repudiators.

(a) How many ways are there to select a committee of 10 senate members with the same number of Demonstrators and Repudiators?

We want to choose 5 Demonstrators and 5 Repudiators. The number of ways to do this is {44} \choose {5} and 56 \choose 5 respectively. Therefore, the number of ways to select the committee is given by:

{{44}\choose {5}} \times {{56}\choose{5}}=\frac{44!}{39!5!}\times\frac{56!}{51!5!}=\frac{44!56!}{51!39!5!5!}=\frac{44\times43\times42\times41\times40\times56\times55\times54\times53\times52}{5!5!}=\\\\=\frac{44\times43\times42\times41\times8\times56\times11\times54\times53\times52}{4!4!}= \frac{11\times43\times42\times41\times2\times56\times11\times54\times53\times52}{3!3!}=\\\\\frac{11\times43\times14\times41\times2\times56\times11\times18\times53\times52}{2!2!}=

11\times43\times14\times41\times28\times11\times18\times53\times52=4,148,350,734,528

(b) Suppose that each party must select a speaker and a vice speaker. How many ways are there for the two speakers and two vice speakers to be selected?

  • <u>If the speaker and vice speaker are chosen between all senators:</u> In this case, the answer will be

44\times43\times56\times55=5,827,360.

This is because there are (in the case of Demonstrators) 44 possibilities to choose an speaker and after choosing one, there would be 43 possibilities to choose a vice speaker. The same situation happens in the case of Repudiators.

  • <u>If the speaker and vice speaker are chosen between the committee:</u> In this case, the answer will be

5\times4\times5\times4=400.

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One way to measure whether the trees in the Wade Tract are uniformly distributed is to examine the average location in the north
tatyana61 [14]

Answer:

Null hypothesis be H₀ : μ = 100

Alternative hypothesis Hₐ : μ < 100

There is sufficient statistical evidence to suggest that the average location of Wade Tract is 100

Step-by-step explanation:

Here we have;

Let our null hypothesis be H₀ : μ = 100 Average location of trees in the Wade Tract is 100

Our alternative hypothesis becomes Hₐ : μ < 100 at 95% confidence level

Proposed average location in Wade Tract, μ = 100

Sample mean, \bar x = 99.74

Standard deviation, s = 58

Sample size, n = 584

The t test formula is therefore;

t=\frac{\bar{x}-\mu }{\frac{s }{\sqrt{n}}}

Therefore, with df = 584 -1 = 583, and α = (1 - 0.95)/2 = 0.025

We have t_{\alpha /2} = -1.65

Plugging in the values into the t test formula, we have t = -0.108338, from which the p-value is given as p = 0.4569 which is much more than the value of α, therefore, we accept the null hypothesis as follows;

There is sufficient statistical evidence to suggest that the average location of Wade Tract = 100.

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Step-by-step explanation:

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q = 1 - p = 1 - 0.7 = 0.3

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Standard deviation, u = √npq = √50×0.7×0.3 = 3.24

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The formula for normal distribution is expressed as

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The z value will be

z = (25- 35)/3.24 = - 10/3.24 = -3.09

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