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In a certain parliament the government consists of 75 New Socialites and the opposition consists of 25 Preservatives. Preservatives never change their mind, always voting against government policy without a second thought; New Socialites vote randomly, but with probability \(p\) that they will vote for their party leader's policies. Following a decision by the New Socialites' leader to drop certain manifesto commitments, \(N\) of his party decide to vote consistently with the opposition. The leader's advisors reluctantly admit that an election must be called if \(N\) is such that, at any vote on government policy, the chance of a simple majority in favour would be less than \(80 \%\). Given that \(p=0.8\), estimate the lowest value of \(N\) that wonld nrecinitate an election

Short Answer

Expert verified
The lowest value of \(N\) that would precipitate an election is 30.

Step by step solution

01

- Understand the Problem

We need to find the value of \(N\) such that the probability of the government getting a majority is less than 80%. The government starts with 75 New Socialites, but \(N\) of them switch sides, reducing the number of random voters.
02

- Define the Number of Random Voters

Initially, there are 75 New Socialites who vote randomly. After \(N\) New Socialites switch sides, the number of random voters left will be \(75 - N\).
03

- Majority Votes Required

A simple majority requires more than half the votes. Since there are 100 total votes (75 New Socialites + 25 Preservatives), the government needs more than 50 votes to have a majority.
04

- Calculate the Probability of Majority

Let's denote the number of votes the government gets from the remaining \(75 - N\) random voters as \(X\). \(X\) follows a binomial distribution with parameters \(75 - N\) and \(p = 0.8\). We need to find \(N\) such that the probability \(P(X > 50 - N) < 0.8\).
05

- Use Normal Approximation

For large \(n\), a binomial distribution can be approximated by a normal distribution with mean \(\mu = np\) and variance \(\sigma^2 = np(1-p)\). Therefore, \(X\) can be approximated by a normal distribution with mean \(\mu = (75 - N) \cdot 0.8\) and variance \(\sigma^2 = (75 - N) \cdot 0.8 \cdot 0.2\).
06

- Set Up the Inequality for Probability

We need \( P(X > 50 - N) < 0.8 \). Using the normal distribution approximation, this inequality translates to finding \(N\) such that: \[ P\left(\frac{X - (75 - N) \cdot 0.8}{\sqrt{(75 - N) \cdot 0.16}} > \frac{50 - N - (75 - N) \cdot 0.8}{\sqrt{(75 - N) \cdot 0.16}} \right) < 0.8 \]
07

- Simplify the Inequality

We standardize the normal variable: \[ P\left(Z > \frac{50 - N - 0.8(75 - N)}{\sqrt{0.16 (75 - N)}}\right) < 0.8 \] Using the standard normal distribution table, the requirement \(P(Z > z) < 0.2\) translates to \(z = -0.8416\).
08

- Solve the Equation for N

Set up the equation: \[ \frac{50 - N - 60 + 0.8N}{\sqrt{12}} = -0.8416 \] Simplifying this, we get: \[ -10 + 0.2N = -0.8416 \cdot \sqrt{12} \] \[ 0.2N = 5.81792 \] \[ N \approx 29.09 \] Since \(N\) must be an integer, the lowest value of \(N\) that satisfies the condition is 30.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Normal Approximation
In the context of binomial distribution, the normal approximation is a method we use when dealing with large sample sizes. Here’s why it's useful: Suppose we have a large number of trials, say hundreds. Calculating the exact probabilities by hand for each outcome using the binomial formula can be very cumbersome and time-consuming. However, if the number of trials is large and the probability of success is neither too close to 0 nor 1, the binomial distribution can be approximated by a normal distribution. This makes the calculations easier and quicker.

The key idea is to match the mean and standard deviation of the binomial distribution to those of a normal distribution. For a binomial distribution with parameters n (number of trials) and p (probability of success), the mean (expected value) \(\backslash mu\) is calculated as:

\( \backslash mu = np \)

and the standard deviation \(\backslash sigma\) is:

\( \backslash sigma = \backslash sqrt{np(1 - p)} \).

Once these are known, we can use the normal distribution to approximate probabilities. This significantly simplifies the problem-solving process, as was done in the original exercise to find the value of N.
Probability Theory
Probability theory involves the study of random events and is essential for understanding distributions like binomial and normal. In probability theory, we define an event’s likelihood using values between 0 and 1. A probability of 0 means an event will not happen, while a probability of 1 means it will certainly happen.

To solve problems involving probability, one typically follows these steps:
  • Define the random variable
  • Identify the distribution of the variable
  • Calculate the relevant probabilities

In the exercise, the random variable X represented the number of New Socialites voting for their party policies and followed a binomial distribution. By leveraging probability theory, we identified the likelihood of different voting outcomes and used these probabilities to infer when an election would become necessary.
Statistical Analysis
Statistical analysis includes methods and techniques for collecting, analyzing, interpreting, and presenting data. It encompasses both descriptive statistics (like mean and standard deviation) and inferential statistics (like confidence intervals and hypothesis tests).

In our exercise, we performed statistical analysis by:
  • Defining our variables and parameters
  • Approximating the distribution of our variable using the normal approximation
  • Solving for the threshold value of N by interpreting the results in the context of real-world decisions

This approach allowed us to use effective statistical techniques to make informed judgments about the government's stability based on probabilistic outcomes.
Decision Making Under Uncertainty
Decision making under uncertainty involves making choices without knowing the exact outcomes with certainty, often relying on probability and statistical analysis. This is a critical capability in fields like economics, finance, and public policy.

In our exercise, the need to determine whether an election should be called is an example of decision making under uncertainty. The New Socialites' leader can't predict exactly how many members will switch sides, but can estimate this likelihood using probability.

By calculating the probability that the government will secure a majority vote, and using normal approximation to make these calculations manageable, we arrive at an informed threshold (N = 30). This informed decision minimizes uncertainty and establishes a clear criterion for when an election needs to be called.

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Most popular questions from this chapter

A shopper buys 36 items at random in a supermarket where, because of the sales tax imposed, the final digit (the number of pence) in the price is uniformly and randomly distributed from 0 to \(9 .\) Instead of adding up the bill exactly she rounds each item to the nearest 10 pence, rounding up or down with equal probability if the price ends in a ' 5 '. Should she suspect a mistake if the cashier asks her for 23 pence more than she estimated?

A continuous random variable \(X\) has a probability density function \(f(x)\); the corresponding cumulative probability function is \(F(x) .\) Show that the random variable \(Y=F(X)\) is uniformly distributed between 0 and 1 .

Show that, as the number of trials \(n\) becomes large but \(n p_{i}=\lambda_{i}, i=1,2, \ldots, k-1\) remains finite, the multinomial probability distribution (26.146), $$ M_{n}\left(x_{1}, x_{2}, \ldots, x_{k}\right)=\frac{n !}{x_{1} ! x_{2} ! \cdots x_{k} !} p_{1}^{x_{1}} p_{2}^{x_{2}} \cdots p_{k}^{x_{k}} $$ can be approximated by a multiple Poisson distribution (with \(k-1\) factors) $$ M_{n}^{\prime}\left(x_{1}, x_{2}, \ldots, x_{k-1}\right)=\prod_{i=1}^{k-1} \frac{e^{-\lambda_{i}} \lambda_{i}^{x_{i}}}{x_{i} !} $$ (Write \(\sum_{i}^{k-1} p_{i}=\delta\) and express all terms involving subscript \(k\) in terms of \(n\) and \(\delta\), either exactly or approximately. You will need to use \(n ! \approx n^{f}[(n-\epsilon) !]\) and \((1-a / n)^{n} \approx e^{-a}\) for large \(\left.n_{1}\right)\) (a) Verify that the terms of \(M_{n}^{\prime}\) when summed over all values of \(x_{1}, x_{2}, \ldots, x_{k-1}\) add up to unity. (b) If \(k=7\) and \(\lambda_{i}=9\) for all \(i=1,2, \ldots, 6\), estimate, using the appropriate Gaussian approximation, the chance that at least three of \(x_{1}, x_{2}, \ldots, x_{6}\) will be 15 or greater.

By shading Venn diagrams, determine which of the following are valid relationships between events. For those that are, prove them using de Morgan's laws. (a) \(\overline{(\bar{X} \cup Y)}=X \cap \bar{Y}\). (b) \(\bar{X} \cup \bar{Y}=\overline{(X \cup Y)}\) (c) \((X \cup Y) \cap Z=(X \cup Z) \cap Y\). (d) \(X \cup \underline{(Y \cap Z)}=(X \cup Y) \cap Z\). (e) \(X \cup \overline{(Y \cap Z)}=(X \cup \bar{Y}) \cup \bar{Z}\)

Kittens from different litters do not get on with each other and fighting breaks out whenever two kittens from different litters are present together. A cage initially contains \(x\) kittens from one litter and \(y\) from another. To quell the fighting, kittens are removed at random, one at a time, until peace is restored. Show, by induction, that the expected number of kittens finally remaining is $$ N(x, y)=\frac{x}{y+1}+\frac{y}{x+1} $$

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