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Chapter 26: Problem 32

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

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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.

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

For a non-negative integer random variable \(X\), in addition to the probability generating function \(\Phi_{X}(t)\) defined in equation (26.71) it is possible to define the probability generating function $$ \Psi_{X}(t)=\sum_{n=0}^{\infty} g_{n} t^{n} $$ where \(g_{n}\) is the probability that \(X>n\). (a) Prove that \(\Phi_{X}\) and \(\Psi_{X}\) are related by $$ \Psi_{X}(t)=\frac{1-\Phi_{X}(t)}{1-t} $$ (b) Show that \(E[X]\) is given by \(\Psi_{X}(1)\) and that the variance of \(X\) can be expressed as \(2 \Psi_{X}^{\prime}(1)+\Psi_{X}(1)-\left[\Psi_{X}(1)\right]^{2}\) (c) For a particular random variable \(X\), the probability that \(X>n\) is equal to \(\alpha^{n+1}\) with \(0<\alpha<1\). Use the results in \((\mathrm{b})\) to show that \(V[X]=\alpha(1-\alpha)^{-2}\).

A continuous random variable \(X\) is uniformly distributed over the interval \([-c, c]\). A sample of \(2 n+1\) values of \(X\) is selected at random and the random variable \(Z\) is defined as the median of that sample. Show that \(Z\) is distributed over \([-c, c]\) with probability density function, $$ f_{n}(z)=\frac{(2 n+1) !}{(n !)^{2}(2 c)^{2 n+1}}\left(c^{2}-z^{2}\right)^{n} $$ Find the variance of \(Z\).

Villages \(A, B, C\) and \(D\) are connected by overhead telephone lines joining \(A B\), \(A C, B C, B D\) and \(C D .\) As a result of severe gales, there is a probability \(p\) (the same for each link) that any particular link is broken. (a) Show that the probability that a call can be made from \(A\) to \(B\) is $$ 1-2 p^{2}+p^{3} $$ (b) Show that the probability that a call can be made from \(D\) to \(A\) is $$ 1-2 p^{2}-2 p^{3}+5 p^{4}-2 p^{5} $$

An electronics assembly firm buys its microchips from three different suppliers; half of them are bought from firm \(X\), whilst firms \(Y\) and \(Z\) supply \(30 \%\) and \(20 \%\) respectively. The suppliers use different quality- control procedures and the percentages of defective chips are \(2 \%, 4 \%\) and \(4 \%\) for \(X, Y\) and \(Z\) respectively. The probabilities that a defective chip will fail two or more assembly-line tests are \(40 \%, 60 \%\) and \(80 \%\) respectively, whilst all defective chips have a \(10 \%\) chance of escaping detection. An assembler finds a chip that fails only one test. What is the probability that it came from supplier \(X\) ?

A discrete random variable \(X\) takes integer values \(n=0,1, \ldots, N\) with probabilities \(p_{n} .\) A second random variable \(Y\) is defined as \(Y=(X-\mu)^{2}\), where \(\mu\) is the expectation value of \(X\). Prove that the covariance of \(X\) and \(Y\) is given by $$ \operatorname{Cov}[X, Y]=\sum_{n=0}^{N} n^{3} p_{n}-3 \mu \sum_{n=0}^{N} n^{2} p_{n}+2 \mu^{3} $$ Now suppose that \(X\) takes all its possible values with equal probability and hence demonstrate that two random variables can be uncorrelated even though one is defined in terms of the other.
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