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Chapter 16: Problem 16

Write the binomial formula for the probability of \(x\) successes in 100 Bernoulli trials with probability \(p=\frac{1}{5}\) of success. Compare the normal and the Poisson approximations if \(x=25\); if \(x=21\). (The exact binomial probability is \(0.0439\) for \(x=25\), and \(0.09+6\) for \(x=21\).)

Short Answer

Expert verified
Use binomial formula to calculate exact probabilities. Use normal approximation for z-scores and Poisson for direct computation. Compare these to exact values given.

Step by step solution

01

Write the Binomial Formula

The binomial formula is given by: \[ P(X = x) = \binom{n}{x} p^x (1-p)^{n-x} \] where: \( n = 100 \) (total trials), \( p = \frac{1}{5} \) (probability of success), and \( x \) is the number of successes.
02

Binomial Formula for Given Values

Substitute the given values into the binomial formula: \[ P(X = 25) = \binom{100}{25} \bigg(\frac{1}{5}\bigg)^{25} \bigg(\frac{4}{5}\bigg)^{75} \] \[ P(X = 21) = \binom{100}{21} \bigg(\frac{1}{5}\bigg)^{21} \bigg(\frac{4}{5}\bigg)^{79} \]
03

Normal Approximation

For the normal approximation, a binomial distribution can be approximated to a normal distribution if \( np \) and \( n(1-p) \) are both greater than 5. Here, mean \( \text{μ} = np \) and standard deviation \( \text{σ} = \sqrt{np(1-p)} \).For \( n = 100 \) and \( p = \frac{1}{5} \): Mean: \( \text{μ} = 100 \times \frac{1}{5} = 20 \) Standard Deviation: \( \text{σ} = \sqrt{100 \times \frac{1}{5} \times \frac{4}{5}} = \sqrt{16} = 4 \)Use these to convert \( x \) scores to z-scores:For \( x = 25 \): \[ z = \frac{25 - 20}{4} = 1.25 \]For \( x = 21 \): \[ z = \frac{21 - 20}{4} = 0.25 \]
04

Poisson Approximation

For the Poisson approximation, use \( λ = np \) as the mean of the Poisson distribution. Here, \( λ = 20 \).The Poisson probability formula is: \[ P(X = x) = \frac{λ^x e^{-λ}}{x!} \]Using this:For \( x = 25 \): \[ P(X = 25) \approx \frac{20^{25} e^{-20}}{25!} \]For \( x = 21 \): \[ P(X = 21) \approx \frac{20^{21} e^{-20}}{21!} \]
05

Compare Approximations to Exact

Compare the results with the given exact binomial probabilities: Exact for \( x = 25 \): \( 0.0439 \). Exact for \( x = 21 \): \( 0.096 \).Evaluate the normal and Poisson approximations to see which one is closer to these exact values.

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Key Concepts

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

Bernoulli trials
A Bernoulli trial is a random experiment where there are exactly two possible outcomes: success and failure. The probability of success is denoted by \( p \), while the probability of failure is \( 1 - p \).
Bernoulli trials are the building blocks for understanding binomial distributions. Imagine flipping a coin or checking if a light bulb works; these are typical examples of Bernoulli trials.
When multiple Bernoulli trials are performed, we get a binomial experiment. It helps calculate the probability of achieving a certain number of successes in a set number of trials.
binomial formula
The binomial formula calculates the probability of achieving exactly \( x \) successes in \( n \) Bernoulli trials, each with a success probability \( p \). The formula is given by:
\[ P(X = x) = \binom{n}{x} p^x (1-p)^{n-x} \]
where:
  • \( \binom{n}{x} \) represents the number of combinations of \( n \) items taken \( x \) at a time.
  • \( p^x \) is the probability of having exactly \( x \) successes.
  • \( (1-p)^{n-x} \) is the probability of having \( n - x \) failures.

Applying this to our problem where \( n = 100 \), \( p = \frac{1}{5} \), and \( x \) values of 25 and 21, you substitute these values into the formula to get:
\[ P(X = 25) = \binom{100}{25} \bigg(\frac{1}{5}\bigg)^{25} \bigg(\frac{4}{5}\bigg)^{75} \]
\[ P(X = 21) = \binom{100}{21} \bigg(\frac{1}{5}\bigg)^{21} \bigg(\frac{4}{5}\bigg)^{79} \]
normal approximation
The normal approximation is used when the number of trials \( n \) is large, and both \( np \) and \( n(1-p) \) are greater than 5. This approach approximates a binomial distribution with a normal distribution using the mean \( \mu \) and standard deviation \( \sigma \).
The mean and standard deviation are calculated as:
  • Mean: \( \mu = np \)
  • Standard Deviation: \( \sigma = \sqrt{np(1-p)} \)
For our problem:
\( \mu = 100 \times \frac{1}{5} = 20 \)
\( \sigma = \sqrt{100 \times \frac{1}{5} \times \frac{4}{5}} = 4 \)
We then convert each \( x \) value to a z-score:
\( z = \frac{x - \mu}{\sigma} \)
  • For \( x = 25 \), \( z = \frac{25 - 20}{4} = 1.25 \).
  • For \( x = 21 \), \( z = \frac{21 - 20}{4} = 0.25 \).
These z-scores can be used to find probabilities from the standard normal distribution table.
Poisson approximation
The Poisson approximation is used when the number of trials \( n \) is large, the probability of success \( p \) is small, and the product \( np \) is moderate. The mean \( \lambda \) of the Poisson distribution is set to \( np \).
The Poisson probability formula is:
\[ P(X = x) = \frac{\lambda^x e^{-\lambda}}{x!} \] where \( \lambda = np \).
For our problem: \( \lambda = 100 \times \frac{1}{5} = 20 \).
We then use this formula for the given \( x \) values:
  • For \( x = 25 \): \[ P(X = 25) \approx \frac{20^{25} e^{-20}}{25!} \]
  • For \( x = 21 \): \[ P(X = 21) \approx \frac{20^{21} e^{-20}}{21!} \]
probability calculations
Probability calculations involve using appropriate formulas and methods to find the likelihood of various outcomes.
In our exercise, we used the binomial formula, normal approximation, and Poisson approximation to compute probabilities for specific scenarios. Each method has its own appropriate use case and provides different levels of accuracy.
Comparing exact calculations with the approximations shows how well these methods perform. For instance:
  • Exact for \( x = 25 \): 0.0439
  • Exact for \( x = 21 \): 0.096
Comparing these with the results of normal and Poisson approximations can help understand the effectiveness and limitations of each approximation method.

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Most popular questions from this chapter

Define the sample variance by \(s^{2}=(1 / n) \sum_{i=1}^{n}\left(x_{i}-\bar{x}\right)^{2} .\) Show that the expected value of \(s^{2}\) is \([(n-1) / n] \sigma^{2} .\) Hints: Write $$ \begin{aligned} \left(x_{i}-\bar{x}\right)^{2} &=\left[\left(x_{i}-\mu\right)-(\bar{x}-\mu)\right]^{2} \\ &=\left(x_{i}-\mu\right)^{2}-2\left(x_{i}-\mu\right)(\bar{x}-\mu)+(\bar{x}-\mu)^{2} \end{aligned} $$ Find the average value of the first term from the definition of \(\sigma^{2}\) and the average value of the third term from Problem 2, To find the average value of the middle term write $$ (\bar{x}-\mu)=\left(\frac{x_{1}+x_{2}+\cdots+x_{n}}{n}-\mu\right)=\frac{1}{n}\left[\left(x_{1}-\mu\right)+\left(x_{2}-\mu\right)+\cdots+\left(x_{n}-\mu\right)\right] $$ show by Problem \(7.12\) that $$ E\left[\left(x_{i}-\mu\right)\left(x_{j}-\mu\right)\right]=E\left(x_{i}-\mu\right) E\left(x_{j}-\mu\right)=0 \quad \text { for } \quad t \neq j $$ and evaluate \(E\left[\left(x_{i}-\mu\right)^{2}\right]\) (same as the first term). Collect terms to find $$ E\left(s^{2}\right)=\frac{n-1}{n} \sigma^{2} $$

What is the probability that you and a friend have different birthdays? (For simplicity, let a year have 365 days.) What is the probability that three people have three different birthdays? Show that the probability that \(n\) people have \(n\) different birthdays is $$ p=\left(1-\frac{1}{365}\right)\left(1-\frac{2}{365}\right)\left(1-\frac{3}{365}\right) \cdots\left(1-\frac{n-1}{365}\right) $$ Estimate this for \(n \notin 365\) by calculating in \(p\) [recall that \(\ln (1+x)\) is approximately \(x\) for \(x<1]\). Find the smallest (integral) \(n\) for which \(p<\frac{1}{2}\). Hence show that for a group of 23 people or more, the probability is greater than \(\frac{1}{2}\) that two of them have the same birthdit? (Try it with a group of friends or a list of people such as the presidents of the United States.)

(a) One box contains one die and another box contains two dice. You select a box at random and take out and toss whatever is in it (that is, toss both dice if you have picked box 2 ). Let \(x=\) number of 3 's showing. Set up the sample space and associated probabilities for \(x\). (b) What is the probability of at least one 3 ? (c) If at least one 3 turns up, what is the probability that you picked the first box? (d) Find \(\bar{x}\) and \(\sigma\).

Recall that two events \(A\) and \(B\) are called independent if \(p(A B)=p(A) \cdot p(B) .\) Similarly, two random variables \(x\) and \(y\) with probability functions \(f(x)\) and \(g(y)\) are called independent if the probability of \(x=x_{i}\) and \(y=y_{j}\) is \(f\left(x_{i}\right) \cdot g\left(y_{j}\right)\) for every pair of values of \(x\) and \(y\), that is, if the joint probability function for \(x, y\) is \(f(x) g(y)\). Show that if \(x\) anct \(y\) are independent, then the expectation or average of \(x y\) is \(E(x y)=E(x) \cdot E(y)=\mu_{x} \mu_{y}\).

Suppose you receive an average of 4 phone calls per day. What is the probability that on a given day you receive no phone calls? Just one call? Exactly 4 calls?
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