For the given values of \(n\) and \(p\) computer plot graphs of the binomial density function for the probability of \(x\) successes in \(n\) Bernoulli trials with probability \(p\) of success. $$n=50, p=4 / 5$$

Short Answer

Expert verified
Plot the graph using a tool like Python's 'matplotlib' with the binom.pmf function for n=50 and p=0.8.

Step by step solution

01

Understand the Binomial Distribution

The Binomial Distribution represents the probability of having exactly x successes in n independent Bernoulli trials with a success probability of p. The probability mass function (pmf) can be given by: \[ P(X = x) = \binom{n}{x} p^x (1-p)^{n-x} \] where \( \binom{n}{x} \) is the binomial coefficient.
02

Define the Parameters

For our problem, we are given \( n = 50 \) (the number of trials), and \( p = \frac{4}{5} \) (the probability of success). Therefore, \( p = 0.8 \) and the probability of failure is \( 1 - p = 0.2 \).
03

Calculate the Probabilities

To plot the binomial distribution, probabilities for each possible value of \( x \) (where \( x \) ranges from 0 to 50) must be calculated. Using the binomial pmf formula, the probability \( P(X = x) \) for each \( x \) can be calculated.
04

Use Software or a Graphing Tool

To plot the graph, software like Python, R, or even a graphing calculator can be utilized. In Python, for example, the 'scipy.stats' library can be used to compute these probabilities and plot the graph using 'matplotlib'.
05

Implement the Code

Here is a sample code snippet in Python:```pythonimport numpy as npimport matplotlib.pyplot as pltfrom scipy.stats import binomn = 50p = 0.8x = np.arange(0, n+1)probabilities = binom.pmf(x, n, p)plt.bar(x, probabilities)plt.xlabel('Number of Successes (x)')plt.ylabel('Probability P(X = x)')plt.title('Binomial Distribution (n=50, p=0.8)')plt.show()```This code will display a bar plot of the binomial distribution.

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Key Concepts

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

Bernoulli Trials
Bernoulli trials are a sequence of independent experiments where each experiment has two possible outcomes: success or failure. Each trial has the same success probability, denoted by \( p \), and, consequently, the failure probability is \( 1 - p \). One of the simplest examples of a Bernoulli trial is flipping a coin, where getting heads can be considered a success and tails a failure.
Bernoulli trials are the building blocks for the binomial distribution.
In the context of the provided exercise, we have 50 Bernoulli trials, each with a probability of success of 0.8.
Probability Mass Function
The probability mass function (pmf) of a binomial distribution gives the probability of observing exactly \( x \) successes in \( n \) Bernoulli trials. The pmf for the binomial distribution is given by the formula:
\[ P(X = x) = \binom{n}{x} p^x (1-p)^{n-x} \]
This formula calculates the likelihood of getting a specific number of successes in a fixed number of trials.
For example, in our problem where \( n = 50 \) and \( p = 0.8 \), we can use this formula to find the probabilities for each possible value of \( x \) (from 0 to 50).
  • \( \binom{n}{x} \) is the binomial coefficient
  • \( p^x \) is the probability of success raised to the power of \( x \)
  • \( (1-p)^{n-x} \) is the probability of failure raised to the power of \( (n - x) \)
This function helps us build and understand the shape of the binomial distribution.
Binomial Coefficient
The binomial coefficient \( \binom{n}{x} \) represents the number of ways to choose \( x \) successes out of \( n \) trials. It is also known as a 'combination' and is mathematically given by:
\[ \binom{n}{x} = \frac{n!}{x!(n-x)!} \]
where \( ! \) denotes factorial (the product of all positive integers up to that number).
In our problem:
  • \( \binom{50}{0} = 1 \)
  • \( \binom{50}{1} = 50 \)
  • \( \binom{50}{25} = 1,264,322,234,167,310 \)
The binomial coefficient is significant because it tells us how many different ways we can achieve a specific number of successes in our set of trials.
It essentially weights the probability of each possible outcome by the number of ways that outcome can occur.
Scipy Library
The scipy library in Python is a powerful tool for scientific and technical computing. It includes modules for optimization, integration, interpolation, eigenvalue problems, algebraic equations, and many other tasks.
One of the modules, 'scipy.stats', offers a huge range of statistical distributions and functions, including the binomial distribution.
In our exercise, the `scipy.stats.binom` class is used to compute the probability mass function for the binomial distribution. Here’s how it works:
  • Import necessary modules: `from scipy.stats import binom`
  • Define the parameters for the binomial distribution: `n = 50`, `p = 0.8`
  • Calculate probabilities for each possible number of successes: `probabilities = binom.pmf(x, n, p)`
Finally, `matplotlib` is used to create a bar plot of these probabilities, giving a visual representation of the binomial distribution.
The scipy library makes it straightforward to perform sophisticated statistical analyses and create visualizations with relatively simple code.

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Most popular questions from this chapter

(a) A candy vending machine is out of order. The probability that you get a candy bar (with or without return of your money) is \(\frac{1}{2},\) the probability that you get your money back (with or without candy) is \(\frac{1}{3}\), and the probability that you get both the candy and your money back is \(\frac{1}{12}\). What is the probability that you get nothing at all? Suggestion: Sketch a geometric diagram similar to Figure 3.1, indicate regions representing the various possibilities and their probabilities; then set up a four-point sample space and the associated probabilities of the points. (b) Suppose you try again to get a candy bar as in part (a). Set up the 16 -point sample space corresponding to the possible results of your two attempts to buy a candy bar, and find the probability that you get two candy bars (and no money back); that you get no candy and lose your money both times; that you just get your money back both times.

(a) A weighted coin has probability \(\frac{2}{3}\) of coming up heads and probability \(\frac{1}{3}\) of coming up tails. The coin is tossed twice. Let \(x=\) number of heads. Set up the sample space for \(x\) and the associated probabilities. (b) Find \(\bar{x}\) and \(\sigma\) (c) If in (a) you know that there was at least one tail, what is the probability that both were tails?

Given that a particle is inside a sphere of radius \(1,\) and that it has equal probabilities of being found in any two volume elements of the same size, find the cumulative distribution function \(F(r)\) for the spherical coordinate \(r,\) and from it find the density function \(f(r) .\) Hint: \(F(r)\) is the probability that the particle is inside a sphere of radius \(r .\) Find \(\overline{r}\) and \(\sigma\).

(a) Suppose you have two quarters and a dime in your left pocket and two dimes and three quarters in your right pocket. You select a pocket at random and from it a coin at random. What is the probability that it is a dime? (b) Let \(x\) be the amount of money you select. Find \(E(x)\) (c) Suppose you selected a dime in (a). What is the probability that it came from your right pocket? (d) Suppose you do not replace the dime, but select another coin which is also a dime. What is the probability that this second coin came from your right pocket?

Show that the expectation of the sum of two random variables defined over the same sample space is the sum of the expectations. Hint: Let \(p_{1}, p_{2}, \cdots, p_{n}\) be the probabilities associated with the \(n\) sample points; let \(x_{1}, x_{2}, \cdots, x_{n},\) and \(y_{1}, y_{2}\) \(\cdots, y_{n},\) be the values of the random variables \(x\) and \(y\) for the \(n\) sample points. Write out \(E(x), E(y),\) and \(E(x+y)\).

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