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Graph the histogram of the given binomial distribution. Check your answer using technology. $$ n=5, p=\frac{1}{4}, q=\frac{3}{4} $$

Short Answer

Expert verified
The histogram for the given binomial distribution with n=5 and p=1/4 has the following probabilities for each possible outcome: - P(X=0) ≈ 0.2373 - P(X=1) ≈ 0.3955 - P(X=2) ≈ 0.2637 - P(X=3) ≈ 0.0879 - P(X=4) ≈ 0.0146 - P(X=5) ≈ 0.0010 The histogram's x-axis represents the number of successes, and the y-axis represents the probability for each outcome. Check the answer using technology such as Excel, R, Python, Desmos, or Wolfram Alpha.

Step by step solution

01

Identify the binomial distribution parameters

To graph the histogram, we need to identify two parameters of the binomial distribution: 1. The number of trials (n) = 5 2. The probability of success (p) = 1/4 And we can calculate the probability of failure as: 3. Probability of failure (q) = 1 - p = 3/4
02

Calculate the probability for each outcome

We need to find the probability of each possible outcome (from 0 to 5 successes) in the binomial distribution: \(P(X=k) = \binom{n}{k}p^k q^{n-k}\), where k represents the number of successes and n is the number of trials. 1. P(X=0) = \(C(5,0) \times (\frac{1}{4})^0 \times (\frac{3}{4})^5\) 2. P(X=1) = \(C(5,1) \times (\frac{1}{4})^1 \times (\frac{3}{4})^4\) 3. P(X=2) = \(C(5,2) \times (\frac{1}{4})^2 \times (\frac{3}{4})^3\) 4. P(X=3) = \(C(5,3) \times (\frac{1}{4})^3 \times (\frac{3}{4})^2\) 5. P(X=4) = \(C(5,4) \times (\frac{1}{4})^4 \times (\frac{3}{4})^1\) 6. P(X=5) = \(C(5,5) \times (\frac{1}{4})^5 \times (\frac{3}{4})^0\)
03

Calculate the probabilities for each outcome

Now, let's compute the probabilities using the binomial formula mentioned above: 1. P(X=0) ≈ 0.2373 2. P(X=1) ≈ 0.3955 3. P(X=2) ≈ 0.2637 4. P(X=3) ≈ 0.0879 5. P(X=4) ≈ 0.0146 6. P(X=5) ≈ 0.0010
04

Graph the histogram

Using the calculated probabilities, we can now graph the histogram. On the x-axis, we will have the number of successes, and on the y-axis, we will have the probability for each outcome. 1. Plot the point (0, 0.2373) and label it "0 successes" 2. Plot the point (1, 0.3955) and label it "1 success" 3. Plot the point (2, 0.2637) and label it "2 successes" 4. Plot the point (3, 0.0879) and label it "3 successes" 5. Plot the point (4, 0.0146) and label it "4 successes" 6. Plot the point (5, 0.0010) and label it "5 successes" Finally, draw the histogram by connecting each point with a vertical line, and filling the area under the curve.
05

Check your answer using technology

To verify the correctness of the calculated probabilities and the histogram, we can use technology. You can use software such as Excel, R, Python (using libraries like NumPy, SciPy, or Matplotlib), or online tools like Desmos or Wolfram Alpha to calculate the binomial probabilities and graph the histogram. Remember to input the parameters of the binomial distribution (n=5 and p=1/4) and compare your results with the calculated histogram. If the results match, then the histogram is accurate.

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Key Concepts

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

Probability
Probability is a fundamental concept in statistics and mathematics. It helps us measure and describe uncertainty. In a binomial distribution, we focus on two types of probabilities: the probability of success \( (p) \) and the probability of failure \( (q) \). For a single experiment, if the chance of success is \( p \), then the chance of failure is \( q = 1 - p \).
For instance, in our example, the probability of success \( p \) for a single trial is \( \frac{1}{4} \), meaning there is a 25% chance of success. Consequently, the probability of failure \( q \) is \( \frac{3}{4} \) or 75%.
When dealing with multiple trials, such as the 5 trials in our problem, we use these probabilities to calculate the chance of obtaining a certain number of successes out of those trials. This calculation is key in forming the base for creating a binomial distribution, which will show us all potential outcomes and their respective probabilities.
Histogram
A histogram is a helpful graph that visually displays the distribution of data. For a binomial distribution, a histogram shows the probability of each possible outcome along the x-axis, while the probability value is displayed on the y-axis.
In creating a histogram for our binomial distribution example, we plot the number of successes (ranging from 0 to 5) against their corresponding probabilities. Each bar in the histogram represents a different number of successes:
  • 0 successes with a probability of approximately 0.2373
  • 1 success with a probability of approximately 0.3955
  • 2 successes with a probability of approximately 0.2637
  • 3 successes with a probability of approximately 0.0879
  • 4 successes with a probability of approximately 0.0146
  • 5 successes with a probability of approximately 0.0010
By visualizing these probabilities, the histogram helps us grasp how likely each outcome is, allowing for an intuitive understanding of the distribution's overall shape.
Binomial Coefficient
The binomial coefficient is a crucial part of the binomial distribution formula. It tells us the number of ways to choose \( k \) successes in \( n \) trials, considering the order doesn't matter. It is denoted as \( \binom{n}{k} \), which reads as 'n choose k'.
The formula for the binomial coefficient is:\[ \binom{n}{k} = \frac{n!}{k!(n-k)!} \]Where \( n! \) (n factorial) is the product of all positive integers up to \( n \).
In our example, with \( n = 5 \) trials, calculating the binomial coefficient helps us find each outcome's probability when plugged into the binomial probability formula.For example, \( \binom{5}{2} \) is used to calculate the probability of getting 2 successes. This value, when combined with the corresponding powers of \( p \) and \( q \), allows us to determine the probability of observing exactly 2 successful outcomes in the 5 trials. The use of binomial coefficients ensures that all potential orders of success are considered in the probability calculation.

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