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Consider a binomial experiment with 20 trials and probability 0.45 of success on a single trial. (a) Use the binomial distribution to find the probability of exactly 10 successes. (b) Use the normal distribution to approximate the probability of exactly 10 successes. (c) Compare the results of parts (a) and (b).

Short Answer

Expert verified
Binomial: 0.2001, Normal approximation: 0.161, with binomial more precise.

Step by step solution

01

Defining Binomial Distribution

The scenario is a binomial experiment with number of trials \( n = 20 \) and probability of success on a single trial \( p = 0.45 \). We need to find the probability of exactly 10 successes using the binomial distribution.
02

Calculate Binomial Probability

Apply the binomial probability formula: \[ P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \] where \( k = 10 \), \( n = 20 \), and \( p = 0.45 \). First, calculate \( \binom{20}{10} = \frac{20!}{10!10!} = 184756 \). Then compute the probability: \[ P(X = 10) = 184756 \times 0.45^{10} \times 0.55^{10} \approx 0.2001 \].
03

Approximating with Normal Distribution

We can approximate the binomial distribution with a normal distribution when \( n \) is large. First, calculate the mean \( \mu = np = 20 \times 0.45 = 9 \) and variance \( \sigma^2 = np(1-p) = 20 \times 0.45 \times 0.55 = 4.95 \). The standard deviation \( \sigma = \sqrt{4.95} \approx 2.22 \).
04

Standard Normal Variable and Continuity Correction

Use continuity correction to approximate the probability of exactly 10 successes. We find \( P(9.5 < X < 10.5) \) using the standard normal distribution: \[ z_1 = \frac{9.5 - 9}{2.22} \approx 0.225 , \quad z_2 = \frac{10.5 - 9}{2.22} \approx 0.675 \].
05

Calculate Probabilities from Z-scores

Find probabilities using standard normal distribution tables (or software). \( P(Z < 0.225) \approx 0.589 \) and \( P(Z < 0.675) \approx 0.750 \). Thus, \( P(9.5 < X < 10.5) \approx 0.750 - 0.589 = 0.161 \).
06

Compare Results

The binomial distribution yields \( P(X = 10) \approx 0.2001 \), while the normal approximation gives \( P(X = 10) \approx 0.161 \). The normal distribution approximation is slightly lower due to the inherent approximation errors, particularly noticeable with smaller sample sizes.

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Key Concepts

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

Probability of Success
In statistical terms, the probability of success refers to the likelihood of a certain event happening in an experiment with multiple trials. In the context of a binomial distribution, which is a common probability distribution used for binary outcomes such as success or failure, each trial has only two possible outcomes. Here, the probability of success is denoted by \( p \). It represents the chance of a successful outcome in a single trial.

In this example, we are examining a binomial experiment with 20 trials, where the probability of success \( p \) on each trial is 0.45. This value is crucial in predicting how often we might expect to achieve a particular number of successes, such as exactly 10 out of 20, using the binomial formula.

To apply the binomial probability formula:
  • Recognize that each trial is independent, meaning the outcome of one trial doesn't affect the others.
  • The formula \( P(X = k) = \binom{n}{k} p^k (1-p)^{n-k} \) helps calculate the probability of having exactly \( k \) successes out of \( n \) trials.
This concept allows statisticians to model binary experiments, such as testing whether a new drug works or if a coin lands on heads.
Normal Distribution Approximation
Normal distribution, often referred to as a bell curve because of its shape, provides an approximate model for the distribution of a set of data with a large number of trials. For a binomial distribution with a large number of trials (generally \( n > 30 \)), a normal distribution can be used to approximate the probability of an exact number of successes. This approach simplifies calculations especially when dealing with complex probabilities.

To use normal distribution approximation, the mean \( \mu \) and the standard deviation \( \sigma \) of the distribution need to be calculated:
  • The mean \( \mu = np \).
  • The variance \( \sigma^2 = np(1-p) \).
  • The standard deviation \( \sigma = \sqrt{np(1-p)} \).
These calculations help in transforming the binomial probabilities to an equivalent normal distribution problem which is then easier to solve. It transforms the problem of calculating an exact probability into finding areas under the curve of the normal distribution. Such approximations are vital when running statistical analyses on large sets of data or simulating real-world phenomena.
Contingency Correction
When approximating a binomial distribution to a normal distribution, a continuity correction is often applied to improve the accuracy of the approximation. The continuity correction adjusts the discrete binomial outcomes to better fit into the continuous nature of the normal distribution.

The continuity correction involves adjusting the binomial probability equation by 0.5 units in accordance with the problem’s limits. In the given example, when we calculate the probability of exactly 10 successes, the approximation adjusts to cover \( P(9.5 < X < 10.5) \) instead of precisely \( P(X = 10) \).

Using the calculated z-scores:
  • Compute \( z_1 = \frac{9.5 - 9}{\sigma} \), converting the left edge of the interval.
  • And \( z_2 = \frac{10.5 - 9}{\sigma} \), converting the right edge of the interval.
  • These z-scores then map to probabilities using standard normal distribution tables or software.
The correction process ensures that when plotting the probability on the continuous normal curve, the edges of the interval are precisely aligned, improving the fidelity of the approximation.

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