Question

Let \(x\) be a binomial random variable with \(n=100\) and \(p=.2 .\) Find approximations to these probabilities: a. \(P(x>22)\) b. \(P(x \geq 22)\) c. \(P(20

Short answer

Answer: a) P(x > 22) ≈ 0.3085, b) P(x ≥ 22) ≈ 0.3537, c) P(20 < x < 25) ≈ 0.3214, d) P(x ≤ 25) ≈ 0.9157.

Step-by-step solution

Calculate Mean and Variance

For a binomial distribution with parameters n and p, the mean and variance are given by: Mean: µ = n * p Variance: σ² = n * p * (1 - p) In this case, with n = 100 and p = 0.2, we have: Mean: µ = 100 * 0.2 = 20 Variance: σ² = 100 * 0.2 * (1 - 0.2) = 16 Now, we can also compute the standard deviation: Standard Deviation: σ = sqrt(σ²) = sqrt(16) = 4

Apply Normal Approximation

With the binomial parameters, mean and variance, we can use the normal approximation for these probabilities. We will convert the x-values to z-scores, then use the z-table to find the corresponding probabilities. Reminder: the z-score transformation is z = (x - µ) / σ a. To calculate P(x > 22), first find the z-score for x = 22: z = (22 - 20) / 4 = 0.5 Using the z-table, we find that P(z > 0.5) = 1 - P(z <= 0.5) = 1 - 0.6915 = 0.3085. Therefore, P(x > 22) ≈ 0.3085. b. To calculate P(x ≥ 22), we can simply consider P(x > 21.5) since we're using a continuous distribution to approximate a discrete one: z = (21.5 - 20) / 4 = 0.375 Using the z-table, we find that P(z > 0.375) = 1 - P(z <= 0.375) = 1 - 0.6463 = 0.3537. Therefore, P(x ≥ 22) ≈ 0.3537. c. To calculate P(20 < x < 25), we need to find P(20.5 < x < 24.5): z1 = (20.5 - 20) / 4 = 0.125 z2 = (24.5 - 20) / 4 = 1.125 Using the z-table, we find that P(z <= 1.125) = 0.8708 and P(z <= 0.125) = 0.5494. Therefore, P(20 < x < 25) ≈ 0.8708 - 0.5494 = 0.3214. d. To calculate P(x ≤ 25), we can simply consider P(x < 25.5) since we're using a continuous distribution to approximate a discrete one: z = (25.5 - 20) / 4 = 1.375 Using the z-table, we find that P(z <= 1.375) = 0.9157. Therefore, P(x ≤ 25) ≈ 0.9157. So, the approximated probabilities are: a. P(x > 22) ≈ 0.3085 b. P(x ≥ 22) ≈ 0.3537 c. P(20 < x < 25) ≈ 0.3214 d. P(x ≤ 25) ≈ 0.9157

Key concepts

Normal Approximation to Binomial Distribution

The normal approximation to the binomial distribution is a useful technique for simplifying probability calculations when dealing with a large number of trials. Essentially, it allows us to use the normal distribution as a stand-in for the binomial distribution, provided certain conditions are met. Typically, these conditions include having a sufficiently large sample size (usually, a rule of thumb is that both np and n(1-p) are greater than 5) and a probability of success that isn't too close to 0 or 1.

Using the normal approximation simplifies the calculations significantly because the normal distribution has been extensively studied and its properties are well-known. Additionally, it's continuous, which makes integration easier when compared to the discrete nature of binomial distributions. For our exercise involving binomial random variable with parameters n=100 and p=0.2, we can safely use the normal approximation because our sample size is large and p is not near 0 or 1.

Understanding the Continuity Correction

When applying the normal approximation to a binomial distribution, the continuity correction should be considered for more accurate results. This correction accounts for the difference between the discrete binomial distribution and the continuous normal distribution. It involves adjusting the value of x by 0.5 in the direction that includes the mean of the distribution. This is why, in our step-by-step solution, we used values like 21.5 and 24.5 for the calculations instead of the original integer values.

Probability Calculations using Normal Approximation

Probability calculations are at the heart of understanding distributions. When dealing with the binomial distribution and large n values, the normal approximation comes in handy to calculate probabilities for different ranges of outcomes. In our textbook example, we were interested in calculating probabilities for different scenarios relative to our binomial random variable x.

To calculate these probabilities, we use the normal approximation and convert our x values to z-scores. The z-score tells us how many standard deviations an element is from the mean. Once we have the z-score, we can use standard normal distribution tables or software to find the probability corresponding to that score.

For instance, in part (a) of the exercise, to find the probability that x is greater than 22, we calculated the z-score and then found our probability as approximately 0.3085. This process, which appears in each part of the given exercise, is commonly used in statistics when it's impractical to calculate probabilities using the binomial formula—especially for large n values.

Z-Score Transformation

A z-score transformation is a method to standardize values in a dataset by converting them to a standard scale where the mean is 0 and the standard deviation is 1. This is a crucial step when applying the normal approximation to binomial distribution.

The formula for calculating a z-score is: \( z = \frac{x - \mu}{\sigma} \) where \( x \) is a value from the dataset, \( \mu \) is the mean, and \( \sigma \) is the standard deviation. In the context of our binomial example, \( x \) represents the number of successes, \( \mu = np \) is the expected number of successes, and \( \sigma \) is the square root of the variance (np(1-p)).

Z-scores enable us to compare individual scores or occurrences to the general distribution, regardless of the original scale of measurement. Moreover, they are a foundational element used in many statistical procedures, from hypothesis testing to confidence intervals. In our classroom context, once we have the z-scores for each part of the exercise, we can easily look up or calculate the associated probabilities from the standard normal distribution, which is exactly what we did to provide the probabilities required for solving our exercise.