Sampling Distribution of Sample Proportion: Videos & Practice Problems

In probability theory, the binomial formula is a powerful tool for calculating the likelihood of a certain number of successes in a fixed number of trials, where each trial has two possible outcomes. However, as the number of trials (n) increases, using the binomial formula can become cumbersome. Fortunately, when both np and nq are greater than or equal to five, we can approximate binomial probabilities using the normal distribution, which simplifies the calculations significantly.

To illustrate this, consider a scenario where the probability of a person voting for a specific candidate in a two-person election is 56%. If we want to find the probability that more than 60 out of a sample of 100 people vote for this candidate, calculating this directly with the binomial formula would require evaluating the probabilities for 40 different outcomes (from 60 to 100). Instead, we can use the normal approximation.

First, we check the conditions for using the normal approximation. Here, n is 100, p is 0.56, and q (which is 1 - p) is 0.44. We calculate:

np = 100 * 0.56 = 56 (which is greater than 5)

nq = 100 * 0.44 = 44 (which is also greater than 5)

Since both conditions are satisfied, we can proceed to find the z-score. The mean of the binomial distribution is given by μ = np, and the standard deviation is σ = √(npq). Thus, we have:

μ = 100 * 0.56 = 56

σ = √(100 * 0.56 * 0.44) ≈ 4.9

Next, we calculate the z-score using the formula:

z = (x - μ) / σ

However, since we are approximating a discrete distribution with a continuous one, we apply a continuity correction. For the probability of more than 60 votes, we adjust our value of x to 60.5 (adding 0.5). Therefore, we calculate:

z = (60.5 - 56) / 4.9 ≈ 0.907

Now, we need to find the probability that z is greater than 0.907. This can be done using a z-table or a calculator. The resulting probability is approximately 0.182, indicating that there is an 18.2% chance that more than 60 people out of 100 will vote for the candidate.

This method of using the normal distribution to approximate binomial probabilities not only saves time but also enhances our understanding of how these two distributions relate to one another. As we continue to explore these concepts, we will gain further practice and insight into their applications.

In 2023, approximately 35,000,000 cars were recalled out of about 94,000,000 produced. A used car dealer has 76 cars from this production year, and we want to determine the probability that more than half of these cars were recalled using a normal distribution to approximate a binomial probability.

First, we calculate the probability of a car being recalled, denoted as \( p \). This is found by dividing the number of recalled cars by the total number of cars produced:

Consequently, the probability of a car not being recalled, \( q \), is:

To ensure that we can use the normal approximation, we check that both \( n \cdot p \) and \( n \cdot q \) are greater than or equal to 5. For our case with \( n = 76 \):

Since both conditions are satisfied, we can proceed with the normal approximation. We need to find the probability that more than half of the 76 cars were recalled, which translates to finding \( P(X > 38) \). Since we are using a continuous normal distribution to approximate a discrete binomial distribution, we apply a continuity correction, adjusting our target to \( P(X > 38.5) \).

Next, we calculate the z-score using the formula:

Substituting in our values:

Calculating \( n \cdot p \):

Now, calculating the standard deviation:

Now substituting back into the z-score formula:

Using the z-score, we find the probability \( P(Z > 2.70) \). This can be determined using a z-table or calculator, yielding:

Thus, the probability that more than half of the dealer's 76 cars were recalled is approximately 0.0035, or 0.35%. This indicates a very low likelihood of more than half of the cars being recalled.

In statistical analysis, the binomial variable \( x \) can be approximated by a normal distribution when both \( np \) and \( nq \) are greater than or equal to five. When examining a binomial experiment, instead of focusing solely on the number of successes, we can calculate the sample proportion, denoted as \( \hat{p} \), which is the ratio of successes \( x \) to the total number of trials \( n \). This sample proportion \( \hat{p} \) also follows a distribution that resembles the binomial distribution and is approximately normal under the same conditions of \( np \) and \( nq \).

To illustrate this, consider a scenario where several students flip a coin 10 times each. The number of heads obtained represents our successes. To analyze the sampling distribution of \( \hat{p} \), we first recognize that \( \hat{p} = \frac{x}{n} \). By calculating \( \hat{p} \) for various outcomes (e.g., 3 heads out of 10 trials gives \( \hat{p} = 0.3 \)), we can observe that the distribution of these sample proportions mirrors the shape of the original binomial distribution, albeit with a rescaled x-axis.

To find the mean and standard deviation of the sampling distribution for \( \hat{p} \), we can leverage the properties of the binomial distribution. The mean of \( \hat{p} \) is simply equal to \( p \), the probability of success. Thus, if \( p = 0.5 \), the mean \( \mu_{\hat{p}} \) is also \( 0.5 \). The standard deviation \( \sigma_{\hat{p}} \) can be calculated using the formula:

$$ \sigma_{\hat{p}} = \sqrt{\frac{pq}{n}} $$

where \( q = 1 - p \). In our example, substituting \( p = 0.5 \) and \( n = 10 \) yields:

$$ \sigma_{\hat{p}} = \sqrt{\frac{0.5 \times 0.5}{10}} = 0.158 $$

This demonstrates that the mean and standard deviation of the sample proportion \( \hat{p} \) are derived from the binomial distribution parameters, adjusted for the number of trials. The key takeaway is that the distribution of \( \hat{p} \) can be assumed to be approximately normal, facilitating further statistical analysis, such as constructing confidence intervals for proportions. Understanding these concepts is crucial as they form the foundation for more advanced statistical methods.