Question

Random samples of size \(n=500\) were selected from a binomial population with \(p=.1\). a. Is it appropriate to use the normal distribution to approximate the sampling distribution of \(\hat{p} ?\) Check to make sure the necessary conditions are met. Using the results of part a, find these probabilities: b. \(\hat{p}>.12\) c. \(\hat{p}<.10\) d. \(\hat{p}\) lies within .02 of \(p\)

Short answer

Answer: The approximate probabilities for the given scenarios are: a. P(\(\hat{p} > 0.12\)) ≈ 0.000004 b. P(\(\hat{p} < 0.10\)) = 0.5 c. P(\(\hat{p}\) lies within 0.02 of p) ≈ 0.999488

Step-by-step solution

Check if the normal distribution can be used

The conditions to use the normal distribution as an approximation for a binomial distribution are: 1. np >= 10 2. n(1-p) >= 10 We are given n = 500 and p = 0.1. Let's check if the conditions are met. np = 500 * 0.1 = 50 >= 10 n(1-p) = 500 * (1-0.1) = 500 * 0.9 = 450 >= 10 Both conditions are met, so we can use the normal distribution as an approximation.

Calculate the mean and standard deviation of the normal distribution

Using the Central Limit Theorem, we can create a normal distribution with: Mean (μ) = np = 500 * 0.1 = 50 Variance (σ^2) = np(1-p) = 500 * 0.1 * 0.9 = 45 Standard deviation (σ) = sqrt(45) ≈ 6.71

Find the z-scores for each probability

To find the probabilities, we need to calculate the z-score for each scenario. The formula for calculating the z-score is: z = (x - μ) / σ a. Probability of \(\hat{p} > 0.12\) z = (0.12 * 500 - 50) / 6.71 ≈ 4.48 b. Probability of \(\hat{p} < 0.10\) z = (0.10 * 500 - 50) / 6.71 ≈ 0 c. Probability of \(\hat{p}\) lies within 0.02 of p Lower bound: (0.08 * 500 - 50) / 6.71 ≈ -3.59 Upper bound: (0.12 * 500 - 50) / 6.71 ≈ 4.48

Calculate probabilities using the z-scores

Now, we can use the z-scores to find the corresponding probabilities using a standard normal table or calculator. a. Probability of \(\hat{p} > 0.12\) P(Z > 4.48) ≈ 0.000004 b. Probability of \(\hat{p} < 0.10\) P(Z < 0) = 0.5 c. Probability of \(\hat{p}\) lies within 0.02 of p P(-3.59 < Z < 4.48) ≈ 0.999665 - 0.000177 = 0.999488 The approximate probabilities for the given scenarios are: a. P(\(\hat{p} > 0.12\)) ≈ 0.000004 b. P(\(\hat{p} < 0.10\)) = 0.5 c. P(\(\hat{p}\) lies within 0.02 of p) ≈ 0.999488

Key concepts

Central Limit Theorem

The Central Limit Theorem (CLT) is a fundamental principle in statistics. It states that, given a sufficiently large sample size, the sampling distribution of the sample mean will approximate a normal distribution. This approximation holds true regardless of the shape of the population distribution.

For a sample to be considered large enough, the rule of thumb is usually a sample size () greater than 30. However, when working with binomial distributions, as in our exercise, the normal approximation is acceptable if the conditions \( np \geq 10 \) and \( n(1-p) \geq 10 \) are met. This ensures that the sample size is sufficiently large for the approximation to be accurate.

In our exercise with \( n = 500 \) and \( p = 0.1 \), both conditions are satisfied with \( np = 50 \) and \( n(1-p) = 450 \). This allows us to utilize the CLT to approximate the binomial distribution with a normal distribution for calculating probabilities around \( \hat{p} \).

To summarize, the Central Limit Theorem provides a bridge to use normal distribution properties to approximate binomial probabilities when sample sizes are large enough. This helps to simplify complex probability calculations, like those required in our exercise.

Binomial Distribution

The binomial distribution is a discrete probability distribution that models the number of successes in a fixed number of independent Bernoulli trials, each with the same probability of success \( p \). It is particularly useful when you are interested in the number of times an event occurs within a specified number of trials.

For a binomial distribution, key parameters include \( n \), the number of trials, and \( p \), the probability of success in each trial. In our exercise, \( n \) is 500 and \( p \) is 0.1. This leads us to calculate certain statistics, like the mean \( \mu = np \) and the variance \( \sigma^2 = np(1-p) \). In this case, the mean is 50 and the variance is 45.

Despite binomial distribution's discrete nature, under certain conditions, it can be approximated by the normal distribution, especially when dealing with large sample sizes. This step reduces the complexity of probability calculations. As previously discussed, our conditions (\( np \geq 10 \) and \( n(1-p) \geq 10 \)) are met, making it suitable to apply the normal approximation.

Ultimately, understanding the binomial distribution's properties allows us to effectively transition to the normal distribution approximation when analyzing probabilities, which is crucial for solving binomial problems like those in the given exercise.

Standard Normal Table

The Standard Normal Table, also known as the Z-table, is a mathematical table used to find the probabilities associated with a normal distribution. It assists in translating the z-scores, which measure how many standard deviations an element is from the mean, into probability values.

To use this table, you first need to calculate the z-score using the formula: \[z = \frac{x - \mu}{\sigma}\]where \( x \) is the value of interest, \( \mu \) is the mean, and \( \sigma \) is the standard deviation.

In our exercise, we used z-scores to determine probabilities for different scenarios around \( \hat{p} \), the sample proportion. For example, to assess the probability of \( \hat{p} > 0.12\), the z-score was computed as approximately 4.48, representing an extreme value within the distribution.

The Standard Normal Table then helps us determine that the probability of a z-score larger than 4.48 is almost negligible (around 0.000004), as it's well into the tail of the distribution.

The table is crucial for quick probability calculations, especially when working with large datasets or complex problems, thus playing a pivotal role in statistical analysis as demonstrated in the exercise.