Assume That We Observe Xi For I=1 To N Are IID Draws From A

Question

Assume That We Observe Xi For I1 To N Are Iid Draws From a Bernou Assume that we observe Xi, for i=1 to n, are independent and identically distributed (i.i.d.) draws from a Bernoulli distribution with parameter p = 0.3. Use the Central Limit Theorem (CLT) to approximate the probabilities for the following scenarios: a) In a sample of size 100, what is the probability that the sample mean of Xi is greater than or equal to 0.35? b) In a sample of size 400, what is the probability that the sample mean of Xi is less than 0.27?

Dr. Jack HW Helper · Accepted Answer

The Central Limit Theorem (CLT) is a fundamental statistical principle stating that, given a sufficiently large sample size, the sampling distribution of the sample mean will approximate a normal distribution regardless of the shape of the population distribution, provided the population has a finite mean and variance. This theorem is especially useful for calculating approximate probabilities concerning sample means, such as those involving Bernoulli distributions. In this case, each observation Xi follows a Bernoulli distribution with parameter p = 0.3, meaning that each Xi can take values 1 (with probability 0.3) or 0 (with probability 0.7). The population mean (expected value) is thus E[Xi] = p = 0.3, and the population variance is Var[Xi] = p(1 - p) = 0.3 * 0.7 = 0.21. The sample mean, denoted as $\bar{X}$, is given by the sum of the observations divided by the sample size n. Because the Xi are i.i.d., the mean of the sampling distribution of $\bar{X}$ is E[$\bar{X}$] = p, and its variance is Var[$\bar{X}$] = p(1 - p) / n. Applying the CLT, for large n, $\bar{X}$ approximately follows a normal distribution with mean p and standard deviation $\sigma_{\bar{X}} = \sqrt{p(1 - p)/n}$. Part a: Probability that $\bar{X} \geq 0.35$ with n = 100 Using the CLT, the mean is $\mu = 0.3$, and the standard deviation is:\n $$ \sigma_{\bar{X}} = \sqrt{\frac{p(1-p)}{n}} = \sqrt{\frac{0.3 \times 0.7}{100}} = \sqrt{\frac{0.21}{100}} = \sqrt{0.0021} \approx 0.0458. $$ We want to find $P(\bar{X} \geq 0.35)$. Standardizing this, we compute the z-score: $$ z = \frac{0.35 - \mu}{\sigma_{\bar{X}}} = \frac{0.35 - 0.3}{0.0458} \approx \frac{0.05}{0.0458} \approx 1.09. $$ Using standard normal distribution tables or approximate calculations, the probability that $Z \geq 1.09$ is: $$ P(Z \geq 1.09) = 1 - P(Z \leq 1.09) \approx 1 - 0.8621 = 0.1379. $$ Therefore, the approximate probability that the sample mean exceeds or equals 0.35 in a sample of 100 observations is app

Assume That We Observe Xi For I=1 To N Are IID Draws From A

Assume That We Observe Xi For I1 To N Are Iid Draws From a Bernou

Paper For Above instruction

Part a: Probability that \(\bar{X} \geq 0.35\) with n = 100

Part b: Probability that \(\bar{X}

Implications of Using the CLT in Bernoulli Distributions

Conclusion

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