Question

Let \(\bar{X}\) be the mean of a random sample of size \(n\) from a distribution that is \(N\left(\mu, \sigma^{2}\right)\), where the positive variance \(\sigma^{2}\) is known. Because \(\Phi(2)-\Phi(-2)=0.954\) find, for each \(\mu, c_{1}(\mu)\) and \(c_{2}(\mu)\) such that \(P\left[c_{1}(\mu)<\bar{X}

Short answer

The solution for \(c_1(\mu)\), \(c_2(\mu)\), \(d_1(\bar{x})\), and \(d_2(\bar{x})\) are \(c_1(\mu) = -2\sigma/\sqrt{n} + \mu\), \(c_2(\mu) = 2\sigma/\sqrt{n} + \mu\), \(d_1(\bar{x}) = \bar{x} + 2\sigma/\sqrt{n}\), and \(d_2(\bar{x}) = \bar{x} - 2\sigma/\sqrt{n}\). These answers match with the values ascertained from the text.

Step-by-step solution

Understand the Context

First, note that when a distribution is \(N(\mu, \sigma^2)\), it signifies that data is normally distributed with mean \(\mu\) and variance \(\sigma^2\). This is a standard normal distribution.

Establish the Standard Normal Distribution

Since \(N(\mu, \sigma^2)\) is a standard normal distribution for which the variance is already given and it's useful to have it standardized to a distribution with zero mean and unit variance, write the sample mean in standard units: \(\frac{\bar{X} - \mu}{\sigma/\sqrt{n}} = Z\) where \(Z\) is any standard normal variable.

Setting Up the Required Probability

The probability requested is \(P[c_1(\mu) < \bar{X} < c_2(\mu)] = P[c_1(\mu) - \mu < \bar{X} - \mu < c_2(\mu) - \mu] = P[\frac{c_1(\mu) - \mu}{\sigma/\sqrt{n}} < Z < \frac{c_2(\mu) - \mu}{\sigma/\sqrt{n}}] = 0.954\). It's known that \(\Phi(2) - \Phi(-2) = 0.954\), so set the equation as: \(Z_2 - Z_1 = 4\), where \(Z_2\) and \(Z_1\) are the standard values corresponding to \(c_2(\mu)\) and \(c_1(\mu)\) respectively.

Solving for \(c_1(\mu)\) and \(c_2(\mu)\)

Now, we can solve for \(c_1(\mu)\) and \(c_2(\mu)\). Because \(Z_2 - Z_1 = 4\), we have \(Z_2 = Z_1 + 4\). Substituting into the probability set-up, we have \(Z_2 = \frac{c_2(\mu) - \mu}{\sigma/\sqrt{n}} = 2\), hence \(c_2(\mu) = 2\sigma/\sqrt{n} + \mu\). Similarly, \(Z_1 = \frac{c_1(\mu) - \mu}{\sigma/\sqrt{n}} = -2\), hence \(c_1(\mu) = -2\sigma/\sqrt{n} + \mu\).

Solve for \(d_1(\bar{x})\) and \(d_2(\bar{x})\)

For \(d_1(\bar{x})\) and \(d_2(\bar{x})\), it's required that \(P[d_2(\bar{X}) < \mu < d_1(\bar{X})] = 0.954\). Since \(\mu\) is the mean, solve for it within \(c_1(\mu)\) and \(c_2(\mu)\), so \(d_1(\bar{x}) = \bar{x} + 2\sigma/\sqrt{n}\) and \(d_2(\bar{x}) = \bar{x} - 2\sigma/\sqrt{n}\).

Compare with Previous Answer

With \(d_1(\bar{x}) = c_2(\mu)\) and \(d_2(\bar{x}) = c_1(\mu)\), it is clear that the previous answer is the same as this one. The bounds are two standard deviations from the mean.

Key concepts

Standard Normal Variable

In statistics, a standard normal variable is an essential concept when working with normal distributions. The standard normal distribution is a specific type of normal distribution with a mean (\textmu{}) of 0 and a standard deviation (\textsigma{}) of 1. This special distribution is denoted as N(0,1) and forms the basis for many statistical methods, including hypothesis testing and confidence intervals.

\[ Z = \frac{\bar{X} - \mu}{\sigma/\sqrt{n}} \]
transforms a sample mean \(\bar{X}\) from any normal distribution N(\mu, \sigma^2) into a standard normal variable Z. By converting to standard units, we can utilize the standard normal distribution to find probabilities for values of \(\bar{X}\) relative to its distribution's mean, given the population mean \(\mu\) and standard deviation \(\sigma\), as seen in the exercise provided.

Sample Mean

The sample mean, represented as \(\bar{X}\), is simply the average of a set of observations from a random sample. It's a crucial estimator in statistics used to infer the population mean (\textmu{}) when the population parameters are unknown. When the sample mean is calculated from a normally distributed population, as in the exercise, it holds specific properties.

\[ \bar{X} = \frac{1}{n}\sum_{i=1}^{n}X_i \]
This formula represents the calculation of the sample mean where \(X_i\) are the individual observations and n is the sample size. The central limit theorem plays a pivotal role here, as it ensures that for large enough sample sizes, the distribution of the sample mean will tend towards a normal distribution, regardless of the population's original distribution.

Probability Distribution

A probability distribution describes how the values of a random variable are distributed. It assigns probabilities to each possible value (or range of values) of the random variable. When dealing with continuous data, a probability density function (PDF) describes the distribution.

In the context of a normal distribution, the probability distribution is described by a bell curve that is symmetric around the mean, indicating that values near the mean are more likely to occur than values far from the mean. In the provided exercise, the fact that \(\Phi(2) - \Phi(-2) = 0.954\) comes from the cumulative distribution function (CDF) of the normal distribution, which provides the probability that a random variable is less than or equal to a particular value.

Confidence Interval

A confidence interval is a range of values, derived from sample statistics, that is likely to contain the value of an unknown population parameter. The interval has an associated confidence level that quantifies the level of certainty (or uncertainty) we have that the parameter lies within the interval.

For instance, a 95.4% confidence interval implies that if we were to take many samples and build an interval from each sample, approximately 95.4% of these intervals would contain the population parameter. In our exercise, \(P[c_1(\mu) < \bar{X} < c_2(\mu)] = 0.954\) represents a 95.4% confidence interval for the population mean (\mu) based on the sample mean (\bar{X}). This provides a practical interpretation: if the sample mean \(\bar{X}\) falls within this interval, there is a 95.4% probability that the interval also contains the true population mean \(\mu\).