Question

Suppose a random sample of \(n=25\) observations is selected from a population that is normally distributed with mean equal to 106 and standard deviation equal to 12 . a. Give the mean and the standard deviation of the sampling distribution of the sample mean \(\bar{x}\). b. Find the probability that \(\bar{x}\) exceeds \(110 .\) c. Find the probability that the sample mean deviates from the population mean \(\mu=106\) by no more than 4

Short answer

What is the probability that the sample mean exceeds 110? What is the probability that the sample mean deviates from the population mean by no more than 4? Answer: The mean and standard deviation of the sampling distribution of the sample mean are 106 and 2.4, respectively. The probability that the sample mean exceeds 110 is 0.5475. The probability that the sample mean deviates from the population mean by no more than 4 is 0.5000.

Step-by-step solution

Determine the Mean and Standard Deviation of the Sampling Distribution

We know the population mean is 106 (\(\mu = 106\)), and the population standard deviation is 12 (\(\sigma = 12\)). Since we are working with a sample of 25 observations (\(n = 25\)), we can determine the mean and standard deviation of the sampling distribution of the sample mean using the following formulas: Mean: \(\mu_{\bar{x}} = \mu = 106\) Standard Deviation: \(\sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{12}{\sqrt{25}} = \frac{12}{5} = 2.4\)

Convert to Standard Normal Distribution

Standardize the sample mean (\(\bar{x}\)) by using the Z-score formula: \(Z = \frac{\bar{x} - \mu_{\bar{x}}}{\sigma_{\bar{x}}}\)

Calculate Probabilities

a. The mean and standard deviation of the sampling distribution of the sample mean are 106 and 2.4, respectively. b. To find the probability that the sample mean exceeds 110, first standardize 110 using the Z-score formula: \(Z = \frac{110 - 106}{2.4} = \frac{4}{2.4} = 1.667\) Now, look up the probability in the standard normal table (Z-table) for 1.667. The table value is 0.4525. Since we want the probability that the sample mean exceeds 110 (\(\bar{x} > 110\)), we need to look at the area to the right of 1.667. The total area under the standard normal curve is 1, so to find the area to the right of 1.667, we subtract the value we found from 1: \(1 - 0.4525 = 0.5475\) So, the probability that the sample mean exceeds 110 is 0.5475. c. To find the probability that the sample mean deviates from the population mean by no more than 4, we can look at the interval between 102 and 110. First, standardize these values using the Z-score formula: \(Z_{102} = \frac{102 - 106}{2.4} = \frac{-4}{2.4} = -1.667\) \(Z_{110} = \frac{110 - 106}{2.4} = \frac{4}{2.4} = 1.667\) Now, look up the probabilities in the standard normal table (Z-table) for -1.667 and 1.667. The table values are 0.4525 and 0.9525, respectively. To find the probability of the sample mean deviating from the population mean by no more than 4, we find the difference between the probabilities: \(0.9525 - 0.4525 = 0.5000\) So, the probability that the sample mean deviates from the population mean by no more than 4 is 0.5000.

Key concepts

Population Mean

The population mean, often denoted as \(\mu\), is the average of all the values in a population. It represents a central point or typical value in the data set. In our exercise, the population mean is 106. This means if you could look at the entire data from which our sample is drawn, on average, each value would center around 106.

Understanding the population mean is crucial because it becomes the point of reference against which we measure our samples. Whether we're dealing with the characteristics of a human population or results from scientific experiments, this mean gives us a fundamental baseline. It's also the anchor around which the concept of the sampling distribution revolves. Each sample mean we calculate will vary slightly, but overall they will hover around the population mean, showcasing its role as the true central tendency of the dataset.

Standard Deviation

The standard deviation is a measure that quantifies the amount of variation or dispersion in a set of data values. In simpler terms, it indicates how much the individual data points typically differ from the mean. In our context, the population standard deviation is 12.

When discussing sampling distributions, the standard deviation of the population helps us calculate the standard deviation of the sample mean, often referred to as the standard error. This is computed using the formula \( \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} \), where \( \sigma \) is the population standard deviation and \( n \) is the sample size. For our exercise, where \( n = 25 \), the standard deviation of the sampling distribution becomes 2.4. This smaller value compared to the population standard deviation reflects the composed, averaged nature of the sample means, attributing less variability.

Normal Distribution

The normal distribution is a probability distribution that is symmetric about the mean, indicating that data near the mean are more frequent in occurrence than data far from the mean. It is often referred to as a bell curve due to its distinct shape.

In our exercise, it is mentioned that the population follows a normal distribution. This assumption is essential as it allows us to use the normal distribution properties to make inferences about the sample mean. For a normally distributed population, the sampling distribution of the sample mean will also be normal, which simplifies the process of conducting various statistical tests. Even if the original population weren't normal, thanks to the Central Limit Theorem, the distribution of the sample mean would be approximately normal if the sample size is sufficiently large (typically \( n > 30 \)), making our calculations reliable and straightforward.

Z-score

A Z-score, sometimes called a standard score, indicates how many standard deviations an element is from the mean. The Z-score is found using the formula:\[ Z = \frac{\bar{x} - \mu_{\bar{x}}}{\sigma_{\bar{x}}} \]where \( \bar{x} \) is the sample mean, \( \mu_{\bar{x}} \) is the mean of the sampling distribution, and \( \sigma_{\bar{x}} \) is the standard deviation of the sampling distribution.

In our exercise, Z-scores help determine the probability of observing a sample mean at a certain distance from the population mean. For instance, to find the likelihood that the sample mean exceeds 110, we first calculate its Z-score as 1.667. This Z-score shows how many standard deviations away the value is from the mean of the sampling distribution. We then use this score to find probabilities using a Z-table or standard normal distribution chart.

Z-scores provide a standardized way to compare observations from different distributions and thus are invaluable in statistical analysis, allowing us to make probabilistic inferences and tests efficiently.