Problem 32 - Basic Statistics for Business and Economics with Student

Question

CRA CDs, Inc., wants the mean lengths of the "cuts" on a CD to be 135 seconds (2 minutes and 15 seconds). This will allow the disk jockeys to have plenty of time for commercials within each 10 -minute segment. Assume the distribution of the length of the cuts follows the normal distribution with a population standard deviation of 8 seconds. Suppose we select a sample of 16 cuts from various CDs sold by CRA CDs, Inc. a. What can we say about the shape of the distribution of the sample mean? b. What is the standard error of the mean? c. What percent of the sample means will be greater than 140 seconds? d. What percent of the sample means will be greater than 128 seconds? e. What percent of the sample means will be greater than 128 but less than 140 seconds?

Step-by-Step Solution

Answer

a. Normal; b. 2 seconds; c. 0.62%; d. 99.98%; e. 99.36%.

Step 1: Understanding the Distribution

Since the lengths of the cuts follow a normal distribution, the sample mean will also be normally distributed due to the properties of the normal distribution and the Central Limit Theorem (CLT). With a sample size of 16, which is greater than 30, the sample mean will have a normal distribution.

Step 2: Calculate the Standard Error of the Mean

The standard error of the mean (SEM) can be calculated using the formula \( SEM = \frac{\sigma}{\sqrt{n}} \), where \( \sigma \) is the population standard deviation and \( n \) is the sample size. Here, \( \sigma = 8 \) seconds and \( n = 16 \). Thus, \( SEM = \frac{8}{\sqrt{16}} = \frac{8}{4} = 2 \) seconds.

Step 3: Calculate Percent of Sample Means Greater than 140 Seconds

First, calculate the z-score for 140 seconds using \( z = \frac{X - \mu}{SEM} \), where \( X = 140 \) seconds and \( \mu = 135 \) seconds. The z-score is \( z = \frac{140 - 135}{2} = \frac{5}{2} = 2.5 \). Using the standard normal distribution table, the probability \( P(Z > 2.5) \) is approximately 0.0062 or 0.62%.

Step 4: Calculate Percent of Sample Means Greater than 128 Seconds

Calculate the z-score for 128 seconds: \( z = \frac{128 - 135}{2} = \frac{-7}{2} = -3.5 \). Using the standard normal distribution table, the probability \( P(Z > -3.5) \) is approximately 1 - 0.0002, which is about 0.9998 or 99.98%.

Step 5: Calculate Percent of Sample Means Between 128 and 140 Seconds

To find the probability that the sample mean is between 128 and 140 seconds, subtract the probability of being less than 128 seconds from the probability of being less than 140 seconds. The probability \( P(Z < 2.5) \) is approximately 0.9938 and \( P(Z < -3.5) \) is approximately 0.0002. Thus, the probability in between is \( 0.9938 - 0.0002 = 0.9936 \) or 99.36%.

Key Concepts

Understanding Normal Distribution

The normal distribution is a fundamental concept in statistics. It is a continuous probability distribution that is symmetric around the mean. Most of the data points cluster around the central peak, and the probabilities for values taper off equally at both ends. This distribution is often called a "bell curve" because of its bell-shaped appearance.

Being a normal distribution, it is characterized by two parameters: the mean (\( \mu \)), which determines the center, and the standard deviation (\( \sigma \)), which dictates the spread of the distribution. In practical terms, this means values are more likely to occur near the mean than far from it. The normal distribution is crucial because many natural phenomena approximate this pattern, making it a key feature of statistical analyses. Understanding this concept helps to interpret data like the mean length of CD cuts in our exercise.

In the given problem, it is stated that the lengths of the cuts on a CD follow a normal distribution, which informs us about the shape of the distribution and the techniques we can use to analyze it.

Standard Error of the Mean

The standard error of the mean (SEM) is an important metric in statistics that measures the dispersion or spread of a sample mean relative to the true population mean. This term essentially helps us understand how much variation we might expect in the average of a large number of samples taken from the same population.

To calculate the SEM, you use the formula:

• \( SEM = \frac{\sigma}{\sqrt{n}} \)

where \( \sigma \) is the standard deviation of the population and \( n \) is the size of the sample. For the CD cuts, with a population standard deviation of 8 seconds and a sample size of 16, the SEM would be calculated as:

• \( SEM = \frac{8}{\sqrt{16}} = \frac{8}{4} = 2 \) seconds

This tells us that the average length of the cuts in our sample will typically differ from the population mean by about 2 seconds.

Z-Score Calculation

Z-score calculation is a method used to understand the relationship of a particular data point to the mean of a group of data, expressed in terms of standard deviations from the mean. It helps determine how far away an individual data point is from the average, making it a powerful tool for understanding variability in data points in relation to the distribution.

The z-score can be calculated using the formula:

• \( z = \frac{X - \mu}{SEM} \)

where \( X \) is the data point in question, \( \mu \) is the mean of the distribution, and \( SEM \) is the standard error of the mean. For example, if we wanted to find out how unusual a sample mean of 140 seconds is, given a population mean of 135 seconds and SEM of 2, we would calculate:

• \( z = \frac{140 - 135}{2} = 2.5 \)

This tells us that 140 seconds is 2.5 standard errors above the mean. Using a standard normal distribution table, we can determine that only about 0.62% of sample means are greater than 140 seconds, indicating this is a fairly rare occurrence. Understanding z-scores helps us make decisions based on how likely an event is within the context of a normal distribution.