Problem 17 - Basic Statistics for Business and Economics with Student

Question

The rent for a one-bedroom apartment in Southern California follows the normal distribution with a mean of \(\$ 2,200\) per month and a standard deviation of \(\$ 250\) per month. The distribution of the monthly costs does not follow the normal distribution. In fact, it is positively skewed. What is the probability of selecting a sample of 50 one-bedroom apartments and finding the mean to be at least \(\$ 1,950\) per month?

Step-by-Step Solution

Answer

The probability is almost 1 (or 100%).

Step 1: Calculate Standard Error

First, we need to calculate the standard error of the mean for a sample of 50 apartments. The standard error (SE) is given by the formula \( SE = \frac{\sigma}{\sqrt{n}} \), where \( \sigma = \$250 \) is the standard deviation, and \( n = 50 \) is the sample size.\[SE = \frac{250}{\sqrt{50}} \approx 35.36\]

Step 2: Calculate the Z-Score

Next, we calculate the Z-score for the sample mean of \( \\(1,950 \). The Z-score formula is \( Z = \frac{\bar{x} - \mu}{SE} \), where \( \bar{x} = \\)1,950 \) is the sample mean, and \( \mu = \$2,200 \) is the population mean.\[ Z = \frac{1950 - 2200}{35.36} \approx -7.07 \]

Step 3: Find the Probability from the Z-Score

Using the calculated Z-score, we refer to the standard normal distribution table to find the probability of a Z-score less than \(-7.07\). The probability associated with such a large negative Z-score is practically 0.

Step 4: Determine the Complement Probability

Since we want the probability of the sample mean being at least \( \$1,950 \), we look for the complement of the probability we found in Step 3. Thus, the probability is \( 1 - 0 = 1 \).

Key Concepts

Understanding the Normal Distribution Curve

The normal distribution curve is a fundamental concept in statistics. It's also known as the bell curve due to its symmetrical bell-shaped appearance. This curve is characterized by its mean, median, and mode, which all coincide at the center of the distribution. The spread of the curve is determined by the standard deviation, which measures how spread out the data points are around the mean. In a normal distribution:

• About 68% of data falls within one standard deviation of the mean.
• About 95% falls within two standard deviations.
• About 99.7% falls within three standard deviations.

For example, if we consider apartment rents as normally distributed, most rents will be close to the average rent, and very few will be extremely high or low. This helps in understanding how data is dispersed around the average value in any dataset following a normal distribution.

Mastering Z-Score Calculation

Calculating the Z-score is a critical step in understanding how far away a particular data point (or sample mean) is from the mean of the data set in terms of standard deviations. The Z-score equation is: \[ Z = \frac{\bar{x} - \mu}{SE} \]Where:

• \( \bar{x} \) is the sample mean.
• \( \mu \) is the population mean.
• \( SE \) is the standard error of the mean, calculated as \( \frac{\sigma}{\sqrt{n}} \), with \( \sigma \) being the standard deviation and \( n \) the sample size.

The Z-score tells us how many standard deviations a sample mean is from the population mean. A Z-score of -7.07, as calculated in this problem, indicates that the sample mean is 7.07 standard deviations below the population mean. Such a large negative Z-score is unusual, highlighting the rarity of finding a sample mean so different from the population mean.

Probability Assessment for Sample Means

Probability assessment in the context of normal distribution involves finding how likely an event is. After calculating the Z-score, the next step is finding its corresponding probability using a standard normal distribution table.The probability we look for is typically the area under the curve to the left of the Z-score if it is negative, or to the right if it is positive. In our scenario, the Z-score was close to -7.07, and the probability associated with such a score is practically zero, meaning it's highly improbable. Since the question asks for the probability of a sample mean being at least \\( 1,950, we calculate the complement (1 minus the probability calculated) because the original probability covers the likelihood of not reaching \\) 1,950. Hence, we found that there's almost a 100% probability that the sample mean is at least \$ 1,950.