Problem 40 - Basic Statistics for Business and Economics

Question

The number of passengers on the Carnival Sensation during one-week cruises in the Caribbean follows the normal distribution. The mean number of passengers per cruise is 1,820 and the standard deviation is \(120 .\) a. What percent of the cruises will have between 1,820 and 1,970 passengers? b. What percent of the cruises will have 1,970 passengers or more? c. What percent of the cruises will have 1,600 or fewer passengers? d. How many passengers are on the cruises with the fewest 25 percent of passengers?

Step-by-Step Solution

Answer

a. 39.44%, b. 10.56%, c. 3.34%, d. ~1,738 passengers.

Step 1: Identify Mean and Standard Deviation

The problem states that the mean number of passengers per cruise is 1,820, and the standard deviation is 120. These values will be used to calculate z-scores for the normal distribution.

Step 2: Calculating Z-Scores

To find the percentage of passengers between given numbers, we must calculate the z-scores. The z-score is calculated using the formula \( z = \frac{X - \mu}{\sigma} \), where \( X \) is the value, \( \mu \) is the mean, and \( \sigma \) is the standard deviation.

Step 3: Calculate Z-Score for 1,970 Passengers

For 1,970 passengers: \( z = \frac{1,970 - 1,820}{120} = \frac{150}{120} = 1.25 \).

Step 4: Using the Z-Table for 1,970 Passengers

Using a z-table or calculator, find the area to the left of z = 1.25. This gives us the percentage of cruises with 1,970 or fewer passengers. The area to the left is approximately 89.44%.

Step 5: Calculate Z-Scores for 1,820 Passengers

For 1,820 passengers: \( z = \frac{1,820 - 1,820}{120} = 0 \).

Step 6: Find Percentage for 1,820 to 1,970 Passengers

The percentage of cruises between z = 0 and z = 1.25 is \( 89.44\% - 50\% = 39.44\% \).

Step 7: Compute Percent of Cruises with 1,970 or More Passengers

Subtract the area to the left of z = 1.25 from 100%. Therefore, \( 100\% - 89.44\% = 10.56\% \).

Step 8: Calculate Z-Score for 1,600 Passengers

For 1,600 passengers: \( z = \frac{1,600 - 1,820}{120} = \frac{-220}{120} = -1.83 \).

Step 9: Using the Z-Table for 1,600 Passengers

The z-table gives an area of approximately 3.34% to the left of z = -1.83. Therefore, 3.34% of cruises have 1,600 or fewer passengers.

Step 10: Determine Passengers in the Lowest 25%

Find the z-score corresponding to the 25th percentile (z = -0.675). Calculate \( X \) using the z-score formula: \( -0.675 = \frac{X - 1,820}{120} \). Solve for \( X \): \( X \approx 1,738 \).

Key Concepts

Z-Score

The z-score is a fundamental concept in statistics, especially when dealing with normal distributions. It helps us understand how far and in what direction a data point is from the mean of the data set. For any given data point, the z-score indicates how many standard deviations it is away from the mean. This allows us to compare different data points within a distribution. To calculate a z-score, we use the formula:\[ z = \frac{X - \mu}{\sigma} \]where:

• \(X\) represents the value of the data point.
• \(\mu\) is the mean of the data set.
• \(\sigma\) is the standard deviation.

By standardizing data points into z-scores, we can easily compare and interpret datums based on their position in the distribution. For example, if a cruise had 1,970 passengers with a z-score of 1.25, it indicates that it's 1.25 standard deviations above the mean of 1,820 passengers.

Mean and Standard Deviation

The mean and standard deviation are key statistical measures that summarize a data set's characteristics. The mean, often referred to as the average, is the central value of a data set. It is calculated by adding all the individual data points and then dividing the sum by the number of data points. For the Carnival Sensation cruises, the mean number of passengers is 1,820, representing the central tendency of passenger distribution across cruises. On the other hand, standard deviation measures the amount of variation or dispersion in a set of values. A low standard deviation means that the data points tend to be close to the mean, whereas a high standard deviation indicates wider spread around the mean. In the context of the cruise problem, the standard deviation is 120. This number tells us that the number of passengers typically varies by about 120 from the mean of 1,820. Knowing the standard deviation helps us understand how typical or unusual a particular passenger count is on a cruise.

Percentile Calculation

Percentile calculation is a statistical method used to understand the relative standing of a value within a data set. A percentile indicates the value below which a given percentage of observations in a data set fall. For example, if a cruise passenger count falls at the 90th percentile, it means that 90% of cruises have fewer passengers, and only 10% have more. This concept is essential for understanding distributions and identifying thresholds in data sets.To calculate the number of passengers in the lowest 25% of cruises, we first determine the z-score that corresponds to the 25th percentile. In the context of a normal distribution, this z-score is approximately -0.675. We can then plug this value into the z-score formula to solve for the actual number of passengers:\[-0.675 = \frac{X - 1,820}{120}\]Rearranging and solving for \(X\), we find:\[X \approx 1,738\]Thus, cruises with fewer than 1,738 passengers represent the bottom 25% of the distribution. This calculation is crucial for assessing how different data points rank within the overall data set.