Problem 13 - Calculus Concepts

Question

Luggage Weight Suppose the weight of pieces of passenger luggage for domestic airline flights follows a normal distribution with \(\mu=40\) pounds and \(\sigma=10.63\) pounds. a. Calculate the probability that a piece of luggage weighs less than 45 pounds. b. Calculate the probability that the total weight of the luggage for 80 passengers on a particular flight is between 1200 and 2400 pounds. (Assume each passenger has one piece of luggage.) c. Calculate where the probability density function for the weight of passenger luggage is decreasing most rapidly.

Step-by-Step Solution

Answer

a) 0.6808, b) Probability is essentially zero, c) Inflection at 50.63 lbs.

Step 1: Understand the Given Normal Distribution

The weight of luggage is normally distributed with a mean \( \mu = 40 \) pounds and a standard deviation \( \sigma = 10.63 \) pounds. This implies that the distribution is defined as \( N(40, 10.63^2) \).

Step 2: Find Probability for Luggage Weighing Less than 45 Pounds

To find the probability that a luggage piece weighs less than 45 pounds, calculate the z-score using the formula \( z = \frac{x - \mu}{\sigma} \):\[ z = \frac{45 - 40}{10.63} = \frac{5}{10.63} \approx 0.47 \]Using the standard normal distribution table, find the probability corresponding to \( z = 0.47 \), which is approximately 0.6808.

Step 3: Calculate Standard Error for Total Weight of 80 Pieces

When considering the total weight of luggage for 80 passengers, the standard deviation of the sample sum is \( \sigma_{total} = \sqrt{n} \cdot \sigma \) where \( n = 80 \). Thus:\[ \sigma_{total} = \sqrt{80} \cdot 10.63 \approx 94.86 \]

Step 4: Calculate Z-Scores for Luggage Weight Range of 80 Passengers

Convert the range 1200 to 2400 pounds into z-scores:\[ z_{1200} = \frac{1200 - (80 \cdot 40)}{94.86} = \frac{1200 - 3200}{94.86} \approx -21.08 \]\[ z_{2400} = \frac{2400 - (80 \cdot 40)}{94.86} = \frac{2400 - 3200}{94.86} \approx -8.45 \]

Step 5: Determine the Probability for the Weight Range

The z-scores calculated are extremely low, indicating the probabilities are near zero. Hence, it's virtually impossible (probability is essentially zero) for the total weight to fall between 1200 and 2400 pounds given typical circumstances.

Step 6: Locate the Inflection Point of the Normal Distribution

The point where the probability density function is decreasing most rapidly is the inflection point of the normal distribution, located one standard deviation from the mean.Find this point:\[ x_{inflection} = 40 + 10.63 \approx 50.63 \] (this is where the curve changes concavity).

Key Concepts

Mean and Standard Deviation

In statistics, the mean and standard deviation are key measures to understand data distribution, especially when examining normal distribution. The mean, denoted by \( \mu \), represents the average, or central value, around which the data is spread. In the context of airline luggage, this tells us that, on average, each piece of luggage weighs 40 pounds.

The standard deviation, denoted by \( \sigma \), measures the amount of variation or dispersion in a set of values. For the luggage problem, a standard deviation of 10.63 pounds means most luggage weights tend to vary 10.63 pounds from the 40 pound mean.

Understanding these two parameters is crucial because they define the shape and spread of the normal distribution:

• The further from the mean, the less likely a value is found.
• Approximately 68% of data falls within one standard deviation of the mean.
• A normal distribution is symmetrical around the mean.

Probability Calculations

Calculating probabilities in a normal distribution often involves measuring the likelihood of a value falling within a specific range. This requires converting the value to a z-score, a concept we'll cover later. Let's look at how these probability calculations work in practice.

In the luggage weight problem, part (a) involves finding the probability that luggage weighs less than 45 pounds. By calculating the z-score first, we then refer to a standard normal distribution table to find this probability, about 0.6808. This means there's roughly a 68.08% chance a piece of luggage weighs less than 45 pounds.

For part (b), which looks at a collective weight for 80 passengers, we calculate the standard error first. This is because when you sum many independent variables like luggage weights, their variability accumulates. Here, the standard error for all 80 pieces reflects how much the total weight fluctuates around the expected mean.

These probability calculations help quantify expectations when dealing with real-world uncertainties, guiding decisions with measurable risk assessments.

Z-Score

The z-score is a statistical measurement that describes a value's position relative to the mean of a group of values, measured in terms of standard deviation. Essentially, it tells you how many standard deviations an element is from the mean.

To calculate a z-score, you use the formula:

• \( z = \frac{x - \mu}{\sigma} \)

Where \( x \) is the value of interest, \( \mu \) is the mean, and \( \sigma \) is the standard deviation. In the luggage example, to find the z-score for a piece of luggage weighing 45 pounds, you plug into the formula resulting in a z-score of approximately 0.47. This indicates 45 pounds is 0.47 standard deviations above the mean luggage weight.

Z-scores are integral in the step involving probability tables, enabling translation of any normal distribution to a standard normal distribution. This normalization facilitates easier calculations by transforming any problem on a common scale.