Question

An airplane with room for 100 passengers has a total baggage limit of 6,000 pounds. Suppose that the weight of the baggage checked by an individual passenger, \(x,\) has a mean of 50 pounds and a standard deviation of 20 pounds. If 100 passengers will board a flight, what is the approximate probability that the total weight of their baggage will exceed the limit? (Hint: With \(n=100,\) the total weight exceeds the limit when the mean weight \(\bar{x}\) exceeds \(6,000 / 100 .)\)

Short answer

The approximate probability that the total weight of their baggage will exceed the limit is 0. This result, however, is an approximation and in real life scenarios this probability could be a very small positive number rather than zero.

Step-by-step solution

Calculate The Mean And Standard Deviation Of The Sum

Given that the mean weight of an individual's baggage is 50 pounds and the standard deviation is 20 pounds, the mean total weight with 100 passengers is given by \(n * \mu = 100 * 50 = 5000\) pounds. The standard deviation of the total weight is given by \(\sqrt{n} * \sigma = 10 * 20 = 200\) pounds.

Standardize The Limit

The z-score is the standard score which informs us how many standard deviations an element is from the mean. We can calculate it using the formula \(Z = \frac{X - \mu}{\sigma}\). With \(X = 6000\) (the limit), \(\mu = 5000\) (mean total weight), and \(\sigma = 200\) (standard deviation of the total weight), we get \(Z = \frac{6000 - 5000}{200} = 5\). This means that the limit is 5 standard deviations above the mean.

Find The Probability

This final step involves finding the probability that the z-score of the total weight of baggage exceeds 5. This is the same as the probability that the total weight of baggage exceeds the limit. Looking up the z-score of 5 in a standard normal distribution table, we find that the area to the left of 5 is practically 1. The probability that a z-score is above 5 (total weight of baggage exceeding the limit) is therefore approximately \(1 - 1 = 0\).

Key concepts

z-score

The concept of a z-score is fundamental in the realm of statistics, serving as a bridge between individual data points and the broader context of a data set's distribution. A z-score, simply put, is a measure that describes the position of a specific data point in relation to the mean of the data set, using the standard deviation as the unit of measurement.

For example, in our airplane baggage scenario, each bag's weight is a data point within our distribution. By converting the absolute weight to a z-score, we enable the comparison of this weight against the distribution's average weight. When you calculate the z-score using the formula \( z = \frac{x - \mu}{\sigma} \), you're positioning the value x in the context of the entire distribution span. A z-score of 1 means the value is one standard deviation above the mean, whereas a z-score of -1 points to one standard deviation below the mean.

standard deviation

Standard deviation is an invaluable statistic for quantifying the level of dispersion or spread in a collection of data. It indicates to what extent the individual data points diverge from the mean of the data set. In less technical terms, the standard deviation reveals how tightly clustered or widely spread the data is.

In the situation of the airplane's baggage limit, the standard deviation provides insight into the variability we might anticipate in the passengers' individual baggage weights. A smaller standard deviation implies that most baggage weights are close to the average weight, suggesting consistency among passenger bags. Conversely, a larger standard deviation suggests a higher degree of unpredictability, with some bags much heavier or lighter than the average. Understanding the concept of standard deviation is crucial because it directly impacts the z-score calculation and our subsequent interpretation of where individual baggage weights stand in relation to the overall limit.

Central Limit Theorem

The Central Limit Theorem (CLT) is a key principle in statistics that provides a pathway to understanding the expected distribution of a sample mean when an experiment is repeated numerous times. It states that if you take sufficiently large random samples from a population, the distribution of the sample means will tend towards a normal distribution, regardless of the population's initial distribution shape.

This theorem underlies the reason why we can treat the total baggage weight in our airplane problem as a normally distributed variable, despite not knowing the actual distribution of individual baggage weights. Essentially, as we consider the average weight of 100 passengers' baggage, the CLT comes into play, guaranteeing that with such a large sample size, the distribution of these averages will approximate normality. This assumption is pivotal in calculating probabilities associated with exceeding the baggage limit.

normal distribution

A normal distribution, often symbolized by the classic 'bell curve,' is a probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean. It is defined by its mean and its variance or standard deviation. In practice, many variables in nature, industry, and research follow a roughly normal distribution, which considerably simplifies analysis.

Specific to our exercise, the total weight of the passengers' baggage is presumed to form a normal distribution, given the large sample size as indicated by the CLT. Utilizing a normal distribution, we can easily calculate the probability of events, such as the total baggage weight exceeding a certain limit, by referencing areas under the curve. This representation is particularly useful because it allows us to utilize z-scores and standard normal distribution tables to find probabilities that would otherwise require complex calculations.