Problem 9 - A Modern Introduction to Probability and Statistics: Understanding Why and How

Question

A factory produces links for heavy metal chains. The research lab of the factory models the length (in \(\mathrm{cm}\) ) of a link by the random variable \(X\), with expected value \(\mathrm{E}[X]=5\) and variance \(\operatorname{Var}(X)=0.04\). The length of a link is defined in such a way that the length of a chain is equal to the sum of the lengths of its links. The factory sells chains of 50 meters; to be on the safe side 1002 links are used for such chains. The factory guarantees that the chain is not shorter than 50 meters. If by chance a chain is too short, the customer is reimbursed, and a new chain is given for free. a. Give an estimate of the probability that for a chain of at least 50 meters more than 1002 links are needed. For what percentage of the chains does the factory have to reimburse clients and provide free chains? b. The sales department of the factory notices that it has to hand out a lot of free chains and asks the research lab what is wrong. After further investigations the research lab reports to the sales department that the expectation value 5 is incorrect, and that the correct value is \(4.99(\mathrm{~cm})\). Do you think that it was necessary to report such a minor change of this value?

Step-by-Step Solution

Answer

With the original mean, 5.7% of chains needed more than 1002 links. With the corrected mean, about 44% of chains need more links, highlighting the importance of precise estimations.

Step 1: Convert Chain Requirement

The chain should be at least 50 meters, which translates to 5000 cm. The factory uses 1002 links for this purpose. If 1002 links are not enough, more links will be needed.

Step 2: Calculate Total Expected Length with Original Mean

The expected total length of the chain can be calculated using the initial mean expected length for each link:\[ \mathrm{E}[S_n] = n \times \mathrm{E}[X] = 1002 \times 5 = 5010 \text{ cm} \]where \( n \) is the number of links (1002).

Step 3: Use Central Limit Theorem (CLT)

The total length \( S_n \) for 1002 links is approximately normal with a mean of \( 5010 \) cm. Since the variance of each link is \( 0.04 \) cm, the variance of the total length is \( 1002 \times 0.04 = 40.08 \). Hence, \( \operatorname{sd}(S_n) = \sqrt{40.08} \approx 6.33 \) cm.

Step 4: Apply Normal Distribution for Original Mean

The probability that 1002 links is not enough is the same as \( P(S_n < 5000) \) for \( S_n \sim \mathcal{N}(5010, 40.08) \). Calculate the standard score (z-score):\[ z = \frac{5000 - 5010}{6.33} \approx -1.58 \]Using a z-table, \( P(Z < -1.58) \approx 0.057 \) or 5.7% of the chains.

Step 5: Recalculate Probability with Corrected Mean

With the corrected mean \( 4.99 \), recalculate the expected total length:\[ \mathrm{E}[S_n] = n \times \mathrm{E}[X'] = 1002 \times 4.99 = 5000.98 \text{ cm} \]Calculate the new z-score for \( S_n < 5000 \):\[ z = \frac{5000 - 5000.98}{6.33} \approx -0.1547 \]The probability \( P(Z < -0.1547) \approx 0.44 \), which is about 44%.

Step 6: Assess Impact of Differences in Mean Length

Even a small decrease from 5 cm to 4.99 cm per link has a large impact, raising the probability of needing reimbursement from 5.7% to 44%. It shows that even minor errors in expected values at large scales can result in significant operational issues.

Key Concepts

Probability Distribution

A probability distribution is a fundamental concept in statistics that describes how the values of a random variable are distributed. It shows the possible values a random variable can take and the likelihood of each value occurring. In the context of the problem with the chain links, the random variable, denoted as \( X \), represents the length of an individual link measured in centimeters. This distribution helps in calculating probabilities related to different configurations of the chain lengths by allowing us to understand how likely certain lengths of the entire chain are.

• Probability distributions can be discrete or continuous. Discrete distributions have specific values (like rolling dice), while continuous ones cover a range (like measuring time or weight).
• In our problem, we deal with a continuous distribution because link lengths are measured on a continuous scale, in centimeters.

Properly understanding the probability distribution of \( X \) helps in making predictions about the chain as a whole, especially when using a large number of links.

Expected Value

The expected value, often denoted as \( E[X] \), is a measure of the central tendency of a random variable. It tells us what the average or mean value is expected to be over a large number of repetitions, which is useful for predicting long-term outcomes.

• The expected value is calculated as the sum of all possible values of a random variable, each multiplied by its probability of occurrence.
• In the chain link scenario, the expected value of each link length was initially given as 5 cm.

This means that if you measure a large number of links, on average, they should approximate this length.

When it was discovered that the actual expected value is slightly less, at 4.99 cm, this small change significantly impacted the overall expected length of the entire 1002-link chain due to the scaling effect. This is a striking example of how small errors in expected values can be magnified over large quantities.

Variance

Variance is a statistical measure that provides insight into the spread or dispersion of a set of values around the mean of a random variable. It informs us about how much the values differ from the expected value.

• The formula for variance is \( \operatorname{Var}(X) = E[(X - E[X])^2] \), which takes the average of the squared deviations from the mean.
• For our problem, the variance of each link's length is 0.04 cm².

Variance is essential in predicting how far from the expected total length the actual length might be for a chain of links. This helps in understanding the risk of a chain being shorter or longer than anticipated.

Lower variance means the lengths are more consistent, while higher variance indicates more fluctuation. Variance plays a crucial role in applying the Central Limit Theorem, which allows us to use normal distribution properties for sums of random variables like our chain links.

z-score

The z-score is a measure that describes the position of a data point in terms of standard deviations from the mean. It can be used to compare different data sets or to find probabilities in a standard normal distribution.

• Calculating a z-score involves subtracting the mean from the data point and dividing by the standard deviation: \( z = \frac{X - \mu}{\sigma} \).
• In our example, the z-score helps determine the probability that a chain will not meet the required length.

With the original expected length per link being 5 cm, the z-score calculation showed that there was about a 5.7% probability that a chain would be too short.

However, with the corrected expected length of 4.99 cm, the z-score shifted, resulting in a much higher probability of 44% that the chain would be insufficient. This illustrates how sensitive z-scores and probabilities are to changes in the mean or standard deviation, impacting practical outcomes like reimbursements.