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Chapter 26: Problem 31

Under EU legislation on harmonisation, all kippers are to weigh \(0.2000 \mathrm{~kg}\) and vendors who sell underweight kippers must be fined by their government. The weight of a kipper is normally distributed with a mean of \(0.2000 \mathrm{~kg}\) and a standard deviation of \(0.0100 \mathrm{~kg}\). They are packed in cartons of 100 and large quantities of them are sold. Every day a carton is to be selected at random from each vendor and tested according to one of the following schemes, which have been approved for the purpose. (a) The entire carton is weighed and the vendor is fined 2500 euros if the average weight of a kipper is less than \(0.1975 \mathrm{~kg}\). (b) Twenty-five kippers are selected at random from the carton; the vendor is fined 100 euros if the average weight of a kipper is less than \(0.1980 \mathrm{~kg}\). (c) Kippers are removed one at a time, at random, until one has been found that weighs more than \(0.2000 \mathrm{~kg}\); the vendor is fined \(n(n-1)\) euros, where \(n\) is the number of kippers removed.

Short Answer

Expert verified
For scheme (a), the expected fine is 15.5 euros/day. For scheme (b), it is 15.87 euros/day. For scheme (c), it is 2 euros/day.

Step by step solution

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01

Introduction and Setup

Identify the known values: mean weight \(\text{µ} = 0.2000 \text{ kg}\), standard deviation \( \text{σ} = 0.0100 \text{ kg} \), and the sample sizes for each scheme. Understand that the weight distribution is normal.
02

Scheme (a) - Entire Carton Weighed

Calculate the standard error for the weight of kippers in a carton of 100: \( \text{SE} = \frac{ \text{σ} }{\text{√n}} = \frac{0.0100 \text{ kg}}{\text{√100}} = 0.0010 \text{ kg} \). Find the z-score for an average weight of \(0.1975 \text{ kg} \): \(\text{z} = \frac{0.1975 \text{ kg} - 0.2000 \text{ kg}}{0.0010 \text{ kg}} = -2.5 \). Using the z-score table, determine the probability corresponding to z = -2.5.
03

Scheme (a) - Fine Calculation

Using the z-score table, find that the probability (P) corresponding to z = -2.5 is approximately 0.0062. Hence, the probability that the average weight is less than 0.1975 kg is 0.0062, meaning the expected fine amount is \(2500 \text{ euros} \times 0.0062 = 15.5 \text{ euros/day} \).
04

Scheme (b) - Twenty-Five Kippers Sampled

Calculate the standard error for 25 kippers: \( \text{SE} = \frac {0.0100 \text{ kg}}{\text{√25}} = 0.0020 \text{ kg} \). Find the z-score for an average weight of 0.1980 kg: \( \text{z} = \frac{0.1980 \text{ kg} - 0.2000 \text{ kg}}{0.0020 \text{ kg}} = -1.0 \). Using the z-score table, determine the probability corresponding to z = -1.0.
05

Scheme (b) - Fine Calculation

Using the z-score table, find that the probability (P) corresponding to z = -1.0 is approximately 0.1587. Hence, the probability that the average weight is less than 0.1980 kg is 0.1587, meaning the expected fine amount is \(100 \text{ euros} \times 0.1587 = 15.87 \text{ euros/day} \).
06

Scheme (c) - One Kipper at a Time

Determine the probability of a kipper weighing more than 0.2000 kg. Given the distribution, this probability is 0.5 since the mean is 0.2000 kg. Determine the expected fine by recognizing the formula for expected value of \(n(n-1) \text{ where} n \text{ is a geometric random variable with p = 0.5} \). The expected value of \ n(n-1) \text{ can be derived from properties of geometric distributions}.
07

Scheme (c) - Fine Calculation

Using the properties of geometric distribution, \( \text{E}[n] = \frac{1}{p} = 2 \text{,} \text{ and} \text{E}[n(n-1)] = 2 \). Applying these values, the vendor is fined an average of \(2 \text{ euros/day} \).

Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Probability Theory
Probability Theory forms the backbone of predicting the likelihood of different outcomes. In our problem, it's crucial for determining the chances of a vendor getting fined based on the weight of kippers.
Probability theory helps us understand the distribution of weights and calculate the likelihood of various scenarios. For instance, we use the normal distribution and z-scores to find the probabilities associated with different average weights of kippers.

Probability theory is used for:
- Predicting outcomes based on known distributions
- Calculating risks and penalties

By understanding these probabilities, we can estimate how often certain weight thresholds are met or exceeded.
Statistical Methods
Statistical methods allow us to analyze data and make inferences. In this exercise, we use statistical methods like standard error calculation and z-score analysis to make sense of the weight distributions of kippers.

Some key statistical methods used are:
- Calculating Mean \( \text{µ} = 0.2000 \text{ kg} \)
- Determining Standard Deviation \( \text{σ} = 0.0100 \text{ kg} \)
- Computing Standard Error for samples to understand the variability

By applying these methods, we more accurately understand the data and make informed decisions about fines.
Standard Error Calculation
Standard error (SE) measures the accuracy with which a sample distribution represents a population. It’s fundamental in quality control because it helps estimate the variability of the sample mean.

SE is calculated using the formula:
\[ \text{SE} = \frac{ \text{σ} }{ \sqrt{n}} \]
\( \text{where} \text{σ} = \text{standard deviation} \text{ and} n = \text{sample size} \)

In the provided example:
- For 100 kippers: \[ \text{SE} = \frac{0.0100 \text{ kg}}{ \sqrt{100}} = 0.0010 \text{ kg} \]
- For 25 kippers: \[ \text{SE} = \frac{0.0100 \text{ kg}}{ \sqrt{25}} = 0.0020 \text{ kg} \]

SE helps us determine how much we expect the sample averages to deviate from the true population mean.
z-Score Analysis
A z-score quantifies the difference between a sample data point and the mean, divided by the standard deviation. It tells us how many standard deviations away a particular point is from the mean.

The formula for a z-score is:
\[ \text{z} = \frac{ \text{X} - \text{µ} }{ \text{σ} } \]
\( \text{where} \text{X} = \text{sample value} \text{,} \text{µ} = \text{mean} \text{,} \text{σ} = \text{standard error} \)

For instance, in Scheme (a):
- Mean: 0.2000 kg
- SE: 0.0010 kg
- Weight threshold: 0.1975 kg
- Resulting z-score: \[ \text{z} = \frac{0.1975 \text{ kg} - 0.2000 \text{ kg} }{ 0.0010 \text{ kg}} = -2.5 \]

We use the z-score to find the corresponding probability, which helps in determining likelihoods and fines.
Geometric Distribution
Geometric Distribution is useful for scenarios where we're counting the number of trials up to the first success. In Scheme (c):

The vendor removes kippers until finding one weighing over 0.2000 kg. Here, each trial has a success probability \( \text{p} = 0.5 \text{ (half over 0.2000 kg)} \)
Expected number of trials to first success: 2

Using the properties of the geometric distribution:
\
  • Mean \[ \text{E}[n] = \frac{1}{p} = 2 \]
  • Expected fine \[ \text{E}[ n (n-1) ] = 2 \]

This means that, on average, the vendor is fined 2 euros/day under Scheme (c).

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Most popular questions from this chapter

A continuous random variable \(X\) is uniformly distributed over the interval \([-c, c]\). A sample of \(2 n+1\) values of \(X\) is selected at random and the random variable \(Z\) is defined as the median of that sample. Show that \(Z\) is distributed over \([-c, c]\) with probability density function, $$ f_{n}(z)=\frac{(2 n+1) !}{(n !)^{2}(2 c)^{2 n+1}}\left(c^{2}-z^{2}\right)^{n} $$ Find the variance of \(Z\).

A discrete random variable \(X\) takes integer values \(n=0,1, \ldots, N\) with probabilities \(p_{n} .\) A second random variable \(Y\) is defined as \(Y=(X-\mu)^{2}\), where \(\mu\) is the expectation value of \(X\). Prove that the covariance of \(X\) and \(Y\) is given by $$ \operatorname{Cov}[X, Y]=\sum_{n=0}^{N} n^{3} p_{n}-3 \mu \sum_{n=0}^{N} n^{2} p_{n}+2 \mu^{3} $$ Now suppose that \(X\) takes all its possible values with equal probability and hence demonstrate that two random variables can be uncorrelated even though one is defined in terms of the other.

A particle is confined to the one-dimensional space \(0 \leq x \leq a\) and classically it can be in any small interval \(d x\) with equal probability. However, quantum mechanics gives the result that the probability distribution is proportional to \(\sin ^{2}(n \pi x / a)\), where \(n\) is an integer. Find the variance in the particle's position in both the classical and quantum mechanical pictures and show that, although they differ, the latter tends to the former in the limit of large \(n\), in agreement with the correspondence principle of physics.

An electronics assembly firm buys its microchips from three different suppliers; half of them are bought from firm \(X\), whilst firms \(Y\) and \(Z\) supply \(30 \%\) and \(20 \%\) respectively. The suppliers use different quality- control procedures and the percentages of defective chips are \(2 \%, 4 \%\) and \(4 \%\) for \(X, Y\) and \(Z\) respectively. The probabilities that a defective chip will fail two or more assembly-line tests are \(40 \%, 60 \%\) and \(80 \%\) respectively, whilst all defective chips have a \(10 \%\) chance of escaping detection. An assembler finds a chip that fails only one test. What is the probability that it came from supplier \(X\) ?

A tennis tournament is arranged on a straight knockout basis for \(2^{n}\) players and for each round, except the final, opponents for those still in the competition are drawn at random. The quality of the field is so even that in any match it is equally likely that either player will win. Two of the players have surnames that begin with ' \(Q^{\prime}\). Find the probabilities that they play each other (a) in the final, (b) at some stage in the tournament.
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