91Ó°ÊÓ

Many bulk products-such as iron ore, coal, and raw sugar- -are sampled for quality by a method that requires many small samples to be taken periodically as the material is moving along a conveyor belt. The small samples are then combined and mixed to form one composite sample. Let \(Y_{i}\) denote the volume of the \(i\) th small sample from a particular lot and suppose that \(Y_{1}, Y_{2}, \ldots, Y_{n}\) constitute a random sample, with each \(Y_{i}\) value having mean \(\mu\) (in cubic inches) and variance \(\sigma^{2}\). The average volume \(\mu\) of the samples can be set by adjusting the size of the sampling device. Suppose that the variance \(\sigma^{2}\) of the volumes of the samples is known to be approximately 4. The total volume of the composite sample must exceed 200 cubic inches with probability approximately .95 when \(n=50\) small samples are selected. Determine a setting for \(\mu\) that will allow the sampling requirements to be satisfied.

Step by step solution

01

Understand the Total Volume Requirement

We are given that the total volume of the composite sample, which is the sum of all small samples, must exceed 200 cubic inches with a probability of 0.95. This total volume can be expressed as \( T = \sum_{i=1}^{n} Y_i \).

02

Express Total Volume in Terms of Mean and Variance

Since each small sample has the same mean \( \mu \) and variance \( \sigma^2 \), according to the properties of a random sample, the total mean volume will be \( n\mu \) and the total variance will be \( n\sigma^2 \). For \( n = 50 \), this gives a total mean of \( 50\mu \) and total variance of \( 50 \times 4 = 200 \).

03

Apply Central Limit Theorem

For a large value of \( n \), the Central Limit Theorem states that \( T \) is approximately normally distributed with mean \( n\mu \) and variance \( n\sigma^2 \), i.e., \( T \sim N(50\mu, 200) \).

04

Set Up Probability Requirement

We need the probability that the total volume \( T \) exceeds 200 cubic inches to be 0.95, i.e., \( P(T > 200) = 0.95 \). This can be rephrased using the standard normal distribution as \( P(Z > \frac{200 - 50\mu}{\sqrt{200}}) = 0.95 \), where \( Z \) is a standard normal variable.

05

Use Z-Score Table

From the standard normal distribution table, find the \( Z \) value corresponding to a probability of 0.05 (since the question is phrased as \( P(T > 200) = 0.95 \), which implies \( P(T \leq 200) = 0.05 \)). The \( Z \) value is approximately -1.645.

06

Plug in Values and Solve for \( \mu \)

Substitute the known \( Z \)-value, mean, and variance into the equation and solve for \( \mu \):\[-1.645 = \frac{200 - 50\mu}{\sqrt{200}}\]Simplifying:\[-1.645 \times \sqrt{200} = 200 - 50\mu\]\[-1.645 \times 14.14 \approx 200 - 50\mu\]\[-23.23 = 200 - 50\mu\]\[50\mu = 223.23\]\[\mu = 4.4646\]

07

Confirm the Calculations

Double-check the calculations for any errors to ensure accuracy. The calculations show \( \mu \approx 4.46 \). Therefore, setting \( \mu \approx 4.46 \) satisfies the requirement.

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Key Concepts

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

Composite Sampling

Composite sampling is a method often used in quality control scenarios, particularly for bulk products. The idea is to take multiple smaller samples from a larger batch and combine them to form a composite sample. This approach helps in assessing the overall quality of the product or batch.

In practice, smaller samples are collected periodically from a moving conveyor belt. Each sample is individually small, but by blending these small samples together, one can obtain a more meaningful single sample that represents the entire batch.

By using a composite sample, we can:

• An estimate the average quality of the entire batch more reliably.
• Reduce costs and labor expenses traditionally associated with testing numerous individual samples.
• Identify potential uniformities or anomalies within a batch.

The concept is widely applicable wherever bulk material inspection is necessary and serves as an efficient method to ensure compliance with quality standards.

Random Sample

A random sample refers to a subset of individuals or items selected from a larger population, chosen in such a way that every individual or item has an equal chance of being selected. In the context of this exercise, the small samples \(Y_1, Y_2, ..., Y_n\) are considered a random sample from the entire batch of material.

This randomness is vital in order to:

• Minimize any potential biases that might occur if selection were based on convenience or accessibility.
• Ensure that the sample is a good representation of the entire batch, leading to more accurate quality assessments.

Random sampling is key in achieving representative data and is foundational in statistics as it underpins the validity of statistical inference.

Variance Calculation

Variance provides a measure of dispersion or spread in a set of data. It quantifies how much the individual data points differ from the mean of the data set. In this exercise, each small sample has a known variance, denoted as \(\sigma^2 = 4\).

Variance is calculated as the average of the squared differences from the mean. Mathematically, for a data set with samples \(Y_1, Y_2, ..., Y_n\) and mean \(\mu\), the variance is given by:\[\sigma^2 = \frac{1}{n} \sum_{i=1}^{n} (Y_i - \mu)^2\]
Variance is critical for understanding the variability inherent in any dataset. High variance indicates that data points are spread out from the mean, while low variance suggests that they are clustered closely around the mean.

Z-score

A Z-score is a statistical measure that describes a value's relation to the mean of a group of values, measured in terms of standard deviations. In the context of this problem, it is used to determine the probability of the total composite volume exceeding a certain threshold.

The Z-score formula is:\[Z = \frac{X - \mu}{\sigma}\]where \(X\) is the value in question, \(\mu\) is the mean, and \(\sigma\) is the standard deviation.

In this exercise, the Z-score helps in translating the requirement that the total volume exceeds 200 cubic inches into a probability statement. By consulting a Z-score table, one can determine the needed settings for the sampling mean based on this probability requirement.

• A positive Z-score means the value is above the mean.
• A negative Z-score indicates the value is below the mean.
• A Z-score of 0 represents a value equal to the mean.

Understanding and utilizing Z-scores is essential for assessing probabilities and ensuring that statistical requirements are met.