/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 80 The lengths of plate glass parts... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The lengths of plate glass parts are measured to the nearest tenth of a millimeter. The lengths are uniformly distributed with values at every tenth of a millimeter starting at 590.0 and continuing through 590.9. Determine the mean and variance of the lengths.

Short Answer

Expert verified
The mean is 590.45 mm and the variance is 0.0675 mm².

Step by step solution

01

Understanding the Uniform Distribution

The problem indicates that the lengths are uniformly distributed between 590.0 and 590.9. This means that every value within this range is equally likely. The values are defined at intervals of 0.1 millimeter, from 590.0 to 590.9.
02

Identifying the Parameters of the Distribution

In a uniform distribution, the format is generally from a minimum value (\( a \)) to a maximum value (\( b \)). For our distribution, \( a = 590.0 \) and \( b = 590.9 \).
03

Formula for the Mean of a Uniform Distribution

The mean \( \mu \) of a uniform distribution can be calculated using the formula: \[ \mu = \frac{a + b}{2} \]
04

Calculating the Mean

Using the formula \( \mu = \frac{a + b}{2} \), we substitute \( a = 590.0 \) and \( b = 590.9 \): \[ \mu = \frac{590.0 + 590.9}{2} = \frac{1180.9}{2} = 590.45 \]. The mean length is 590.45 mm.
05

Formula for the Variance of a Uniform Distribution

The variance \( \sigma^2 \) can be calculated using the formula: \[ \sigma^2 = \frac{(b - a)^2}{12} \]
06

Calculating the Variance

Substituting \( a = 590.0 \) and \( b = 590.9 \) into the variance formula: \[ \sigma^2 = \frac{(590.9 - 590.0)^2}{12} = \frac{0.9^2}{12} = \frac{0.81}{12} = 0.0675 \]. The variance is 0.0675 \( \text{mm}^2 \).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Mean of Uniform Distribution
In statistics, the term _"mean"_ represents the average value of a set of numbers. In the context of a uniform distribution, where every outcome is equally likely, the mean gives us a central value around which the data is distributed.
For data that is uniformly distributed between two values, the mean can be calculated using the straightforward formula:
  • \[ \mu = \frac{a + b}{2} \]
where \( a \) is the minimum value, and \( b \) is the maximum value.
This formula works because it weights each possible outcome equally, essentially finding the midpoint of the distribution.
Using the example of lengths ranging from 590.0 mm to 590.9 mm, the mean length is calculated as follows:
  • First, we add the two boundary values: 590.0 + 590.9 = 1180.9.
  • Next, we divide by 2 to find the average: \[ \mu = \frac{1180.9}{2} = 590.45 \]
This result, 590.45 mm, represents the mean of the distribution, offering a central point that characterizes the entire range of uniformly distributed lengths.
Variance of Uniform Distribution
Variance is a measure that tells us how much the values in a set differ from the mean. For a uniform distribution, calculating variance helps us understand the spread of the data across the distribution's range.
The specific formula for variance \( \sigma^2 \) in a uniform distribution is:
  • \[ \sigma^2 = \frac{(b - a)^2}{12} \]
Here, \( a \) and \( b \) represent the minimum and maximum values of the distribution, respectively.
The term \((b - a)^2\) gives us the squared range of the distribution, and dividing by 12 accounts for the equal likelihood of all data points, spreading the range evenly.
In our glass plate examples, substituting the values 590.0 mm and 590.9 mm into the formula gives us:
  • First, find the range: \( 590.9 - 590.0 = 0.9 \).
  • Square the range: \( 0.9^2 = 0.81 \).
  • Divide by 12 to find the variance: \[ \sigma^2 = \frac{0.81}{12} = 0.0675 \]
Thus, the variance is 0.0675 \( \text{mm}^2 \), indicating how closely packed the lengths are around the mean.
Uniformly Distributed Data
When data is said to be uniformly distributed, this means each value within a specified range is equally probable. Essentially, no outcome within the range is more or less likely than any other.
Uniform distribution is often visualized as a flat, horizontal line in a probability distribution graph, where the height of the line is constant.
In real-world applications, uniform distributions can occur in scenarios as diverse as the rolling of a fair die or the measurement of consistently manufactured products.
For example, if you are gauging the lengths of glass plates measured every tenth of a millimeter from 590.0 to 590.9, each potential value like 590.0, 590.1, and so on up to 590.9 is measured with equal likelihood.
This means:
  • There is no favoritism in the spread of data points.
  • Each interval within the given range holds the same significance.
  • The distribution creates a scenario where any random observation within the range has the same chance of selection.
Understanding this even disbursement can be useful in many practical and theoretical contexts, providing a straightforward model for calculations such as mean and variance.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The probability of a successful optical alignment in the assembly of an optical data storage product is 0.8 . Assume that the trials are independent. (a) What is the probability that the first successful alignment requires exactly four trials? (b) What is the probability that the first successful alignment requires at most four trials? (c) What is the probability that the first successful alignment requires at least four trials?

The number of flaws in bolts of cloth in textile manufacturing is assumed to be Poisson distributed with a mean of 0.1 flaw per square meter. (a) What is the probability that there are two flaws in one square meter of cloth? (b) What is the probability that there is one flaw in 10 square meters of cloth? (c) What is the probability that there are no flaws in 20 square meters of cloth? (d) What is the probability that there are at least two flaws in 10 square meters of cloth?

Orders arrive at a Web site according to a Poisson process with a mean of 12 per hour. Determine the following: (a) Probability of no orders in five minutes. (b) Probability of 3 or more orders in five minutes. (c) Length of a time interval such that the probability of no orders in an interval of this length is 0.001 .

Suppose that the number of customers who enter a bank in an hour is a Poisson random variable, and suppose that \(P(X=0)=0.05 .\) Determine the mean and variance of \(X .\)

A fault-tolerant system that processes transactions for a financial services firm uses three separate computers. If the operating computer fails, one of the two spares can be immediately switched online. After the second computer fails, the last computer can be immediately switched online. Assume that the probability of a failure during any transaction is \(10^{-8}\) and that the transactions can be considered to be independent events. (a) What is the mean number of transactions before all computers have failed? (b) What is the variance of the number of transactions before all computers have failed?

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.