/*! 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 52 The distribution of resistance f... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 4: Problem 52

The distribution of resistance for resistors of a certain type is known to be normal, with \(10 \%\) of all resistors having a resistance exceeding \(10.256\) ohms and \(5 \%\) having a resistance smaller than \(9.671\) ohms. What are the mean value and standard deviation of the resistance distribution?

Short Answer

Expert verified
The mean is 10 ohms and the standard deviation is 0.2 ohms.

Step by step solution

01

Understanding the Normal Distribution

The problem states that the resistance follows a normal distribution. We need to find its mean (\(\mu\)) and standard deviation (\(\sigma\)). This is a typical problem where we use percentile information to determine these parameters.
02

Interpreting the Given Percentiles

We know from the problem that 10% of the resistors have a resistance above 10.256 ohms, which means this is the 90th percentile of the distribution. Similarly, 5% are below 9.671 ohms, which corresponds to the 5th percentile.
03

Finding Z-scores from Percentiles

Use the standard normal distribution table (Z-table) to find the Z-score for the 90th and 5th percentiles. The Z-score for the 90th percentile is approximately 1.28, and for the 5th percentile, it's approximately -1.645.
04

Establish Equations from Z-scores

These Z-scores are calculated using the formula: \(Z = \frac{x - \mu}{\sigma}\). For the 90th percentile: \(1.28 = \frac{10.256 - \mu}{\sigma}\). For the 5th percentile: \(-1.645 = \frac{9.671 - \mu}{\sigma}\).
05

Solve the System of Equations

Solve the two equations simultaneously to find \(\mu\) and \(\sigma\). **Equation 1:** \(10.256 - \mu = 1.28\sigma\) **Equation 2:** \(9.671 - \mu = -1.645\sigma\)Subtract Equation 2 from Equation 1:\(10.256 - 9.671 = 1.28\sigma + 1.645\sigma\), \(0.585 = 2.925\sigma\). Thus, \(\sigma = \frac{0.585}{2.925} \approx 0.2\).
06

Find the Mean \(\mu\)

Substitute \(\sigma = 0.2\) back into one of the equations, for example, Equation 1:\(10.256 - \mu = 1.28 \times 0.2\)\(10.256 - \mu = 0.256\) \(\mu = 10.256 - 0.256 = 10\).

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 91影视!

Key Concepts

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

Mean Estimation
To estimate the mean of a normally distributed dataset, we need information from known data points or percentiles. In the given exercise, we determine the mean, denoted as \( \mu \), from known percentile data. Percentiles give us a snapshot of the data distribution, showing what percentage of data falls below or above certain values. Here, we're given percentiles which indirectly tell us about the mean:
  • The 90th percentile states that 10% of resistors have more than 10.256 ohms.
  • The 5th percentile shows that 5% have less than 9.671 ohms.
Understanding the distance between these points, and using statistical measures from the normal distribution, allows computation of \( \mu \). The mean estimation involves solving equations generated from provided percentiles, helping us derive the average around which our data clusters.
Standard Deviation Calculation
Standard deviation, denoted as \( \sigma \), measures the dispersion or spread of a set of data points within a normal distribution. It tells us how much each value typically varies from the mean. To find \( \sigma \), we use the Z-scores corresponding to known percentiles of the normal distribution.Given our exercise:
  • Z-score for the 90th percentile is approximately 1.28.
  • Z-score for the 5th percentile is approximately -1.645.
Using these Z-scores in conjunction with related formulas provides equations:
  • For the 90th percentile: \(1.28 = \frac{10.256 - \mu}{\sigma}\)
  • For the 5th percentile: \(-1.645 = \frac{9.671 - \mu}{\sigma}\)
Solving these simultaneously allows us to isolate \( \sigma \), giving clarity on how much variability there is in resistance measurements around the mean.Understanding this variability is integral, particularly in applications where consistency and quality control are vital.
Percentiles
Percentiles are key tools in describing data within a normal distribution. They indicate the relative standing of a value within a dataset, expressed as a percentage of the data below that value. For instance, the 90th percentile means 90% of data points are below this value. In our example:
  • 90th percentile with a resistance of 10.256 ohms means only 10% exceed this value.
  • 5th percentile with a resistance of 9.671 ohms indicates that only 5% are less than this value.
These values help us derive meaningful insights about data distribution, such as identifying the central tendency (mean) and measuring variability (standard deviation). Interpreting percentiles is fundamental for statistics and data analysis as they help locate data points, understand spread, and make informed predictions.

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

Commuting to work requires getting on a bus near home and then transferring to a second bus. If the waiting time (in minutes) at each stop has a uniform distribution with \(A=0\) and \(B=5\), then it can be shown that the total waiting time \(Y\) has the pdf $$ f(y)=\left\\{\begin{array}{cl} \frac{1}{25} y & 0 \leq y<5 \\ \frac{2}{5}-\frac{1}{25} y & 5 \leq y \leq 10 \\ 0 & y<0 \text { or } y>10 \end{array}\right. $$ a. Sketch the pdf of \(Y\). b. Verify that \(\int_{-\infty}^{\infty} f(y) d y=1\). c. What is the probability that total waiting time is at most 3 min? d. What is the probability that total waiting time is at most \(8 \mathrm{~min}\) ? e. What is the probability that total waiting time is between 3 and 8 min? f. What is the probability that total waiting time is either less than 2 min or more than 6 min?

Let \(X\) be the temperature in \({ }^{\circ} \mathrm{C}\) at which a chemical reaction takes place, and let \(Y\) be the temperature in \({ }^{\circ} \mathrm{F}\) (so \(Y=1.8 X+32\) ). a. If the median of the \(X\) distribution is \(\bar{\mu}\), show that \(1.8 \bar{\mu}+32\) is the median of the \(Y\) distribution. b. How is the 90 th percentile of the \(Y\) distribution related to the 90 th percentile of the \(X\) distribution? Verify your conjecture. c. More generally, if \(Y=a X+b\), how is any particular percentile of the \(Y\) distribution related to the corresponding percentile of the \(X\) distribution?

Let \(X\) have the pdf \(f_{X}(x)=2 / x^{3}, x>1\). Find the pdf of \(Y=\sqrt{X}\).

A machine that produces ball bearings has initially been set so that the true average diameter of the bearings it produces is \(500 \mathrm{in}\). A bearing is acceptable if its diameter is within .004 in. of this target value. Suppose, however, that the setting has changed during the course of production, so that the bearings have normally distributed diameters with mean value \(.499\) in. and standard deviation \(.002\) in. What percentage of the bearings produced will not be acceptable?

Let \(X\) have the uniform distribution on \([0,1]\). Find the pdf of \(Y=-\ln (X)\).
See all solutions

Recommended explanations on Math Textbooks

Probability and Statistics

Read Explanation

Statistics

Read Explanation

Applied Mathematics

Read Explanation

Pure Maths

Read Explanation

Decision Maths

Read Explanation

Discrete Mathematics

Read Explanation
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.