/*! 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 3 The resistances of 50 resistors ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 23: Problem 3

The resistances of 50 resistors are measured and the results recorded as follows: $$ \begin{array}{cc} \hline \text { Resistance }(\Omega) & \text { Frequency } \\ \hline 5.0 & 17 \\ 5.5 & 12 \\ 6.0 & 10 \\ 6.5 & 6 \\ 7.0 & 5 \\ \hline \end{array} $$ Calculate the standard deviation of the measurements.

Short Answer

Expert verified
Answer: The standard deviation of the resistance measurements is approximately 0.66惟.

Step by step solution

01

Calculate the mean of the resistances

To calculate the mean resistance, multiply each resistance value by its frequency and sum those products, then divide the sum by the total number of resistances. Mean = \(\frac{(5.0\times17)+(5.5\times12)+(6.0\times10)+(6.5\times6)+(7.0\times5)}{50}\) Mean = \(\frac{85+66+60+39+35}{50}\) Mean = \(\frac{285}{50}\) Mean = \(5.7\) The mean resistance value is 5.7惟.
02

Calculate the variance

To calculate the variance, find the squared difference between each resistance value and the mean (5.7), then multiply the squared difference by the respective frequency, sum all the results and divide by the total number of resistances. Variance = \(\frac{(5.0-5.7)^2\times17+(5.5-5.7)^2\times12+(6.0-5.7)^2\times10+(6.5-5.7)^2\times6+(7.0-5.7)^2\times5}{50}\) Variance = \(\frac{(-0.7)^2\times17+(-0.2)^2\times12+(0.3)^2\times10+(0.8)^2\times6+(1.3)^2\times5}{50}\) Variance = \(\frac{0.49\times17+0.04\times12+0.09\times 10+0.64\times6+1.69\times5}{50}\) Variance = \(\frac{8.33+0.48+0.90+3.84+8.45}{50}\) Variance = \(\frac{22}{50}\) Variance = \(0.44\)
03

Calculate the standard deviation

Finally, the standard deviation is the square root of the variance. Standard Deviation = \(\sqrt{0.44}\) Standard Deviation 鈮� \(0.66\) The standard deviation of the resistance measurements is approximately 0.66惟.

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.

Variance
Variance is a key concept in statistics and measures how much the values in a data set differ from the mean. In other words, it tells us how spread out the numbers are. Calculating variance is the first step in finding standard deviation, which further simplifies the measure of spread.
To calculate the variance:
  • First, subtract the mean from each value in the data set. This shows how far each value is from the average.
  • Next, square each of these differences. Squaring them gets rid of negative values and emphasizes larger differences.
  • Then, multiply these squared differences by their corresponding frequencies.
  • Finally, sum up all these products and divide by the total number of items in the set.
This formula gives us an average of the squared differences within the dataset. In our example, after performing these steps with the resistance values, we found that the variance was 0.44.
Frequency Distribution
Frequency distribution helps us understand how often each value occurs in a dataset. It's particularly useful when dealing with numerous data points as it simplifies analysis by describing the dataset in terms of frequency of occurrences.
Imagine you have a dataset of resistance measurements. Instead of listing each value, frequency distribution summarizes it as:
  • 5.0 惟 occurs 17 times
  • 5.5 惟 occurs 12 times
  • 6.0 惟 occurs 10 times
  • 6.5 惟 occurs 6 times
  • 7.0 惟 occurs 5 times
Presenting data in this manner not only makes it easier to process but also lays the groundwork for calculating statistics such as the mean and variance. Understanding frequency distribution is crucial as it assists in expressing data concisely and identifies the concentration of measurements around different values.
Mean Calculation
The mean, commonly known as the average, gives us a central value that summarizes the entire dataset. It's the sum of all the data points divided by the number of points. In datasets grouped with frequencies, the calculation goes a step further.
To calculate the mean with frequency distribution:
  • Multiply each value by its frequency to account for how often each value appears.
  • Add up all these products.
  • Divide the sum by the total number of data points in the dataset.
In our example with resistances, we found the mean by using these steps and discovered that the average resistance is 5.7 惟. Knowing the mean is important as it acts as a baseline for further computations like variance and standard deviation, offering insight into the dataset's central tendency.

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

Components are made by machines A, B and C. Machine A makes \(35 \%\) of the components, machine B makes \(25 \%\) and machine C makes the rest. Two components are picked at random. Calculate the probability that (a) both are made by machine \(\mathrm{C}\) (b) one is made by machine A and one is made by machine B (c) exactly one is made by machine \(\mathrm{A}\) (d) at least one is made by machine B (e) both are made by the same machine.

The lifespans, \(L\), of 2500 components were monitored and recorded in Table 2 . $$ \begin{array}{rr} \hline \text { Lifespan, } L \text { (hours) } & \text { Frequency } \\ \hline 0 \leq L \leq 5000 & 16 \\ 500012000 & 145 \\ \hline \end{array} $$ Calculate the probability that a component picked at random has a lifespan (a) of between 8000 and 11000 hours (b) of more than 11000 hours (c) of more than 10000 hours given that it has already lasted for at least 8000 hours (d) of between 11000 and 12000 hours given that it has already lasted for at least 9000 hours.

The probability that a motor will malfunction within 5 years of manufacture is \(0.03\). Out of eight motors calculate the probability that within 5 years of manufacture (a) all eight will malfunction (b) six will malfunction (c) none will malfunction.

Studies of a particular type of car tyre show that the mileage for which it can be used legally follows a normal distribution with mean 38000 miles and standard deviation 2500 miles. The manufacturers claim that ' 9 out of 10 of our tyres last more than 35000 miles'. Is the claim justified?

The lengths of components have a normal distribution, with a mean of \(7 \mathrm{~cm}\) and a standard deviation of \(0.03 \mathrm{~cm}\). Calculate the probability that a component chosen at random has a length (a) between \(6.95 \mathrm{~cm}\) and \(7.02 \mathrm{~cm}\) (b) more than \(7.05 \mathrm{~cm}\) (c) less than \(6.96 \mathrm{~cm}\) (d) between \(6.95 \mathrm{~cm}\) and \(6.99 \mathrm{~cm}\).
See all solutions

Recommended explanations on Physics Textbooks

Fields in Physics

Read Explanation

Scientific Method Physics

Read Explanation

Astrophysics

Read Explanation

Torque and Rotational Motion

Read Explanation

Modern Physics

Read Explanation

Further Mechanics and Thermal Physics

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.