/*! 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 Resistors to be used in a circui... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Resistors to be used in a circuit have average resistance 200 ohms and standard deviation 10 ohms. Suppose 25 of these resistors are randomly selected to be used in a circuit. a. What is the probability that the average resistance for the 25 resistors is between 199 and 202 ohms? b. Find the probability that the total resistance does not exceed 5100 ohms. [Hint: see Example 7.9.]

Short Answer

Expert verified
a. Probability is 0.5328. b. Probability is 0.9772.

Step by step solution

01

Understand the Problem

We have a sample of 25 resistors with a known average resistance and standard deviation. We need to find the probability of certain conditions involving the sample mean and total resistance.
02

Define Known Values

The average resistance (mean) \( \mu = 200 \) ohms, and the standard deviation \( \sigma = 10 \) ohms. We have \( n = 25 \) resistors.
03

Calculate the Standard Error of the Mean

The standard error of the mean (SEM) is given by \( \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} = \frac{10}{\sqrt{25}} = 2 \) ohms.
04

Find the Z-Scores for Part a

To find the probability that the average resistance is between 199 and 202 ohms, calculate the Z-scores: \( z_1 = \frac{199 - 200}{2} = -0.5 \) and \( z_2 = \frac{202 - 200}{2} = 1 \).
05

Calculate the Probability for Part a

Using a standard normal distribution table, find the probability for \( z_1 = -0.5 \) which is approximately 0.3085, and \( z_2 = 1 \) which is approximately 0.8413. The probability that the average resistance is between 199 and 202 ohms is \( 0.8413 - 0.3085 = 0.5328 \).
06

Find the Z-Score for Part b

To find the total resistance not exceeding 5100 ohms, consider \( \bar{x}_{max} = \frac{5100}{25} = 204 \) ohms as the average. Then calculate the Z-score: \( z_3 = \frac{204 - 200}{2} = 2 \).
07

Calculate the Probability for Part b

Using the standard normal distribution table, the probability that \( z_3 \) is less than 2 is about 0.9772. This is the probability that the total resistance does not exceed 5100 ohms.

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.

Probability
Probability is an essential part of mathematical statistics that quantifies how likely an event is to occur. In our resistor problem, we are interested in the probability of specific outcomes related to the sample mean and total resistance. When calculating probabilities in scenarios involving continuous random variables, such as the resistance of resistors, we typically utilize the concept of a normal distribution.

To find the probability between two values, such as the average resistance ranging from 199 to 202 ohms, we first determine Z-scores for those values using our sample mean and standard deviation. Once we have these Z-scores, we refer to a standard normal distribution table to find the probabilities. Subtracting these probabilities gives us the probability of our average falling within the specified range.

Probability tables or software can facilitate this process, making it much more accessible by providing precomputed probability values for different Z-scores.
Standard Deviation
The standard deviation is a measure of how spread out numbers are from the mean. In the context of our resistor exercise, it quantifies the variability of resistance values within our sample. In mathematical statistics, standard deviation provides insight into the consistency of the data.

For our set of resistors, a standard deviation of 10 ohms implies that the resistance values typically fall within 10 ohms above or below the average resistance of 200 ohms. However, to effectively use the standard deviation in our problem, we need to calculate the standard error of the mean, which adjusts the standard deviation based on the sample size. This is done by dividing the standard deviation by the square root of the sample size (25 in this case), resulting in a standard error of 2 ohms.

This reduced value assists in precisely estimating probabilities concerning the sample mean.
Normal Distribution
The normal distribution is a key concept in statistics, often referred to as the "bell curve" due to its shape. It represents the distribution of many types of data that cluster around a mean value. In the resistor problem, this distribution is assumed for calculating probabilities related to the sample mean.

A normal distribution is characterized by two parameters: the mean (where the peak of the distribution lies) and the standard deviation (which determines the spread of the curve). In mathematical statistics, it is typically standardized to a mean of zero and a standard deviation of one, known as the standard normal distribution.

Whenever we calculate probabilities using Z-scores, we rely on this standardized form, allowing us to translate our specific situation to a universal scale that can be easily evaluated using normal distribution tables.
Sample Mean
The sample mean is the average of a sample set of data points, and it plays a crucial role in the field of statistics. For our resistors, the sample mean gives us an estimate of the average resistance among the selected 25 resistors. Understanding its properties is key to solving the given problem.

Using the central limit theorem, we know that the distribution of the sample mean will approximate a normal distribution if the sample size is large enough. In our example, even with a sample size of just 25, the sample mean closely follows a normal distribution.

This allows us to apply normal distribution statistics, like finding Z-scores, to determine probabilities for questions like whether the total resistance exceeds a certain value. When each resistance value is added together, as in part b of the problem, we convert this total into a sample mean scenario by dividing by the sample size to facilitate the use of the normal distribution efficiently.

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

From each of two normal populations with identical means and with standard deviations of 6.40 and \(7.20,\) independent random samples of 64 observations are drawn. Find the probability that the difference between the means of the samples exceeds. 6 in absolute value.

Access the applet Normal Approximation to Binomial Distribution (at https://college.cengage.com/nextbook/statistics/wackerly 966371/student/html/index.html). When the applet is started, it displays the details in Example 7.11 and Figure 7.9 . Initially, the display contains only the binomial histogram and the exact value (calculated using the binomial probability function) for \(p(8)=P(Y=8) .\) Scroll down a little and click the button "Toggle Normal Approximation" to overlay the normal density with mean 10 and standard deviation \(\sqrt{6}=2.449\) the same mean and standard deviation as the binomial random variable \(Y\). You will get a graph superior to the one in Figure \(7.9 .\) a. How many probability mass or density functions are displayed? b. Enter 0 in the box labeled "Begin" and press the enter key. What probabilities do you obtain? c. Refer to part (b). On the line where the approximating normal probability is displayed, you see the expression Normal: $$ P(-0.5<=k<=8.5)=0.2701 $$ Why are the .5s in this expression?

The Environmental Protection Agency is concerned with the problem of setting criteria for the amounts of certain toxic chemicals to be allowed in freshwater lakes and rivers. A common measure of toxicity for any pollutant is the concentration of the pollutant that will kill half of the test species in a given amount of time (usually 96 hours for fish species). This measure is called LC50 (lethal concentration killing 50\% of the test species). In many studies, the values contained in the natural logarithm of LC50 measurements are normally distributed, and, hence, the analysis is based on \(\ln (\mathrm{LC} 50)\) data. Studies of the effects of copper on a certain species of fish (say, species A) show the variance of In(LC50) measurements to be around. 4 with concentration measurements in milligrams per liter. If \(n=10\) studies on \(L C 50\) for copper are to be completed, find the probability that the sample mean of \(\ln (\text { LC50 })\) will differ from the true population mean by no more than .5.

Let \(Y_{1}, Y_{2}, \ldots, Y_{5}\) be a random sample of size 5 from a normal population with mean 0 and variance 1 and let \(\bar{Y}=(1 / 5) \sum_{i=1}^{5} Y_{i} .\) Let \(Y_{6}\) be another independent observation from the same population. What is the distribution of a. \(W=\sum_{i=1}^{5} Y_{i}^{2} ?\) Why? b. \(U=\sum_{i=1}^{5}\left(Y_{i}-\bar{Y}\right)^{2} ?\) Why? c. \(\sum_{i=1}^{5}\left(Y_{i}-\bar{Y}\right)^{2}+Y_{6}^{2} ?\) Why?

Just as the difference between two sample means is normally distributed for large samples, so is the difference between two sample proportions. That is, if \(Y_{1}\) and \(Y_{2}\) are independent binomial random variables with parameters \(\left(n_{1}, p_{1}\right)\) and \(\left(n_{2}, p_{2}\right),\) respectively, then \(\left(Y_{1} / n_{1}\right)-\left(Y_{2} / n_{2}\right)\) is approximately normally distributed for large values of \(n_{1}\) and \(n_{2}\) a. Find \(E\left(\frac{Y_{1}}{n_{1}}-\frac{Y_{2}}{n_{2}}\right)\) b. Find \(V\left(\frac{Y_{1}}{n_{1}}-\frac{Y_{2}}{n_{2}}\right)\)

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.