/*! 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 88 The efficiency (in lumens per wa... [FREE SOLUTION] | 91Ó°ÊÓ

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The efficiency (in lumens per watt) of light bulbs of a certain type has population mean 9.5 and standard deviation. \(5,\) according to production specifications. The specifications for a room in which eight of these bulbs are to be installed call for the average efficiency of the eight bulbs to exceed 10. Find the probability that this specification for the room will be met, assuming that efficiency measurements are normally distributed.

Short Answer

Expert verified
The probability that the average efficiency exceeds 10 is approximately 0.3888.

Step by step solution

01

Understand the Problem

To solve the problem, we need to find the probability that the average efficiency of 8 randomly selected light bulbs exceeds 10. The population mean is 9.5 and the standard deviation is 5.
02

Calculate the Sampling Distribution Parameters

Since we are dealing with the average of a sample of 8 bulbs, we use the Central Limit Theorem. The mean of the sample distribution is the same as the population mean, which is 9.5. The standard deviation of the sample distribution (also known as the standard error) is calculated by dividing the population standard deviation by the square root of the sample size: \[ \sigma_{\bar{x}} = \frac{5}{\sqrt{8}} \approx 1.7678 \]
03

Determine the Z-Score

To find the probability that the sample mean exceeds 10, calculate the Z-score using the formula: \[ Z = \frac{\bar{x} - \mu}{\sigma_{\bar{x}}} \]Plug in the numbers:\[ Z = \frac{10 - 9.5}{1.7678} \approx 0.2833 \]
04

Find the Probability Using the Z-Score

Consult the standard normal distribution table or use a calculator to find the probability that a Z-score is less than 0.2833. This gives us the cumulative probability for having means less than 10. Looking it up, we find \( P(Z < 0.2833) \approx 0.6112 \).
05

Determine the Complement for Exceeding 10

Since we want the probability of the mean being greater than 10, we need the complement of the cumulative probability calculated: \[ P(Z > 0.2833) = 1 - 0.6112 = 0.3888 \]

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Key Concepts

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

Normal Distribution
A normal distribution is a type of continuous probability distribution that is symmetrical and bell-shaped. It represents the distribution of many natural phenomena. In mathematics, it is often depicted by the iconic bell curve. This distribution is defined by two main parameters:
  • Mean (\( \mu \)): This is the central peak of the bell curve. The mean indicates the average or central value of the data set. In our exercise, it was given as 9.5 lumens per watt.
  • Standard Deviation (\( \sigma \)): This describes the spread or width of the bell curve. A larger standard deviation means that the data points are more spread out around the mean.
Most of the data within a normal distribution lie close to the mean. About 68% of data falls within one standard deviation of the mean, making it easier to predict probabilities for certain outcomes. In our exercise, it mentioned that the efficiencies are normally distributed, meaning we can apply these principles to solve our problem.
Sampling Distribution
The sampling distribution refers to the probability distribution of a statistic, such as a sample mean, which is obtained from a large number of samples drawn from a specific population. According to the Central Limit Theorem, the sampling distribution of the sample mean will be approximately normal, regardless of the shape of the population distribution, provided the sample size is sufficiently large. In our exercise, we have a sample of 8 light bulbs. The sample mean's distribution mirrors the population mean, which is given as 9.5. We assume normality for this sampling distribution because the population data are normally distributed. This means the results can be generalized to the population with certain reliability, especially regarding predicting probabilities using this sampled data set.
Z-score
A Z-score is a statistical measure that describes a value's relationship to the mean of a group of values, measured in terms of standard deviations from the mean. It indicates how many standard deviations a particular score is from the mean.
  • Formula: The Z-score can be calculated by:\[ Z = \frac{\bar{x} - \mu}{\sigma_{\bar{x}}} \]where \( \bar{x} \) is the sample mean, \( \mu \) is the population mean, and \( \sigma_{\bar{x}} \) is the standard error.
In our exercise, we found that a Z-score of 0.2833 corresponds to the sample mean of 10, which we are interested to investigate. A positive Z-score indicates the sample mean is above the population mean, which helps in assessing the likelihood of occurrences. Using Z-scores, one can easily convert individual scores into a standard form to determine probabilities.
Standard Error
Standard error is a measure of the variation or dispersion of sample means over a series of samples drawn from the same population. It provides an estimate of the sampling distribution's standard deviation.
  • Calculation: It can be computed using the formula:\[ \sigma_{\bar{x}} = \frac{\sigma}{\sqrt{n}} \]where \( \sigma \) is the population standard deviation, and \( n \) is the sample size.
In our exercise, the standard error helps to understand how much the sample mean of 8 light bulbs deviates from the true population mean. With a standard deviation of 5 and a sample size of 8, the standard error was calculated as approximately 1.7678. This small value indicates the sample mean is a close estimate of the population mean, allowing us to predict the probability more accurately.

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Most popular questions from this chapter

A pollster believes that \(20 \%\) of the voters in a certain area favor a bond issue. If 64 voters are randomly sampled from the large number of voters in this area, approximate the probability that the sampled fraction of voters favoring the bond issue will not differ from the true fraction by more than .06.

a. If \(U\) has a \(\chi^{2}\) distribution with \(\nu\) df, find \(E(U)\) and \(V(U)\). b. Using the results of Theorem 7.3 , find \(E\left(S^{2}\right)\) and \(V\left(S^{2}\right)\) when \(Y_{1}, Y_{2}, \ldots, Y_{n}\) is a random sample from a normal distribution with mean \(\mu\) and variance \(\sigma^{2}\).

What does the sampling distribution of the sample mean look like if samples are taken from an approximately normal distribution? Use the applet Sampling Distribution of the Mean (at https://college.cengage.com/nextbook/statistics/wackerly 966371/student/html/index.html) to complete the following. The population to be sampled is approximately normally distributed with \(\mu=16.50\) and \(\sigma=6.03\) (these values are given above the population histogram and denoted \(M\) and S, respectively). a. Use the button "Next Obs" to select a single value from the approximately normal population. Click the button four more times to complete a sample of size \(5 .\) What value did you obtain for the mean of this sample? Locate this value on the bottom histogram (the histogram for the values of \(Y\) ). b. Click the button "Reset" to clear the middle graph. Click the button "Next Obs" five more times to obtain another sample of size 5 from the population. What value did you obtain for the mean of this new sample? Is the value that you obtained equal to the value you obtained in part (a)? Why or why not? c. Use the button "1 Sample" eight more times to obtain a total of ten values of the sample mean. Look at the histogram of these ten means. i. What do you observe? ii. How does the mean of these \(10 \bar{y}\) -values compare to the population mean \(\mu\) ? d. Use the button "1 Sample" until you have obtained and plotted 25 realized values for the sample mean \(\bar{Y}\), each based on a sample of size 5 . i. What do you observe about the shape of the histogram of the 25 values of \(\bar{y}_{i}\) $$ i=1,2, \ldots, 25 ? $$ ii. How does the value of the standard deviation of the \(25 \bar{y}\) values compare with the theoretical value for \(\sigma_{Y}\) obtained in Example 5.27 where we showed that, if \(Y\) is computed based on a sample of size \(n,\) then \(V(\bar{Y})=\sigma^{2} / n ?\) e. Click the button "1000 Samples" a few times, observing changes to the histogram as you generate more and more realized values of the sample mean. What do you observe about the shape of the resulting histogram for the simulated sampling distribution of \(Y\) ? f. Click the button "Toggle Normal" to overlay (in green) the normal distribution with the same mean and standard deviation as the set of values of \(Y\) that you previously generated. Does this normal distribution appear to be a good approximation to the sampling distribution of \(Y ?\)

One-hour carbon monoxide concentrations in air samples from a large city average 12 ppm (parts per million) with standard deviation 9 ppm. a. Do you think that carbon monoxide concentrations in air samples from this city are normally distributed? Why or why not? b. Find the probability that the average concentration in 100 randomly selected samples will exceed 14 ppm.

An airline finds that \(5 \%\) of the persons who make reservations on a certain flight do not show up for the flight. If the airline sells 160 tickets for a flight with only 155 seats, what is the probability that a seat will be available for every person holding a reservation and planning to fly?

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