/*! 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 85 The print on the package of Sylv... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The print on the package of Sylvania CFL \(65 \mathrm{~W}\) replacement bulbs that use only \(16 \mathrm{~W}\) claims that these bulbs have an average life of 8000 hours. Assume that the distribution of lives of all such bulbs is normal with a mean of 8000 hours and a standard deviation of 400 hours. Let \(x\) be the life of a randomly selected such light bulb. a. Find \(x\) so that about \(22.5 \%\) of such light bulbs have lives longer than this value. b. Find \(x\) so that about \(63 \%\) of such light bulbs have lives shorter than this value.

Short Answer

Expert verified
Part a: About \(22.5\%\) of light bulbs have lives longer than 8300 hours. Part b: About \(63\%\) of light bulbs have lives shorter than 8120 hours.

Step by step solution

01

- Convert percentages to z-scores

The given percentages need to be converted to z-scores. The z-score corresponding to \(22.5\%\) in the right tail of the standard normal distribution can be found in a z-table or using an online calculator. Similarly, the z-score corresponding to \(63\%\) cumulative from the left of the standard normal distribution can be calculated.
02

- Calculation for part a

The z-score for the upper \(22.5\%\) is approximately \(0.75\). We then use the formula \(Z = (X-\mu) / \sigma\), where \(Z\) is the z-score, \(X\) is the value we are trying to find, \(\mu\) is the mean, and \(\sigma\) is the standard deviation. From the given data, \(\mu = 8000\) hours and \(\sigma = 400\) hours. The formula can be rearranged to solve for \(X\), giving \(X = Z\sigma + \mu\). Substituting the appropriate values gives \(X = 0.75(400) + 8000\), which simplifies to \(x = 8300\) hours.
03

- Calculation for part b

The z-score for the lower \(63\%\) is about \(0.3\). Again, using the formula \(Z = (X-\mu) / \sigma\), and rearranging for \(X\), we get \(X = Z\sigma + \mu\). Substituting the appropriate values gives \(X= 0.3(400) + 8000\), which simplifies to \(x = 8120\) hours. Hence, about \(63\%\) of such light bulbs have lives shorter than 8120 hours.

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.

Normal Distribution
The normal distribution is a fundamental concept in statistics that describes how the values of a variable are distributed. It is often depicted as a bell-shaped curve that is symmetrical around the mean. This is because most of the data points tend to cluster around the mean, with fewer points appearing as you move further away in either direction.
- Mean: This is the central value around which the data is distributed. In the problem, the mean life of a light bulb is 8000 hours.
- Variance: This describes how spread out the data is from the mean.
- Standard deviation: A measure that indicates the extent of deviation for a group as a whole from the mean. In our case, it is 400 hours.
The normal distribution is used frequently because it applies to many natural phenomena. Understanding it allows us to make predictions and probabilities about a population based on a sample.
Z-Score Calculation
To understand where a particular value stands within a normal distribution, we use the z-score. A z-score indicates how many standard deviations a data point is from the mean. It's a method of standardizing scores to create comparability.
The formula to compute the z-score is: \( Z = \frac{X - \mu}{\sigma} \), where:
  • \(X\) is the data point you're evaluating.
  • \(\mu\) is the mean of the distribution.
  • \(\sigma\) is the standard deviation.
For the given exercise, calculating a z-score allows us to determine what percentage of bulbs fall above or below a certain hour threshold. For instance, a z-score of 0.75 indicates that a data point is 0.75 standard deviations above the mean.
Statistics Problems
Statistics problems like these often involve finding probabilities related to real-world scenarios, such as light bulb lifespan. By distinguishing percentages in a normal distribution, we can solve for individual values.
The steps include:
  • Identifying the given percentage that relates to the cumulative distribution.
  • Use of z-tables or calculators to find corresponding z-scores.
  • Application of these z-scores to find specific data points within the distribution.
These steps guide us in interpreting data and making informed predictions or decisions based on statistical evidence. For this problem, understanding these methods helped solve for the bulb lifespan at given probabilities efficiently.
Standard Deviation
Standard deviation is a vital statistical measure representing the amount of variation or dispersion in a data set. A low standard deviation points to data being close to the mean, while a high standard deviation indicates data is more spread out.
It is calculated as the square root of the variance and helps describe how values differ from the mean:
\[ \sigma = \sqrt{\frac{1}{N} \sum_{i=1}^{N} (X_i - \mu)^2} \]
In the exercise, the standard deviation of 400 hours provides a context of how much the bulb life's hour count typically varies from the mean of 8000 hours. Appreciating this variability is key in assessing the reliability and expectations of the light bulbs.

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 transmission on a model of a specific car has a warranty for 40,000 miles. It is known that the life of such a transmission has a normal distribution with a mean of 72,000 miles and a standard deviation of 13,000 miles. a. What percentage of the transmissions will fail before the end of the warranty period? b. What percentage of the transmissions will be good for more than 100,000 miles?

For a binomial probability distribution, \(n=20\) and \(p=.60\). a. Find the probability \(P(x=14)\) by using the table of binomial probabilities (Table I of Appendix C). b. Find the probability \(P(x=14)\) by using the normal distribution as an approximation to the binomial distribution. What is the difference between this approximation and the exact probability calculated in part a?

The Bank of Connecticut issues Visa and MasterCard credit cards. It is estimated that the balances on all Visa credit cards issued by the Bank of Connecticut have a mean of \(\$ 845\) and a standard deviation of \(\$ 270 .\) Assume that the balances on all these Visa cards follow a normal distribution. a. What is the probability that a randomly selected Visa card issued by this bank has a balance between \(\$ 1000\) and \(\$ 1440\) ? b. What percentage of the Visa cards issued by this bank have a balance of \(\$ 730\) or more?

A charter bus company is advertising a singles outing on a bus that holds 60 passengers. The company has found that, on average, \(10 \%\) of ticket holders do not show up for such trips; hence, the company routinely overbooks such trips. Assume that passengers act independently of one another. a. If the company sells 65 tickets, what is the probability that the bus can hold all the ticket holders who actually show up? In other words, find the probability that 60 or fewer passengers show up. b. What is the largest number of tickets the company can sell and still be at least \(95 \%\) sure that the bus can hold all the ticket holders who actually show up?

For the standard normal distribution, what is the area within two standard deviations of the mean?

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.