/*! 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 6 Find the indicated term of each ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Find the indicated term of each binomial expansion. The time until recharge for a battery in a laptop computer under common conditions is normally distributed with a mean of 260 minutes and a standard deviation of 50 minutes. a. What is the probability that a battery lasts more than four hours? b. What are the quartiles (the \(25 \%\) and \(75 \%\) values) of battery life? c. What value of life in minutes is exceeded with \(95 \%\) probability?

Short Answer

Expert verified
a. 0.6554; b. 226.3 min (25th), 293.7 min (75th); c. 342.25 min

Step by step solution

01

Convert Time to Standard Normal Distribution

First, convert the time of interest, 4 hours (240 minutes), into a standard normal variable (z-score). The formula for the z-score is: \[ z = \frac{X - \mu}{\sigma} \]where \( X \) is the time of interest, \( \mu = 260 \) minutes is the mean, and \( \sigma = 50 \) minutes is the standard deviation. Plugging in the numbers: \[ z = \frac{240 - 260}{50} = \frac{-20}{50} = -0.4 \]
02

Find the Probability from the Z-Table

Use the z-table to find the probability that the z-score is less than -0.4. The z-table shows this probability as approximately 0.3446. Therefore, the probability that the battery lasts less than 240 minutes is 0.3446.
03

Calculate the Probability of Lasting More than 4 Hours

Since we need the probability of the battery lasting more than 240 minutes, calculate: \[ P(X > 240) = 1 - P(X < 240) = 1 - 0.3446 = 0.6554 \]Thus, the probability that the battery lasts more than four hours is approximately 0.6554.
04

Calculate the 25th and 75th Percentiles (Quartiles)

To find the 25th and 75th percentiles, utilize the inverse of the cumulative normal distribution function (often found with statistical software or z-tables). - The z-score for the 25th percentile is approximately -0.674.- The z-score for the 75th percentile is approximately 0.674.Calculate the corresponding times using the formula: \[X = \mu + z \times \sigma \]For the 25th percentile:\[ X_{25} = 260 + (-0.674) \times 50 \approx 226.3 \] minutes.For the 75th percentile:\[ X_{75} = 260 + (0.674) \times 50 \approx 293.7 \] minutes.
05

Calculate the Value Exceeded with 95% Probability

The z-score corresponding to the top 5% (equivalent to 95% being below this z-score) is approximately 1.645. Calculate the time that corresponds to this z-score:\[ X = \mu + z \times \sigma = 260 + 1.645 \times 50 = 342.25 \] Thus, a battery life of approximately 342.25 minutes is exceeded with 95% probability.

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.

Z-score
The Z-score is a fundamental concept when working with normal distributions. It is a measure of how many standard deviations a data point, such as a battery life in our case, is from the mean value. In simple words, the Z-score tells us where a specific value stands in relation to the average. The formula for calculating a Z-score is:\[z = \frac{X - \mu}{\sigma}\]where:
  • \( X \) is the data point for which we want to find the Z-score,
  • \( \mu \) is the mean of the distribution, and
  • \( \sigma \) is the standard deviation of the distribution.

In our example, we calculated a Z-score for a battery life of 240 minutes. By substituting into the formula, we obtained a Z-score of \(-0.4\), indicating that the battery life of 240 minutes is 0.4 standard deviations below the mean of 260 minutes. Understanding Z-scores is crucial for finding probabilities and identifying where data lies relative to a normal distribution.
Probability
Probability helps us determine how likely an event is to occur within a particular distribution. For a normally distributed dataset like battery life, we can use the Z-score to find the probability of certain outcomes.
Once we have a Z-score, we can refer to a Z-table, a tool that tells us the probability of a value falling below a given Z-score. This is known as a cumulative probability. In the battery example, a Z-score of \(-0.4\) had a cumulative probability of 0.3446. That means there's a 34.46% chance that a battery won't last up to 240 minutes.
  • To find the probability of a battery lasting more than a certain time, we subtract the cumulative probability from 1. This gives us the probability that a battery will exceed that time.
  • In our instance, the probability that a battery lasts more than 240 minutes is 0.6554, or 65.54%.
Understanding probability in the context of normal distribution helps us make predictions and decisions based on statistical data.
Quartiles
Quartiles are values that divide a data set into four equal parts, helping us understand the data's distribution. In statistics, the first quartile (Q1) and the third quartile (Q3) show where 25% and 75% of the data points lie, respectively.
Calculating quartiles involves finding specific Z-scores and converting them into actual values from the data set. For a normal distribution:
  • The 25th percentile Z-score is approximately \(-0.674\), and for the laptop battery, this corresponds to a life of about 226.3 minutes.
  • The 75th percentile Z-score is approximately 0.674, translating to a battery life of around 293.7 minutes.

These values help describe the spread of the dataset. The quartiles tell us that 25% of batteries will last less than 226.3 minutes, and 75% will last more than this but less than 293.7 minutes. Analyzing quartiles can help manufacturers and users understand battery performance better.

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

Data from www.centralhudsonlabs.com determined the mean number of insect fragments in 225 -gram chocolate bars was 14.4, but three brands had insect contamination more than twice the average. See the U.S. Food and Drug Administration-Center for Food Safety and Applied Nutrition for Defect Action Levels for food products. Assume that the number of fragments (contaminants) follows a Poisson distribution. a. If you consume a 225-gram bar from a brand at the mean contamination level, what is the probability of no insect contaminants? b. Suppose that you consume a bar that is one-fifth the size tested (45 grams) from a brand at the mean contamination level. What is the probability of no insect contaminants? c. If you consume seven 28.35 -gram (one-ounce) bars this week from a brand at the mean contamination level, what is the probability that you consume one or more insect fragments in more than one bar? \(?\) d. Is the probability of contamination more than twice the mean of 14.4 unusual, or can it be considered typical variation? Explain.

A Web site randomly selects among 10 products to discount each day. The color printer of interest to you is discounted today. a. What is the expected number of days until this product is again discounted? b. What is the probability that this product is first discounted again exactly 10 days from now? c. If the product is not discounted for the next five days, what is the probability that it is first discounted again 15 days from now? d. What is the probability that this product is first discounted again within three or fewer days?

WP The number of surface flaws in plastic panels used in the interior of automobiles has a Poisson distribution with a mean of 0.05 flaw per square foot of plastic panel. Assume that an automobile interior contains 10 square feet of plastic panel. a. What is the probability that there are no surface flaws in an auto's interior? b. If 10 cars are sold to a rental company, what is the probability that none of the 10 cars has any surface flaws? c. If 10 cars are sold to a rental company, what is the probability that at most 1 car has any surface flaws?

The probability that a visitor to a Web site provides contact data for additional information is \(0.01 .\) Assume that 1000 visitors to the site behave independently. Determine the following probabilities: a. No visitor provides contact data. b. Exactly 10 visitors provide contact data. c. More than 3 visitors provide contact data.

The random variable \(X\) has a binomial distribution with \(n=10\) and \(p=0.01\). Determine the following probabilities. a. \(P(X=5)\) b. \(P(X \leq 2)\) c. \(P(X \geq 9)\) d. \(P(3 \leq X<5)\)

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.