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Chapter 5: Problem 18

The length of time that the battery in Hippolyta's cell phone will hold enough charge to operate acceptably is normally distributed with mean 25.6 hours and standard deviation 0.32 hour. Hippolyta forgot to charge her phone yesterday, so that at the moment she first wishes to use it today it has been 26 hours 18 minutes since the phone was last fully charged. Find the probability that the phone will operate properly.

Short Answer

Expert verified
The probability the phone will operate is about 1.48%.

Step by step solution

01

Convert Time to Hours

First, we need to convert the time since the phone was last charged into hours from the hours and minutes given. The problem states 26 hours and 18 minutes. The 18 minutes can be converted to hours by dividing 18 by 60, which equals 0.3 hours. Therefore, the total time since the last charge is 26 + 0.3 = 26.3 hours.
02

Identify Distribution Parameters

We are told that the time the battery holds a charge is normally distributed with a mean of 25.6 hours and a standard deviation of 0.32 hour. This means we are working with a normal distribution: \( \mu = 25.6 \) and \( \sigma = 0.32 \).
03

Calculate Z-Score

The Z-score is a way to standardize our variable so we can use the standard normal distribution table. The formula for the Z-score is: \[ Z = \frac{X - \mu}{\sigma} \] where \( X \) is 26.3 hours. Substituting the values in, we have: \[ Z = \frac{26.3 - 25.6}{0.32} = \frac{0.7}{0.32} \approx 2.1875 \].
04

Find Probability from Z-Score

Using the Z-score calculated, we can look up the value in a standard normal distribution table or use a calculator. The Z-score of 2.1875 corresponds to a probability of approximately 0.9852. However, since we are interested in the probability of the battery lasting beyond 26.3 hours, we need to calculate the complimentary probability: \( 1 - 0.9852 = 0.0148 \). Thus, the probability the phone will still work is approximately 1.48%.

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Key Concepts

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

Understanding the Normal Distribution
In statistics, a normal distribution is a type of continuous probability distribution for a real-valued random variable.
The most common representation of the normal distribution is the bell-shaped curve also known as the "Gaussian" curve.
This distribution is symmetric about its mean (average value), where most of the data tends to cluster around the mean.
  • The mean ( \( \mu \)) is the center point of the distribution.
  • The spread or "width" of the curve is determined by the standard deviation ( \( \sigma \)).
  • A larger standard deviation results in a wider curve, while a smaller one results in a narrower curve.
In the exercise, the time a battery lasts is normally distributed around a mean of 25.6 hours.
This means most batteries will last for about 25.6 hours. However, some might last slightly longer or shorter. Understanding this distribution is essential for predicting probabilities related to the battery life.
Z-Score Calculation Explained
A Z-score is a statistical measurement that describes a value's relation to the mean of a group of values.
Z-scores are expressed in terms of standard deviations from the mean.
If a Z-score is 0, it indicates that the data point's score is identical to the mean score.
Mapping Z-scores to probabilities involves standardizing your data so it can be compared to the standard normal distribution, which has a mean of 0 and a standard deviation of 1.
  • The formula for the Z-score is \[ Z = \frac{X - \mu}{\sigma} \]
  • Where \(X\) is your data point (e.g., 26.3 hours in the problem),
  • To find probabilities, Z-scores are used with the standard normal distribution reference tables or calculators.
In our exercise, we found the Z-score to be approximately 2.1875.
This indicates the battery's duration is about 2.1875 standard deviations higher than the mean battery life.
The Role of Standard Deviation
Standard deviation is a crucial concept in statistics that measures the amount of variation or dispersion in a set of values.
In simpler terms, it indicates how much individual data points differ from the mean.
A small standard deviation suggests that the values are closely clustered around the mean.
  • Mathematically, it's calculated as the square root of the variance.
  • A standard deviation of 0 means every data point is the same.
  • A higher standard deviation means a greater spread of data points.
In our battery life problem, the standard deviation is 0.32 hours, indicating that the times at which the battery fails to operate as required will generally lie within 0.32 hours of the mean. This helps in determining the unusualness or rarity of a data point, like the 26.3 hours of battery life being analyzed.

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Most popular questions from this chapter

A regulation hockey puck must weigh between 5.5 and 6 ounces. The weights \(X\) of pucks made by a particular process are normally distributed with mean 5.75 ounces and standard deviation 0.11 ounce. Find the probability that a puck made by this process will meet the weight standard.

Dogberry's alarm clock is battery operated. The battery could fail with equal probability at any time of the day or night. Every day Dogberry sets his alarm for 6: 30 a.m. and goes to bed at 10: 00 p.m. Find the probability that when the clock battery finally dies, it will do so at the most inconvenient time, between 10:00 p.m. and 6: 30 a.m.

The average finishing time among all high school boys in a particular track event in a certain state is 5 minutes 17 seconds. Times are normally distributed with standard deviation 12 seconds. a. The qualifying time in this event for participation in the state meet is to be set so that only the fastest \(5 \%\) of all runners qualify. Find the qualifying time. (Hint: Convert seconds to minutes.) b. In the western region of the state the times of all boys running in this event are normally distributed with standard deviation 12 seconds, but with mean 5 minutes 22 seconds. Find the proportion of boys from this region who qualify to run in this event in the state meet.

Scores on the common final exam given in a large enrollment multiple section course were normally distributed with mean 69.35 and standard deviation \(12.93 .\) The department has the rule that in order to receive an A in the course his score must be in the top \(10 \%\) of all exam scores. Find the minimum exam score that meets this requirement.

All students in a large enrollment multiple section course take common in- class exams and a common final, and submit common homework assignments. Course grades are assigned based on students' final overall scores, which are approximately normally distributed. The department assigns a C to students whose scores constitute the middle \(2 / 3\) of all scores. If scores this semester had mean 72.5 and standard deviation \(6.14,\) find the interval of scores that will be assigned a \(\mathrm{C}\)
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