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Chapter 5: Problem 52

The lifetime of a certain type of battery is normally distributed with mean value 10 hours and standard deviation 1 hour. There are four batteries in a package. What lifetime value is such that the total lifetime of all batteries in a package exceeds that value for only \(5 \%\) of all packages?

Short Answer

Expert verified
The lifetime value is 43.29 hours.

Step by step solution

01

Understand the Problem

We need to find a lifetime value such that only 5% of packages have a total battery lifetime that exceeds this value. Each battery's life is normally distributed with a mean of 10 hours and a standard deviation of 1 hour.
02

Determine Combined Mean and Standard Deviation

Since there are 4 batteries in a package, the total lifetime mean for all 4 batteries is \(4\times10 = 40\) hours. The standard deviation of the total lifetime is given by \(\sqrt{4} \times 1 = 2\) hours.
03

Find the Z-score for 5% Exceedance

We are looking for a point on the normal distribution curve where 5% of the data lies to the right. This corresponds to the 95th percentile. The Z-score for 95% is 1.645, since 5% is in the upper tail.
04

Calculate the Lifetime Value

Using the Z-score formula, \(X = \mu + Z \times \sigma\), where \(\mu=40\), \(Z=1.645\), and \(\sigma=2\), calculate the target lifetime: \[X = 40 + 1.645 \times 2 = 40 + 3.29 = 43.29\] hours.

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Key Concepts

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

Battery Lifetime Analysis
Battery lifetime analysis is a crucial exercise many undertake to determine the reliability and efficiency of batteries in a given context. The goal is often to predict how long a battery will last under normal usage conditions. In this example, we are dealing with batteries whose lifetimes follow a normal distribution鈥攁 common statistical model used in such analyses.

Batteries, like many natural phenomena, don't have a fixed lifetime. Rather, their lifetimes can vary, which we capture using a probability distribution.
  • The lifetime is characterized by two key parameters: the mean value and the standard deviation.
  • The mean gives us the average expected lifetime, while the standard deviation indicates how much the lifetimes can vary from this average.
Understanding how these factors interplay helps in predicting scenarios like the total lifetime for a package of batteries.
Mean Value
The mean value, often referred to as the expected value, is a fundamental concept in statistics and probability. It represents the average outcome one can expect from a random variable. In the context of battery lifetime, the mean value is 10 hours.

This means that, on average, each battery in the package will last for 10 hours. For four batteries, the total mean lifetime extends to 40 hours as each contributes its average to the total.
  • Mean acts as a central measure, showing the centre point of the data set or distribution.
  • In a normal distribution, it is located at the peak of the bell curve, representing the most probable value.
Calculating the mean is crucial when analyzing large sets of data, as it helps provide a baseline expectation.
Standard Deviation
Standard deviation is a measure of how dispersed or spread out data points are within a set. In battery lifetime analysis, it depicts how much the battery lifes can deviate from the mean. For the given problem, the standard deviation for a single battery is 1 hour.

When considering multiple batteries together, as in a package of four, the standard deviation of their combined lifetime changes. To find the total standard deviation for these four batteries, we use the formula: \[ \sigma_{total} = \sqrt{n} \times \sigma_{individual} \]where \(n\) is the number of batteries.

This results in a total standard deviation of 2 hours for the package. Understanding standard deviation allows us to gauge the variability of battery life and establish confidence levels around the mean.
Percentile Calculations
Percentile calculations in the context of normal distributions help to identify thresholds within the data set. These thresholds tell us the percentage of data points that fall below or exceed a certain value. For instance, the 95th percentile falls where 95% of the data lies to the left and 5% lies to the right.

In the battery example, we seek the point where only 5% of packages have a lifetime exceeding it. This position corresponds to the 95th percentile of the total lifetime distribution.
  • To find this, we calculate the Z-score associated with the 95th percentile, which is 1.645 in this context.
  • We then apply the formula \( X = \mu + Z \times \sigma \) to determine the lifetime value X.
This calculation gives us a threshold of 43.29 hours. Understanding percentiles and how to calculate them helps in determining critical values in data analysis.

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Most popular questions from this chapter

A stockroom currently has 30 components of a certain type, of which 8 were provided by supplier 1,10 by supplier 2 , and 12 by supplier 3. Six of these are to be randomly selected for a particular assembly. Let \(X=\) the number of supplier l's components selected, \(Y=\) the number of supplier 2 's components selected, and \(p(x, y)\) denote the joint pmf of \(X\) and \(Y\). a. What is \(p(3,2)\) ? [Hint: Each sample of size 6 is equally likely to be selected. Therefore, \(p(3,2)=\) (number of outcomes with \(X=3\) and \(Y=2\) )/(total number of outcomes). Now use the product rule for counting to obtain the numerator and denominator.] b. Using the logic of part (a), obtain \(p(x, y)\). (This can be thought of as a multivariate hypergeometric distribution-sampling without replacement from a finite population consisting of more than two categories.)

One piece of PVC pipe is to be inserted inside another piece. The length of the first piece is normally distributed with mean value \(20 \mathrm{in}\). and standard deviation \(5 \mathrm{in}\). The length of the second piece is a normal rv with mean and standard deviation \(15 \mathrm{in}\). and \(.4 \mathrm{in}\)., respectively. The amount of overlap is normally distributed with mean value 1 in. and standard deviation \(1 \mathrm{in}\). Assuming that the lengths and amount of overlap are independent of one another, what is the probability that the total length after insertion is between \(34.5 \mathrm{in}\). and 35 in.?

Let \(X_{1}, X_{2}, X_{3}, X_{4}, X_{5}\), and \(X_{6}\) denote the numbers of blue, brown, green, orange, red, and yellow \(\mathrm{M} \& \mathrm{M}\) candies, respectively, in a sample of size \(n\). Then these \(X_{l}\) 's have a multinomial distribution. According to the M\&M Web site, the color proportions are \(p_{1}=.24, p_{2}=.13, p_{3}=.16\), \(p_{4}=.20, p_{5}=.13\), and \(p_{6}=.14\). a. If \(n=12\), what is the probability that there are exactly two M\&Ms of each color? b. For \(n=20\), what is the probability that there are at most five orange candies? [Hint: Think of an orange candy as a success and any other color as a failure.] c. In a sample of \(20 \mathrm{M} \& \mathrm{Ms}\), what is the probability that the number of candies that are blue, green, or orange is at least 10 ?

Suppose that you have ten lightbulbs, that the lifetime of each is independent of all the other lifetimes, and that each lifetime has an exponential distribution with parameter \(\lambda\). a. What is the probability that all ten bulbs fail before time \(r\) ? b. What is the probability that exactly \(k\) of the ten bulbs fail before time \(t\) ? c. Suppose that nine of the bulbs have lifetimes that are exponentially distributed with parameter \(\lambda\) and that the remaining bulb has a lifetime that is exponentially distributed with parameter \(\theta\) (it is made by another manufacturer). What is the probability that exactly five of the ten bulbs fail before time \(t\) ?

Two airplanes are flying in the same direction in adjacent parallel corridors. At time \(t=0\), the first airplane is \(10 \mathrm{~km}\) ahead of the second one. Suppose the speed of the first plane \((\mathrm{km} / \mathrm{hr})\) is normally distributed with mean 520 and standard deviation 10 and the second plane's speed is also normally distributed with mean and standard deviation 500 and 10 , respectively. a. What is the probability that after 2 hr of flying, the second plane has not caught up to the first plane? b. Determine the probability that the planes are separated by at most \(10 \mathrm{~km}\) after \(2 \mathrm{hr}\).
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