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Chapter 5: Problem 7

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 95th percentile total lifetime is approximately 43.29 hours.

Step by step solution

01

Understand the Problem

The problem asks for a lifetime value such that, for 5% of all packages, the combined lifetime of all four batteries exceeds this value. The lifetime of each battery is normally distributed with a mean of 10 hours and a standard deviation of 1 hour.
02

Define the Total Lifetime Distribution

Since the lifetimes are normally distributed and independent, the total lifetime for all four batteries will also be normally distributed. If each battery has a mean lifetime of 10 hours, the mean lifetime for four batteries is 4 times 10, which is 40 hours. Similarly, since the standard deviation of one battery is 1 hour, for four batteries it is \( ext{SD}_{total} = ext{SD} imes ext{number of batteries} = 1 imes ext{sqrt}(4) = 2\).
03

Calculate the 95th Percentile

We need to find the value that corresponds to the 95th percentile for the total lifetime. Given the total mean of 40 hours and total standard deviation of 2 hours, we use the Z-score table to find the Z-score corresponding to 95%, which is approximately 1.645.
04

Apply the Z-score Formula

The formula to convert a Z-score to a raw score in a normal distribution is \( X = ext{mean} + Z imes ext{SD} \). Substitute the values: \( X = 40 + 1.645 imes 2 = 40 + 3.29 = 43.29\).
05

Interpret the Result

This calculated value means that for only 5% of all packages, the combined lifetime of the batteries exceeds 43.29 hours.

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Key Concepts

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

Z-score
The Z-score is a way to standardize individual data points in a normally distributed dataset. It tells us how many standard deviations a data point is from the mean. This makes it easier to understand the position of the data within the distribution. For example, we convert values into Z-scores to compare them to a standard normal distribution.
Calculating a Z-score involves using the formula: \[ Z = \frac{X - \text{mean}}{\text{SD}} \] where:
  • \( X \) is the value we're examining
  • \( \text{mean} \) is the average of the dataset
  • \( \text{SD} \) is the standard deviation

The Z-score is particularly useful in finding percentiles or probabilities. This means we can determine how likely a certain value, or a range of values, is. For instance, in the exercise, we used the Z-score to find out what total lifetime value for the batteries falls within the 95th percentile.

By using Z-score tables, we can quickly look up the probability of a Z-score occurring, allowing us to make precise statistical inferences.
Standard Deviation
Standard deviation (SD) measures the amount of variation or dispersion in a set of values. Small SD means the data points are close to the mean, while a large SD shows more spread out values. In normal distribution, this helps understand how data points differ from the average.

Mathematically, SD is the square root of the variance. Variance is the average of squared differences from the mean: \[ \text{SD} = \sqrt{\frac{1}{N}\sum_{i=1}^{N} (X_i - \text{mean})^2} \] where:
  • \( N \) is the number of data points
  • \( X_i \) is each data point

In the exercise, the SD of the battery's lifetime was 1 hour. For four batteries, we needed to calculate the total SD using the formula for the sum of independent variables' variances, resulting in 2 hours.

Understanding SD provides insight into how much individual observations might fluctuate around the mean. It plays a crucial role in predictions and setting expectations for any normally distributed data.
Mean
The mean, often known as the average, is the sum of all data points divided by the number of points. It's the central value of a set of numbers and is a core concept in statistics for understanding data distributions.

To calculate the mean, use the formula:\[ \text{mean} = \frac{1}{N} \sum_{i=1}^{N} X_i \] where:
  • \( N \) is the total number of observations
  • \( X_i \) represents each data point

In the battery exercise, the mean lifetime of one battery was 10 hours. Since the batteries are independent, the mean for four batteries simply becomes 4 times the mean of one, thus 40 hours.

The mean provides a simple numerical measure of the center of the distribution. It's essential for predicting outcomes and understanding where data sits within a normal distribution. This understanding can guide decisions based on typical expected outcomes.

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Most popular questions from this chapter

Suppose the amount of liquid dispensed by a certain machine is uniformly distributed with lower limit \(A=8 \mathrm{oz}\) and upper limit \(B=10 \mathrm{oz}\). Describe how you would carry out simulation experiments to compare the sampling distribution of the (sample) fourth spread for sample sizes \(n=5,10,20\), and \(30 .\)

Suppose that for a certain individual, calorie intake at breakfast is a random variable with expected value 500 and standard deviation 50 , calorie intake at lunch is random with expected value 900 and standard deviation 100 , and calorie intake at dinner is a random variable with expected value 2000 and standard deviation 180. Assuming that intakes at different meals are independent of one another, what is the probability that average calorie intake per day over the next \((365\) day) year is at most 3500 ? [Hint: Let \(X_{i}, Y_{\mathrm{j}}\), and \(Z_{i}\) denote the three calorie intakes on day \(i\). Then total intake is given by \(\Sigma\left(X_{i}+Y_{f}+Z_{j}\right)\).]

Let \(X_{1}, X_{2}\), and \(X_{3}\) represent the times necessary to perform three successive repair tasks at a certain service facility. Suppose they are independent, normal rv's with expected values \(\mu_{1}, \mu_{2}\), and \(\mu_{3}\) and variances \(\sigma_{1}^{2}, \sigma_{2}^{2}\), and \(\sigma_{3}^{2}\), respectively. a. If \(\mu=\mu_{2}=\mu_{3}=60\) and \(\sigma_{1}^{2}=\sigma_{2}^{2}=\sigma_{3}^{2}=15\), calculate \(P\left(T_{o} \leq 200\right)\) and \(P\left(150 \leq T_{o} \leq 200\right)\) ? b. Using the \(\mu_{i}\) 's and \(\sigma_{i}\) 's given in part (a), calculate both \(P(55 \leq \bar{X})\) and \(P(58 \leq \bar{X} \leq 62)\). c. Using the \(\mu_{i}\) 's and \(\sigma_{i}\) 's given in part (a), calculate and interpret \(P\left(-10 \leq X_{1}-.5 X_{2}-.5 X_{3} \leq 5\right)\). d. If \(\mu_{1}=40, \mu_{2}=50, \mu_{3}=60, \sigma_{1}^{2}=10, \sigma_{2}^{2}=12\), and \(\sigma_{3}^{2}=14\), calculate \(P\left(X_{1}+X_{2}+X_{3} \leq 160\right)\) and also \(P\left(X_{1}+X_{2} \geq 2 X_{3}\right)\).

A binary communication channel transmits a sequence of "bits" (0s and 1s). Suppose that for any particular bit transmitted, there is a \(10 \%\) chance of a transmission error (a 0 becoming a 1 or a 1 becoming a 0 ). Assume that bit errors occur independently of one another. a. Consider transmitting 1000 bits. What is the approximate probability that at most 125 transmission errors occur? b. Suppose the same 1000 -bit message is sent two different times independently of one another. What is the approximate probability that the number of errors in the first transmission is within 50 of the number of errors in the second?

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 \(f\) ? b. What is the probability that exactly \(k\) of the ten bulbs fail before time \(f\) ? 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 \(t\) ?
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