Question

When used in a particular DVD player, the lifetime of a certain brand of battery is normally distributed with a mean value of 6 hours and a standard deviation of 0.8 hour. Suppose that two new batteries are independently selected and put into the player. The player ceases to function as soon as one of the batteries fails. a. What is the probability that the DVD player functions for at least 4 hours? b. What is the probability that the DVD player functions for at most 7 hours? c. Find a number \(x^{*}\) such that only \(5 \%\) of all DVD players will function without battery replacement for more than \(x^{*}\) hours.

Short answer

The probability that the DVD player functions for at least 4 hours is approximately 0.9938. For it to function for at most 7 hours, the probability is approximately 0.8944. The value \(x^{*}\), such that only 5% of all DVD players will function without battery replacement for more than \(x^{*}\) hours, is approximately 7.32.

Step-by-step solution

Understanding & Calculating Standard Normal Distribution

The lifetime of the battery is normally distributed with a mean, \( \mu \), of 6 hours and a standard deviation, \( \sigma \), of 0.8 hour. In order to find the probability, we need to calculate Z values for each case where \( Z = \frac{X - \mu}{\sigma} \) (X represents the random variable).

Calculate Probability for Part a

In part a, the DVD player needs to function for at least 4 hours. Calculating the Z value: \( Z = \frac{4 - 6}{0.8} = -2.5 \). The probability that Z is greater than -2.5 is the same as 1 minus the probability that Z is less than -2.5 which can be found in the Z-table: \( P(Z \ge -2.5) = 1 - P(Z \le -2.5) = 1 - 0.0062 = 0.9938 \).

Calculate Probability for Part b

In part b, we need to find the probability that the DVD player functions for at most 7 hours. Thus, we calculate the Z value: \( Z = \frac{7 - 6}{0.8} = 1.25 \). The probability that Z is less than or equal to 1.25 can be found in the Z-table: \( P(Z \le 1.25) = 0.8944 \).

Calculate X* for Part c

In part c, we need to find a value such that only 5% of all DVD players will function without battery replacement for more than that number of hours. This is the same as finding the 95th percentile of the distribution. Consulting the Z-table, the Z score corresponding to a cumulative probability of 0.95 is approximately 1.65. Solving the Z-score equation for X gives: \( X = Z*\sigma + \mu = 1.65*0.8 + 6 = 7.32 \).

Key concepts

Standard Normal Distribution

The standard normal distribution is a crucial concept in statistics, representing a bell-shaped curve where most of the data points are concentrated around the mean, and the distribution symmetrically tapers off on both sides. This distribution is categorized by having a mean (average) of 0 and a standard deviation of 1, serving as a reference point to compare all other normal distributions.

When a variable follows a normal distribution with any mean and standard deviation, we can convert it to a standard normal distribution using a process called standardization. This allows us to understand where a particular data point stands in relation to the average and compute probabilities for real-world scenarios, such as the lifetime of batteries in your DVD player exercise.

Z-score Calculation

A Z-score is a statistical measurement that describes a value's relationship to the mean of a group of values, measured in terms of standard deviations away from the mean. The formula for calculating a Z-score is given by: \[\begin{equation}Z = \frac{X - \mu}{\sigma}\end{equation}\] where \(X\) is the value we are looking at, \(\mu\) is the mean, and \(\sigma\) refers to the standard deviation.

In your exercise, the Z-score calculation allows us to transform the battery life hours into a standardized score to further understand probabilities of the DVD player functioning for certain durations. This step is vital in converting real-world questions into a statistical format that can be easily worked with.

Probability Calculation

Calculating probability involves determining the likelihood of a given event occurring. In the context of the normal distribution, probabilities can tell us the proportion of data that falls above, below, or between certain values. After finding the Z-scores for the duration of battery life, we can look up corresponding probabilities in a Z-table or use statistical software. These probabilities help us answer questions like the likelihood of the DVD player operating within a specific time range given its battery life distribution.

If we want to find how likely it is for an event not to happen, we can subtract the Z-table value from 1, which is what we did in part a of the exercise. Understanding how to accurately calculate these probabilities is essential for interpreting statistical data and making informed decisions based on those interpretations.

Percentiles in Normal Distribution

Percentiles are measures that divide a data set into 100 equal parts, making it possible to compare individual scores to the larger distribution. In a normal distribution, percentiles correspond to Z-scores, providing a way to find a value associated with a certain percentage of the data lying below it. This is especially useful in scenarios where we want to identify a specific threshold, such as the battery lifespan needed for the top 5% of DVD players in terms of longevity.

By using the Z-table to locate the Z-score that corresponds to the 95th percentile, as demonstrated in part c of the exercise, we can calculate the specific number of battery hours at which only 5% of DVD players will function. This percentile information is widely used in various fields, from education to quality control and beyond.