Question

The lifetime of a certain brand of battery is normally distributed with a mean value of \(6 \mathrm{hr}\) and a standard deviation of \(0.8 \mathrm{hr}\) when it is used in a particular cassette player. 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 player functions for at least \(4 \mathrm{hr}\) ? b. What is the probability that the cassette player works for at most \(7 \mathrm{hr}\) ? c. Find a number \(z^{*}\) such that only \(5 \%\) of all cassette players will function without battery replacement for more \(\operatorname{than} z^{*} \mathrm{hr}\)

Short answer

The following are the probabilities: a. The probability that the player functions for at least 4 hours is about 0.9938. b. The probability that the player functions for at most 7 hours is 0.8944. c. Only 5% of all cassette players will function without battery replacement for more than about 7.316 hours.

Step-by-step solution

Standardize the Normal Distribution

In order to handle probabilities in a normal distribution, it is advisable to convert the distribution to a standard normal distribution. This is a normal distribution with a mean of 0 and standard deviation of 1. Each value of the actual distribution is converted to a z-score using the formula \(z = (X - \mu) / \sigma\), where \(X\) is the value from the original normal distribution, \(\mu\) is the mean of the distribution, and \(\sigma\) is the standard deviation.

Calculate the Probability for at Least 4 Hours

According to the question, it is desired to know the probability that the battery life lasts at least 4 hours. We calculate the z-score for 4 hours using the formula in Step 1, obtaining -2.5. The probability that a z-score is at least -2.5 is the same as 1 minus the probability that the score is less than -2.5, because all probabilities in a distribution add up to 1. Using the standard normal distribution table, we find the latter probability to be 0.0062. So, the calculated probability is 1 - 0.0062 = 0.9938.

Calculate the Probability for at Most 7 Hours

Now, we need to calculate the probability that the battery life lasts at most 7 hours. The z-score for 7 hours is calculated as (7-6)/0.8 = 1.25. The probability that a z-score is at most 1.25 is given directly by the standard normal distribution table as 0.8944.

Find z* For Which Only 5% of Batteries Last Longer

The question now requires finding a battery life such that only 5% of all batteries last longer. This means that 95% of batteries last less than this time. So, we need to find the z-score that has 95% of the distribution to its left. This z-score is 1.645 according to the z-table. To convert this z-score back to battery life, we use the formula \(X = \mu + \sigma*z\), which gives us \(X = 6 + 0.8*1.645 = 7.316\) hours.

Key concepts

Z-Score

The z-score is a statistical measurement that describes a value's relationship to the mean of a group of values. In a normal distribution, it represents the number of standard deviations a particular value is from the mean. To calculate a z-score, use the formula:
\( z = \frac{X - \mu}{\sigma} \),
where \(X\) is the value being standardized, \(\mu\) is the mean of the distribution, and \(\sigma\) is the standard deviation. In the context of the given exercise, calculating the z-score allows us to determine how typical or atypical a battery's lifetime is compared to the average.
Understanding the z-score is pivotal for performing probability calculations under the normal distribution. The z-score transformation is a method of standardization that makes it possible to compare different sets of data or different points within the same set of data.

Probability

Probability measures the likelihood that an event will occur, expressed as a number between 0 and 1, where 0 indicates impossibility and 1 indicates certainty. In terms of the normal distribution, it refers to the area under the curve corresponding to a specific range of values. For example, to find the probability that something occurs 'at least' or 'more than' a certain value, you look at the area under the curve to the right of that value's z-score. Conversely, for something that occurs 'at most' or 'less than' a certain value, you examine the area to the left.
In the exercise, we calculate the probability of the cassette player working for at least 4 hours and at most 7 hours by finding the area under the normal curve corresponding to those values' z-scores.

Standard Deviation

Standard deviation is a measure of the amount of variation or dispersion in a set of values. A low standard deviation implies that most of the numbers are close to the mean (average), while a high standard deviation indicates that the numbers are more spread out from the mean. The standard deviation in the battery life example (0.8 hours) tells us that typically, the battery lifetimes are clustered within 0.8 hours of the mean.
When we transform battery lifetimes into z-scores, the standard deviation serves as the 'scaling factor' that standardizes the distances from the mean, giving us a dimensionless quantity that we can compare against a standard normal distribution. This property of the standard deviation is essential when we wish to make probabilistic statements about our data.

Mean

The mean is the average of a set of numbers, and it's calculated by adding all the values together and then dividing by the number of values. In a normal distribution, the mean is the central value where the distribution is symmetrically centered. In the supplied exercise, the mean battery life is 6 hours. This number represents the 'central' or 'expected' battery life and is used in determining how far off other values are (like the 4 or 7 hours in questions a and b), as we do this via computing the z-score.
The mean is also crucial when we revert a z-score back to a value in the original distribution, often in scenarios where we want to find the threshold value that corresponds to a certain cumulative probability, as seen in question c of the exercise, where we reverse-engineer the z-score to find the battery life associated with the top 5% of the distribution.