Question

The life span of oil-drilling bits depends on the types of rock and soil that the drill encounters, but it is estimated that the mean length of life is 75 hours. Suppose an oil exploration company purchases drill bits that have a life span that is approximately normally distributed with a mean equal to 75 hours and a standard deviation equal to 12 hours. a. What proportion of the company's drill bits will fail before 60 hours of use? b. What proportion will last at least 60 hours? c. What proportion will have to be replaced after more than 90 hours of use?

Short answer

Answer: Approximately 21.19% of the company's drill bits will fail before 60 hours, 78.81% will last at least 60 hours, and 21.19% will have to be replaced after more than 90 hours of use.

Step-by-step solution

Calculate the z-score for 60 hours and 90 hours

We'll calculate the z-score for both 60 hours and 90 hours with the formula: $$z = \frac{x - \mu}{\sigma}$$ where x is the specific value we're trying to find, μ is the mean, and σ is the standard deviation. For 60 hours: $$z = \frac{60 - 75}{12} = -1.25$$ For 90 hours: $$z = \frac{90 - 75}{12} = 1.25$$

Find the proportion of drill bits that will fail before 60 hours

Using a standard normal table, we look up the z-score of -1.25 and find the proportion corresponding to this value, which is approximately 0.2119. So, approximately 21.19% of the company's drill bits will fail before 60 hours of use.

Find the proportion of drill bits that will last at least 60 hours

To find the proportion of drill bits that will last at least 60 hours, we subtract the proportion that will fail before 60 hours (0.2119) from 1: $$1 - 0.2119 = 0.7881$$ So, 78.81% of the company's drill bits will last at least 60 hours.

Find the proportion of drill bits that will have to be replaced after more than 90 hours

We look up the z-score of 1.25 in a standard normal table and find the corresponding proportion of 0.7881. This proportion is the probability of a drill bit lasting up to 90 hours. To find the probability of it lasting more than 90 hours, we subtract this value from 1: $$1 - 0.7881 = 0.2119$$ So, 21.19% of the company's drill bits will have to be replaced after more than 90 hours of use.

Key concepts

Z-Score Calculation

Understanding the z-score is essential for anyone diving into the world of statistics. A z-score, also known as a standard score, is a numerical measurement that describes a value's relationship to the mean of a group of values, measured in terms of standard deviations from the mean.

When you calculate a z-score, you are essentially figuring out how many standard deviations away from the mean a particular value is. The formula for calculating a z-score is fairly simple: \[ z = \frac{x - \mu}{\sigma} \] where \( x \) is the value in question, \( \mu \) is the mean of the data set, and \( \sigma \) is the standard deviation.

A positive z-score indicates that the value is above the mean, while a negative z-score indicates that the value is below the mean. This calculation is used in various statistical techniques, including standardization of scores and hypothesis testing.

Practical Application of Z-Score

In the original exercise, calculating the z-scores for 60 hours and 90 hours allowed us to determine the proportion of drill bits likely to fail before reaching those usage hours. By doing so, the company can better manage their inventory and expectations regarding product lifespan.

Standard Deviation

Standard deviation is a fundamental concept in statistics that measures the amount of variation or dispersion in a set of values.

A low standard deviation means that the values tend to be close to the mean of the set, whereas a high standard deviation means that the values are spread out over a wider range. The standard deviation is crucial for understanding the spread of a data set and can help in making predictions and decisions. The standard deviation is represented mathematically as \( \sigma \) and is calculated by taking the square root of the variance.

Importance in Probability Distributions

In the context of the original problem, the standard deviation of 12 hours tells us that the lifespans of the drill bits are expected to deviate from the mean by an average of 12 hours. Insights about variability are vital for quality control and risk management within any business operation, including the reliability of oil-drilling bits.

Probability and Statistics

The disciplines of probability and statistics are intricately linked, providing tools for analyzing and making inferences about data. Probability is the study of randomness and uncertainty. It allows us to quantify the likelihood of events occurring and is the foundation upon which statistical theory is built.

Statistics is the science that enables us to collect, analyze, interpret, present, and organize data. It's all about providing a meaningful context to numbers and patterns that we observe. By employing statistical methods, we can make decisions that are informed by data, rather than guesses or assumptions.

Utilization in the Real World

In the exercise given, we applied the principles of probability by using the normal distribution to determine what proportion of oil-drilling bits would fail at different usage hours. This kind of analysis helps companies establish warranties, predict maintenance schedules, and manage customer expectations – all of which are grounded in probabilistic and statistical methods.