/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 69 A technician plans to test a cer... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A technician plans to test a certain type of resin developed in the laboratory to determine the nature of the time it takes before bonding takes place. It is known that the mean time to bonding is 3 hours and the standard deviation is 0.5 hour. It will be considered an undesirable product if the bonding time is either less than 1 hour or more than 4 hours. Comment on the utility of the resin. How often would its performance be considered undesirable? Assume that time to bonding is normally distributed.

Short Answer

Expert verified
The probability of the resin's performance being undesirable, represented by the bonding time being less than 1 hour or more than 4 hours, can be calculated using the standard normal distribution. Once the respective z-scores are found, the corresponding probabilities for these z-scores can be looked up in a standard normal distribution table, and the sum of these probabilities indicates how often the resin's performance would be considered undesirable.

Step by step solution

01

Standardize the desirable bonding times

The bonding times of 1 hour and 4 hours need to be standardized. The standardization or 'z-score' formula is \[Z = \frac{x - \mu}{\sigma}\] where \(Z\) is the standardized value or z-score, \(x\) is the observed value, \(\mu\) is the mean value, and \(\sigma\) is the standard deviation. For \(x = 1\) hour, \[Z_1 = \frac{1 - 3}{0.5} = -4\] and for \(x = 4\) hours, \[Z_2 = \frac{4 - 3}{0.5} = 2\]
02

Find probabilities using standard normal distribution table

Look up the z-scores in a standard normal distribution table, or use statistical software or a calculator with this function, to find the probabilities associated with these z-scores. The result is the probability \(P\) for the bonding time being less than 1 hour (\(Z_1\)) and for it being more than 4 hours (\(Z_2\)). These probabilities are then summed to find the total probability of the resin's performance being undesirable.
03

Interpret the result

The resulting probability indicates how often the resin's performance would be considered undesirable. If the probability is very small, the resin can be considered to be generally useful as the probability of undesirable bonding time is low. If the probability is high, the opposite is true.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Understanding Standard Deviation
Standard deviation is a measure of how spread out the numbers in a data set are. In simpler terms, it tells us how much variation or "dispersion" there is from the average (mean) value.

When examining the bonding times of resin, for instance, the standard deviation helps us understand how much the actual bonding time might differ from the expected average bonding time, which is 3 hours here. In this context, a standard deviation of 0.5 hours means that most bonding occurrences will happen close to the 3-hour mark, give or take around 0.5 hours.
  • If the standard deviation is small, the data points tend to be very close to the mean.
  • If it is large, the data points are spread out over a wider range of values.
This concept is crucial in the study of normal distribution as it affects the shape of the bell curve that represents the data.
Mean as the Average Measure
The mean is essentially the "center" of a data set. It's calculated by adding up all the numbers, then dividing by the count of those numbers. In our resin exercise, the mean or average bonding time given is 3 hours.

Why is mean important? It provides a single value that summarizes the entire data set, giving us a sense of what is "typical." However, if the data has very high or very low values (outliers), the mean might not always be representative. But in a normal distribution scenario, like our resin bonding times, it serves as a crucial point around which observations are spread.
  • The mean helps us predict what happens on average.
  • It is important for determining other statistical measures like variance and standard deviation.
Decoding the Z-score
A z-score tells you how many standard deviations a data point is from the mean. It's a way of standardizing data on a normal distribution, reducing it from its original units to a measure that allows for broad comparisons.

In the resin case, we calculated z-scores for bonding times of 1 hour and 4 hours. Applying the formula \[Z = \frac{x - \mu}{\sigma}\]revealed that 1 hour translates into a z-score of -4, and 4 hours into a z-score of 2. These scores tell us how unusual these observations are.
  • A z-score of 0 indicates the value is exactly the mean.
  • Positive z-scores correspond to values above the mean, while negative scores indicate values below it.
  • Z-scores help identify outliers and calculate probabilities in the standard normal distribution.
Probability in Normal Distribution
Probability is the chance of an event occurring in a statistical experiment. Within the realm of normal distribution, it often aims at understanding how likely data points are to fall within a certain range.

For the resin example, once we calculated our z-scores, we were able to use a standard normal distribution table (or a statistical tool) to find the probabilities for these scores:
  • The probability of bonding happening in less than 1 hour (z-score -4).
  • The probability of bonding in more than 4 hours (z-score 2).
By adding these probabilities, we learned how often the resin might be classified as "undesirable."

This probability calculation is pivotal, as it helps in decision-making about the utility of the resin, giving clear insights into the product's performance consistency.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Suppose that the time, in hours, taken to repair a heat pump is a random variable \(X\) having a gamma distribution with parameters \(\alpha=2\) and \(3=1 / 2\). What is the probability that the next service call will require (a) at most 1 hour to repair the heat pump? (b) at least 2 hours to repair the heat pump?

In the November 1990 issue of Chemical Engineering Progress, a study discussed the percent purity of oxygen from a certain supplier. Assume that the mean was 99.61 with a standard deviation of 0.08 . Assume that the distribution of percent purity was approximately normal. (a) What percentage of the purity values would you expect to be between 99.5 and \(99.7 ?\) (b) What purity value would you expect to exceed exactly \(5 \%\) of the population?

The daily amount of coffee, in liters, dispensed by a machine located in an airport, lobby is a random variable \(X\) having a continuous uniform distribution with \(A=7\) and \(B=10 .\) Find the probability that on a given day the amount of coffee dispensed by this machine will be (a) at most 8.8 liters; (b) more than 7.4 liters but less than 9.5 liters; (c) at least 8.5 liters.

The average rate of water usage (thousands of gallons per hour) by a certain community is known to involve the lognormal distribution with parameters \(\mu=5\) and \(a=2 .\) It is important for planning purposes to get a sense of periods of high usage. What is the probability that, for any given hour, 50,000 gallons of water are used?

The length of life, in hours, of a drill bit in a mechanical operation has a Weibull distribution with \(\alpha=2\) and \(0=50 .\) Find the probability that the bit will fail before 10 hours of usage.

See all solutions

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.