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The diameter of ball-bearings produced by a machine is a random variable having a normal distribution with mean \(6.00 \mathrm{~mm}\) and standard deviation \(0.02 \mathrm{~mm}\). If the diameter tolerance is \(\pm 1 \%\), find the proportion of ball-bearings produced that are out of tolerance. After several years' use, machine wear has the effect of increasing the standard deviation, although the mean diameter remains constant. The manufacturer decides to replace the machine when \(2 \%\) of its output is out of tolerance. What is the standard deviation when this happens?

Short Answer

Expert verified
The standard deviation is approximately 0.0258 mm when 2% of output is out of tolerance.

Step by step solution

01

Define the tolerance range

The diameter tolerance is set to \( \pm 1\% \) of the mean diameter, which is \(6.00\, \text{mm} \). Therefore, the tolerance limits are calculated as follows: \[\text{Lower limit} = 6.00 - 0.01 \times 6.00 = 5.94\, \text{mm}\] \[\text{Upper limit} = 6.00 + 0.01 \times 6.00 = 6.06\, \text{mm}\] Hence, any ball-bearing with a diameter outside this range is considered out of tolerance.
02

Calculate the probability of being out of tolerance

Since the diameter follows a normal distribution, the probability of a ball-bearing being within the tolerance limits is given by the probability that a standard normal variable \(Z\) falls between certain values. We standardize to find:\[P(5.94 < X < 6.06) = P\left(\frac{5.94 - 6.00}{0.02} < Z < \frac{6.06 - 6.00}{0.02}\right)\]Calculating these values gives: \[-3 < Z < 3\]Using a standard normal distribution table, the probability \(P(-3 < Z < 3)\) is approximately equal to 0.9973. Therefore, the proportion of ball-bearings out of tolerance is \(1 - 0.9973 = 0.0027\) or 0.27\%.
03

Determine the standard deviation for out of tolerance threshold

To find the standard deviation when 2\% of the output is out of tolerance, we must solve for \(\sigma\) such that:\[P(X < 5.94) + P(X > 6.06) = 0.02\]Since the probabilities are symmetric, we have:\[P(X < 5.94) = P(X > 6.06) = 0.01\]Converting to a standard normal variable gives:\[P\left(\frac{5.94 - 6.00}{\sigma} < Z < \frac{6.06 - 6.00}{\sigma}\right) = 0.01\]From the standard normal distribution table, \(Z = -2.33\) corresponds to 0.01 in the lower tail, hence:\[\frac{-0.06}{\sigma} = -2.33\, \Rightarrow\, \sigma = \frac{0.06}{2.33} \approx 0.0258\, \text{mm}\]

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Key Concepts

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

Understanding Standard Deviation
The concept of standard deviation is vital in statistics, particularly when discussing normal distribution. It helps us understand how spread out the values in a data set are from the mean.
Values that are closer to the mean indicate a small standard deviation, while values more spread out have a larger standard deviation.
For instance, in our ball-bearing example, the initial standard deviation is 0.02 mm, which shows a small variation in the diameter of most ball-bearings produced. Standard deviation is crucial in quality control and manufacturing because it provides insights into consistency. A smaller standard deviation points to consistent product quality.
  • When the machine starts wearing out, the standard deviation increases.
  • This increase signals more variability in the ball-bearing sizes.
  • Quality assurance processes often replace machinery when the variation affects product quality to unacceptable levels.
Understanding and monitoring standard deviation helps manufacturers maintain high-quality standards.
Exploring Probability
Probability is the measure of the likelihood that an event will occur, a key concept within statistics, particularly when assessing outcomes associated with a normal distribution.
In our problem, we assess the probability that a ball-bearing has a diameter within certain limits.
When this probability is calculated using normal distribution, it considers how likely it is for a value to fall within an interval around the mean. The calculated probability in the problem showed that 99.73% of the ball-bearings fell within the tolerance limits. This value, derived from the standard normal distribution, indicates the overall quality and precision of the machine currently producing the ball-bearings.
  • Probabilities can guide decisions regarding when to carry out maintenance or replace equipment.
  • Here, only when the probability of out-of-tolerance diameters reaches 2%, the machine is deemed for replacement.
Probability helps manufacturers predict when machinery wear might impact product quality.
Defining Tolerance Limits
Tolerance limits specify the acceptable range of dimensions for manufactured products. These limits ensure that the products meet predefined quality standards and function as intended.
In the ball-bearing scenario, tolerance is set at ±1% of the mean diameter.
This means that any ball-bearing has to be between 5.94 mm and 6.06 mm to be considered within tolerance. Defining tolerance limits is a critical aspect of quality control to ensure products perform reliably.
  • Tolerance limits help in identifying when variations in production require corrective actions.
  • Calculating the proportion of output that falls outside these limits helps assess manufacturing accuracy.
Clear tolerance limits guide manufacturers in maintaining consistent production quality and identifying potential issues early, thus preventing defects from reaching consumers. Understanding how to define and use tolerance limits is fundamental in various manufacturing processes.

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