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Chapter 4: Problem 43

The distribution of resistance for resistors of a certain type is known to be normal, with \(10 \%\) of all resistors having a resistance exceeding \(10.256\) ohms and \(5 \%\) having a resistance smaller than \(9.671\) ohms. What are the mean value and standard deviation of the resistance distribution?

Short Answer

Expert verified
The mean is 10 ohms and the standard deviation is 0.2 ohms.

Step by step solution

01

Understand the Problem

We have a normal distribution of resistances, and we know two probabilities: the probability that a resistance is greater than 10.256 ohms is 10%, and the probability that it is less than 9.671 ohms is 5%. We need to find the mean (\(\mu\)) and standard deviation (\(\sigma\)) of this distribution.
02

Identify Z-scores for Known Probabilities

From the standard normal distribution table: - A probability of 10% in the upper tail corresponds to a Z-score of approximately 1.28. - A probability of 5% in the lower tail corresponds to a Z-score of approximately -1.645.
03

Set Up the Equations

For the resistance of 10.256 ohms:\[ z = \frac{10.256 - \mu}{\sigma} = 1.28 \]For the resistance of 9.671 ohms:\[ z = \frac{9.671 - \mu}{\sigma} = -1.645 \]
04

Solve the System of Equations

We have two equations:\[ 10.256 - \mu = 1.28\sigma \] \[ 9.671 - \mu = -1.645\sigma \]Subtract the second equation from the first:\[ (10.256 - 9.671) = (1.28 + 1.645)\sigma \]\[ 0.585 = 2.925\sigma \]Solving for \(\sigma\): \[ \sigma = \frac{0.585}{2.925} \approx 0.2 \]
05

Calculate the Mean

Substituting \(\sigma = 0.2\) back into one of the original equations:\[ 10.256 - \mu = 1.28 \times 0.2 \]\[ 10.256 - \mu = 0.256 \]\[ \mu = 10.256 - 0.256 = 10 \]

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Key Concepts

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

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 the values in a set of data differ from the mean (average) value.
For a normal distribution, standard deviation indicates how much the individual observations deviate from the mean, while taking into account all the data points.
Because we are dealing with a normal distribution of resistors, the calculation of the standard deviation (\(\sigma\)) helps us quantify this variability.
  • In the exercise, we found \(\sigma = 0.2\) ohms for the resistors. This tells us how consistently resistors hover around the mean resistance value.
  • A smaller standard deviation would suggest that most resistors are close to the mean, whereas a larger standard deviation would indicate more dispersion.
The formula for standard deviation in a normal distribution relates to the Z-score, making it an essential part of understanding data spread.
Mean Calculation
The mean is the average of all data points in a set, calculated by adding them all together and dividing by the number of data points. It offers a central value around which the data is distributed.
In a normal distribution, the mean is part of the parameters (mean and standard deviation) that define the entire curve.
In the exercise, we needed to calculate the mean resistance (\(\mu\)) of the resistors.
  • Using the provided probabilities and respective Z-scores, we established equations that connected the mean and standard deviation.
  • By solving these equations, it was determined that the mean resistance is \(\mu = 10\) ohms.
This value serves as a reference point indicating where the majority of resistor values are clustered. A precise mean helps predict and ensure the quality control of resistors consistently meeting targeted resistance values.
Z-scores
Z-scores are a way to standardize scores by expressing them in terms of standard deviations from the mean. They help identify whether a value is below or above the mean, and how extreme that value is within a distribution.
In the context of normal distribution, Z-scores are essential tools for translating probabilities into exact values based on the standard normal distribution table.
  • In this exercise, Z-scores helped us relate given probabilities to the actual resistance values. A resistor with resistance greater than 10.256 ohms had a Z-score of 1.28, indicating it's 1.28 standard deviations above the mean.
  • Likewise, a resistor with resistance less than 9.671 ohms had a Z-score of -1.645, showing it's 1.645 standard deviations below the mean.
Using Z-scores, we managed to link physical resistor values to statistical probabilities, allowing us to construct the equations needed to find the mean and standard deviation.

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Most popular questions from this chapter

Suppose that blood chloride concentration (mmol/L) has a normal distribution with mean 104 and standard deviation 5 (information in the article "Mathematical Model of Chloride Concentration in Human Blood," J. of Med. Engr. and Tech., 2006: 25-30, including a normal probability plot as described in Section 4.6, supports this assumption). a. What is the probability that chloride concentration equals 105 ? Is less than 105 ? Is at most 105 ? b. What is the probability that chloride concentration differs from the mean by more than 1 standard deviation? Does this probability depend on the values of \(\mu\) and \(\sigma\) ? c. How would you characterize the most extreme . \(1 \%\)

If \(X\) is a normal rv with mean 80 and standard deviation 10 , compute the following probabilities by standardizing: a. \(P(X \leq 100)\) b. \(P(X \leq 80)\) c. \(P(65 \leq X \leq 100)\) d. \(P(70 \leq X)\) e. \(P(85 \leq X \leq 95)\) f. \(P(|X-80| \leq 10)\)

The special case of the gamma distribution in which \(\alpha\) is a positive integer \(n\) is called an Erlang distribution. If we replace \(\beta\) by \(1 / \lambda\) in Expression (4.8), the Erlang pdf is $$ f(x, \lambda, n)=\left\\{\begin{array}{cc} \frac{\lambda(\lambda x)^{n-1} e^{-\lambda x}}{(n-1) !} & x \geq 0 \\ 0 & x<0 \end{array}\right. $$ It can be shown that if the times between successive events are independent, each with an exponential distribution with parameter \(\lambda\), then the total time \(X\) that elapses before all of the next \(n\) events occur has pdf \(f(x ; \lambda, n)\). a. What is the expected value of \(X\) ? If the time (in minutes) between arrivals of successive customers is exponentially distributed with \(\lambda=.5\), how much time can be expected to elapse before the tenth customer arrives? b. If customer interarrival time is exponentially distributed with \(\lambda=.5\), what is the probability that the tenth customer (after the one who has just arrived) will arrive within the next \(30 \mathrm{~min}\) ? c. The event \(\\{X \leq t\\}\) occurs iff at least \(n\) events occur in the next \(t\) units of time. Use the fact that the number of events occurring in an interval of length \(t\) has a Poisson distribution with parameter \(\lambda t\) to write an expression (involving Poisson probabilities) for the Erlang cdf \(F(t, \lambda, n)=\) \(P(X \leq t)\)

a. If a normal distribution has \(\mu=30\) and \(\sigma=5\), what is the 91 st percentile of the distribution? b. What is the 6th percentile of the distribution? c. The width of a line etched on an integrated circuit chip is normally distributed with mean \(3.000 \mu \mathrm{m}\) and standard deviation \(.140\). What width value separates the widest \(10 \%\) of all such lines from the other \(90 \%\) ?

Extensive experience with fans of a certain type used in diesel engines has suggested that the exponential distribution provides a good model for time until failure. Suppose the mean time until failure is 25,000 hours. What is the probability that a. A randomly selected fan will last at least 20,000 hours? At most 30,000 hours? Between 20,000 and 30,000 hours? b. The lifetime of a fan exceeds the mean value by more than 2 standard deviations? More than 3 standard deviations?
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