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Suppose the force acting on a column that helps to support a building is a normally distributed random variable \(X\) with mean value \(15.0\) kips and standard deviation \(1.25\) kips. Compute the following probabilities by standardizing and then using Table A.3. a. \(P(X \leq 15)\) b. \(P(X \leq 17.5)\) c. \(P(X \geq 10)\) d. \(P(14 \leq X \leq 18)\) e. \(P(|X-15| \leq 3)\)

Short Answer

Expert verified
a. 0.5, b. 0.9772, c. ~1.0, d. 0.7619, e. 0.9836

Step by step solution

01

Understand the Normal Distribution

The problem states that the force is a normally distributed random variable, denoted by \(X\), with a mean (\(\mu\)) of 15 kips and a standard deviation (\(\sigma\)) of 1.25 kips. We will standardize the variable \(X\) to a standard normal distribution \(Z\) using the formula \(Z = \frac{X - \mu}{\sigma}\).
02

Compute Standardized Values

For each part of the problem, convert the given \(X\) values into \(Z\) scores using the standardization formula. \(Z = \frac{15 - 15}{1.25} = 0\), \(Z = \frac{17.5 - 15}{1.25} = 2\), \(Z = \frac{10 - 15}{1.25} = -4\), \(Z = \frac{14 - 15}{1.25} = -0.8\), and \(Z = \frac{18 - 15}{1.25} = 2.4\). For part (e), use both \(Z = \frac{12 - 15}{1.25} = -2.4\) and \(Z = \frac{18 - 15}{1.25} = 2.4\).
03

Use Z-Table for Probabilities

Use the Z-table to find probabilities. For \(P(X \leq 15)\), \(P(Z \leq 0)=(0.5)\), for \(P(X \leq 17.5)\), \(P(Z \leq 2) \approx 0.9772\), for \(P(X \geq 10)\), \(1 - P(Z < -4)\) (since this Z-value is off-tab limit, assume the probability to be nearly 1). For \(P(14 \leq X \leq 18)\), calculate \(P(-0.8 \leq Z \leq 2.4) \approx 0.9738 - 0.2119 = 0.7619\). Lastly, for \(P(|X-15| \leq 3)\), find \(P(-2.4 \leq Z \leq 2.4) \approx 0.9918 - 0.0082 = 0.9836\).
04

Interpret Results

From the calculations, we have the probabilities for each situation: \(a)\, P(X \leq 15) = 0.5\), \(b)\, P(X \leq 17.5) = 0.9772\), \(c)\, P(X \geq 10) \approx 1.\), \(d)\, P(14 \leq X \leq 18) = 0.7619\), and \(e)\, P(|X-15| \leq 3) = 0.9836\).

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Key Concepts

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

Standardization
Standardization is a crucial part of working with normal distributions, especially when you want to connect with the standard normal distribution. It transforms a normally distributed variable into the standard normal distribution, which simplifies probability calculations.
The process involves converting your variable using the formula:
  • \( Z = \frac{X - \mu}{\sigma} \)
Here, \(X\) represents the original variable, \(\mu\) is the mean of that variable, and \(\sigma\) is its standard deviation. When standardized, \(Z\) will have a mean of 0 and a standard deviation of 1. This conversion is essential for utilizing the Z-table since the table is based on the standard normal distribution.
Through standardization, you change the scale of your data, which allows you to use uniform resources like the Z-table for probability computation.
Z-table
The Z-table is a fundamental tool when dealing with standard normal distributions. It provides the area under the curve to the left of a given Z-value, representing probability.
  • To use the Z-table, locate your Z-score calculated through standardization.
  • Find the corresponding area in the table, which equals the probability.
This table assumes a standard normal distribution, where the curve is centered at zero and the total area under the curve equals one.
For instance, if you standardized a value and obtained a Z-score of 2, you would look this up in the Z-table to find that the probability \(P(Z \leq 2) \approx 0.9772\). In practical exercises, using the Z-table is a straightforward method to find probabilities without complex calculations.
Probability Computation
Computing probabilities in the context of normal distributions often relies on standardization and the Z-table. Once you've standardized your variable and referred to the Z-table, you can easily determine various probabilities.
For instance, to compute a probability like \(P(X \geq 10)\), you first convert it to \(P(Z \geq -4)\). Since values like \(Z = -4\) aren't in the Z-table, you assume the probability approaches 1 because we are significantly beyond the left tail limit of the distribution. This logic simplifies computational tasks.
Multiple cases can arise:
  • For upper tail calculations, use \(1 - P(Z \leq a)\).
  • For ranges, compute \(P(a \leq Z \leq b)\) by \(P(b) - P(a)\).
  • For absolute differences, focus on \(P(-a \leq Z \leq a)\) by utilizing symmetry.
Probability computation thus circles around generating correct Z-values and efficiently reading the Z-table.
Standard Normal Distribution
The standard normal distribution is a special type of normal distribution where the mean is zero and the variance is one. It is represented by the bell-shaped curve centered at zero, with standard deviations marking the spread.
Understanding this distribution is important because it serves as the basis for the Z-table. Any normal distribution can be converted into this form through the process of standardization.
Key properties include:
  • Symmetrical about the mean, meaning probabilities are equal on both sides of the mean.
  • As it changes from -3 to 3 in Z-scores, this accounts for most (about 99.7%) of data within the curve.
  • Allows uncertainties in many real-world scenarios to be modeled effectively.
The standard normal distribution simplifies complex computations into a manageable format by supporting uniformity in probability determination.

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Most popular questions from this chapter

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 \(F(t ; \lambda, n)=P(X \leq t)\).

The Rockwell hardness of a metal is determined by impressing a hardened point into the surface of the metal and then measuring the depth of penetration of the point. Suppose the Rockwell hardness of a particular alloy is normally distributed with mean 70 and standard deviation 3 . a. If a specimen is acceptable only if its hardness is between 67 and 75 , what is the probability that a randomly chosen specimen has an acceptable hardness? b. If the acceptable range of hardness is \((70-c, 70+c)\), for what value of \(c\) would \(95 \%\) of all specimens have acceptable hardness? c. If the acceptable range is as in part (a) and the hardness of each of ten randomly selected specimens is indepen-dently determined, what is the expected number of acceptable specimens among the ten? d. What is the probability that at most eight of ten independently selected specimens have a hardness of less than 73.84?

A system consists of five identical components connected in series as shown: As soon as one component fails, the entire system will fail. Suppose each component has a lifetime that is exponentially distributed with \(\lambda=.01\) and that components fail independently of one another. Define events \(A_{i}=\) \(\\{i\) th component lasts at least \(t\) hours \(\\}, i=1, \ldots, 5\), so that the \(A_{i} \mathrm{~s}\) are independent events. Let \(X=\) the time at which the system fails-that is, the shortest (minimum) lifetime among the five components. a. The event \(\\{X \geq t\\}\) is equivalent to what event involving \(A_{1}, \ldots, A_{5}\) ? b. Using the independence of the \(A_{i}{ }^{\prime}\) s, compute \(P(X \geq t)\). Then obtain \(F(t)=P(X \leq t)\) and the pdf of \(X\). What type of distribution does \(X\) have? c. Suppose there are \(n\) components, each having exponential lifetime with parameter \(\lambda\). What type of distribution does \(X\) have?

Suppose that when a transistor of a certain type is subjected to an accelerated life test, the lifetime \(X\) (in weeks) has a gamma distribution with mean 24 weeks and standard deviation 12 weeks. a. What is the probability that a transistor will last between 12 and 24 weeks? b. What is the probability that a transistor will last at most 24 weeks? Is the median of the lifetime distribution less than 24 ? Why or why not? c. What is the 99th percentile of the lifetime distribution? d. Suppose the test will actually be terminated after \(t\) weeks. What value of \(t\) is such that only \(.5 \%\) of all transistors would still be operating at termination?

Vehicle speed on a particular bridge in China can be modeled as normally distributed ("Fatigue Reliability Assessment for Long-Span Bridges under Combined Dynamic Loads from Winds and Vehicles" \(J\). of Bridge Engr., 2013: 735-747). a. If \(5 \%\) of all vehicles travel less than \(39.12 \mathrm{~m} / \mathrm{h}\) and \(10 \%\) travel more than \(73.24 \mathrm{~m} / \mathrm{h}\), what are the mean and standard deviation of vehicle speed? [Note: The resulting values should agree with those given in the cited article.] b. What is the probability that a randomly selected vehicle's speed is between 50 and \(65 \mathrm{~m} / \mathrm{h}\) ? c. What is the probability that a randomly selected vehicle's speed exceeds the speed limit of \(70 \mathrm{~m} / \mathrm{h}\) ?

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