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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

Based on data from a dart-throwing experiment, the article "Shooting Darts" (Chance, Summer 1997, 16-19) proposed that the horizontal and vertical errors from aiming at a point target should be independent of one another, each with a normal distribution having mean 0 and variance \(\sigma^{2}\). It can then be shown that the pdf of the distance \(V\) from the target to the landing point is $$ f(v)=\frac{v}{\sigma^{2}} \cdot e^{-v^{2} / 2 \sigma^{2}} \quad v>0 $$ a. This pdf is a member of what family introduced in this chapter? b. If \(\sigma=20 \mathrm{~mm}\) (close to the value suggested in the paper), what is the probability that a dart will land within \(25 \mathrm{~mm}\) (roughly \(1 \mathrm{in}\).) of the target?

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The article "Response of \(\mathrm{SiC}_{\mathrm{I}} / \mathrm{Si}_{3} \mathbf{N}_{4}\) Composites Under Static and Cyclic Loading-An Experimental and Statistical Analysis" \((J\). of Engr. Materials and Technology, 1997: 186-193) suggests that tensile strength (MPa) of composites under specified conditions can be modeled by a Weibull distribution with \(\alpha=9\) and \(\beta=180\). a. Sketch a graph of the density function. b. What is the probability that the strength of a randomly selected specimen will exceed 175 ? Will be between 150 and 175 ? c. If two randomly selected specimens are chosen and their strengths are independent of one another, what is the probability that at least one has a strength between 150 and 175 ? d. What strength value separates the weakest \(10 \%\) of all specimens from the remaining \(90 \%\) ?

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