/*! 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 5 Suppose the force acting on a co... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Suppose the force acting on a column that helps to support a building is a normally distributed random variable \(X\) with mean value \(15.0 \mathrm{kips}\) 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. 0.99997, d. 0.7799, e. 0.9836

Step by step solution

01

Standardizing the Random Variable

The random variable has a normal distribution with a mean \( \mu = 15.0 \) kips and a standard deviation \( \sigma = 1.25 \) kips. Thus, we can standardize it using the formula for the z-score: \( Z = \frac{X - \mu}{\sigma} \). This will allow us to use the standard normal distribution table (Table A.3) to find probabilities.
02

Compute Standardized Value for Part a

For part (a), we need to find \( P(X \leq 15) \). The z-score for 15 is: \( Z = \frac{15 - 15}{1.25} = 0 \). The probability \( P(Z \leq 0) \) is equal to 0.5 from the standard normal distribution table.
03

Compute Standardized Value for Part b

For part (b), we need to find \( P(X \leq 17.5) \). The z-score for 17.5 is: \( Z = \frac{17.5 - 15}{1.25} = 2 \). Using the standard normal distribution table, \( P(Z \leq 2) \approx 0.9772 \). Thus, \( P(X \leq 17.5) = 0.9772 \).
04

Compute Standardized Value for Part c

For part (c), we need to find \( P(X \geq 10) \). First, find the z-score for 10: \( Z = \frac{10 - 15}{1.25} = -4 \). Using the table, \( P(Z \leq -4) \approx 0.00003 \). Therefore, \( P(X \geq 10) = 1 - 0.00003 \approx 0.99997 \).
05

Compute Standardized Values for Part d

For part (d), find \( P(14 \leq X \leq 18) \). Compute z-scores for 14 and 18: \( Z_{14} = \frac{14 - 15}{1.25} = -0.8 \) and \( Z_{18} = \frac{18 - 15}{1.25} = 2.4 \). From the table, \( P(Z \leq -0.8) \approx 0.2119 \) and \( P(Z \leq 2.4) \approx 0.9918 \). Thus, \( P(14 \leq X \leq 18) = 0.9918 - 0.2119 = 0.7799 \).
06

Compute Standardized Values for Part e

For part (e), find \( P(|X-15| \leq 3) \), which simplifies to \( P(12 \leq X \leq 18) \). Compute z-scores for 12 and 18: \( Z_{12} = \frac{12 - 15}{1.25} = -2.4 \) and use the previous z-score \( Z_{18} = 2.4 \). From the table, \( P(Z \leq -2.4) \approx 0.0082 \). So \( P(12 \leq X \leq 18) = 0.9918 - 0.0082 = 0.9836 \).

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.

Z-Score
The z-score is a key concept when working with the normal distribution. It helps convert a normal distribution into a standard normal distribution. This standardization allows us to use standard normal distribution tables to calculate probabilities. A z-score tells us how many standard deviations a given data point is from the mean.
For any variable, the z-score is calculated with the formula: \[ Z = \frac{X - \mu}{\sigma} \] Where:
  • \( X \) is the data point of interest.
  • \( \mu \) is the mean of the distribution.
  • \( \sigma \) is the standard deviation.
Once you obtain the z-score, use the standard normal distribution table to find the corresponding probability. This process is crucial in probability calculations for normal distributions.
For example, if the z-score equals 0, the data point is exactly at the mean, indicating a 50% probability of observing a value less than or equal to that data point.
Probability Calculation
After calculating the z-score, effecting a probability calculation requires referring to a standard normal distribution table. This table provides probabilities associated with each z-score.
The standard normal distribution table helps in determining:
  • The area under the curve to the left of a particular z-score (i.e., \( P(Z \leq z) \)).
  • The area to the right is easily found by subtracting from 1 (i.e., \( P(Z \geq z) = 1 - P(Z \leq z) \)).
For instance, if you're calculating \( P(X \leq 17.5) \), first find the z-score for 17.5. Use the table value for this z-score to determine the percentage of data falling below this value.
This probability calculation is extremely useful in fields like finance, engineering, and science where predicting outcomes based on known distributions can assist in decision-making.
Standard Deviation
Standard deviation is a key measure in statistics that quantifies the amount of variation in a set of data. In the context of a normal distribution, it indicates how much the individual data points typically deviate from the mean.
It's calculated with the formula: \[ \sigma = \sqrt{\frac{1}{N} \sum_{i=1}^{N} (X_i - \mu)^2} \]Where:
  • \( N \) is the number of data points.
  • \( X_i \) represents each data point.
  • \( \mu \) is the mean.
Standard deviation is essential for computing z-scores as it serves as the scale factor. A smaller standard deviation indicates that the data points are clustered around the mean, while a larger standard deviation indicates more spread out data.
Mean Value
The mean value or arithmetic mean is a central value of a set of numbers. In probability and statistics, it represents the expected value of a random variable.
To calculate the mean, sum all the values in a dataset and divide by the total number of values: \[ \mu = \frac{1}{N} \sum_{i=1}^{N} X_i \]Where:
  • \( N \) is the total number of data points.
  • \( X_i \) represents each individual data point in the set.
In a normal distribution, the mean is located at the center of its symmetric bell-shaped curve. It is used as a point of reference in many statistical calculations, such as z-scores, and plays a critical role in hypothesis testing and data analysis.

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

The lifetime \(X\) (in hundreds of hours) of a certain type of vacuum tube has a Weibull distribution with parameters \(\alpha=2\) and \(\beta=3\). Compute the following: a. \(E(X)\) and \(V(X)\) b. \(P(X \leq 6)\) c. \(P(1.5 \leq X \leq 6)\) (This Weibull distribution is suggested as a model for time in service in "On the Assessment of Equipment Reliability: Trading Data Collection Costs for Precision," J. of Engr: Manuf., 1991: 105-109.)

Let \(X=\) the time between two successive arrivals at the drive-up window of a local bank. If \(X\) has an exponential distribution with \(\lambda=1\) (which is identical to a standard gamma distribution with \(\alpha=1\) ), compute the following: a. The expected time between two successive arrivals b. The standard deviation of the time between successive arrivals c. \(P(X \leq 4)\) d. \(P(2 \leq X \leq 5)\)

Let \(X\) be the total medical expenses (in 1000 s of dollars) incurred by a particular individual during a given year. Although \(X\) is a discrete random variable, suppose its distribution is quite well approximated by a continuous distribution with pdf \(f(x)=k(1+x / 2.5)^{-7}\) for \(x \geq 0\). a. What is the value of \(k\) ? b. Graph the pdf of \(X\). c. What are the expected value and standard deviation of total medical expenses? d. This individual is covered by an insurance plan that entails a \(\$ 500\) deductible provision (so the first \(\$ 500\) worth of expenses are paid by the individual). Then the plan will pay \(80 \%\) of any additional expenses exceeding \(\$ 500\), and the maximum payment by the individual (including the deductible amount) is \(\$ 2500\). Let \(Y\) denote the amount of this individual's medical expenses paid by the insurance company. What is the expected value of \(Y\) ? [Hint: First figure out what value of \(X\) corresponds to the maximum out-of- pocket expense of \(\$ 2500\). Then write an expression for \(Y\) as a function of \(X\) (which involves several different pieces) and calculate the expected value of this function.]

The temperature reading from a thermocouple placed in a constant-temperature medium is normally distributed with mean \(\mu\), the actual temperature of the medium, and standard deviation \(\sigma\). What would the value of \(\sigma\) have to be to ensure that \(95 \%\) of all readings are within \(1^{\circ}\) of \(\mu\) ?

Nonpoint source loads are chemical masses that travel to the main stem of a river and its tributaries in flows that are distributed over relatively long stream reaches, in contrast to those that enter at well-defined and regulated points. The article "Assessing Uncertainty in Mass Balance Calculation of River Nonpoint Source Loads" (J. of Envir. Engr., 2008: 247-258) suggested that for a certain time period and location, \(X\) = nonpoint source load of total dissolved solids could be modeled with a lognormal distribution having mean value \(10,281 \mathrm{~kg} / \mathrm{day} / \mathrm{km}\) and a coefficient of variation \(C V=.40\left(C V=\sigma_{X} / \mu_{X}\right)\). a. What are the mean value and standard deviation of \(\ln (X) ?\) b. What is the probability that \(X\) is at most 15,000 \(\mathrm{kg} / \mathrm{day} / \mathrm{km}\) ? c. What is the probability that \(X\) exceeds its mean value, and why is this probability not \(.5\) ? d. Is 17,000 the 95 th percentile of the distribution?

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.