/*! 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 23 Fire load \(\left(\mathrm{MJ} / ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Fire load \(\left(\mathrm{MJ} / \mathrm{m}^{2}\right)\) is the heat energy that could be released per square meter of floor area by combustion of contents and the structure itself. The article "Fire Loads in Office Buildings" \((J\). of Structural Engr., 1997: 365-368) gave the following cumulative percentages (read from a graph) for fire loads in a sample of 388 rooms: \(\begin{array}{lrrrrr}\text { Value } & 0 & 150 & 300 & 450 & 600 \\ \text { Cumulative } \% & 0 & 19.3 & 37.6 & 62.7 & 77.5 \\ \text { Value } & 750 & 900 & 1050 & 1200 & 1350 \\ \text { Cumulative } \% & 87.2 & 93.8 & 95.7 & 98.6 & 99.1 \\ \text { Value } & 1500 & 1650 & 1800 & 1950 & \\ \text { Cumulative } \% & 99.5 & 99.6 & 99.8 & 100.0 & \end{array}\) a. Construct a relative frequency histogram and comment on interesting features. b. What proportion of fire loads are less than 600? At least 1200 ? c. What proportion of the loads are between 600 and \(1200 ?\)

Short Answer

Expert verified
77.5% are less than 600, 1.4% are at least 1200, and 21.1% are between 600 and 1200.

Step by step solution

01

Understand the Data

The data provided consists of different fire load values along with their corresponding cumulative percentages. The values represent the amount of heat energy (in MJ/m²) that could be released, and cumulative percentages show which proportion of rooms fall below each of these values.
02

Calculate Relative Frequencies

For each fire load value, calculate the relative frequency by finding the difference in cumulative percentages. This will show the proportion of rooms within each fire load range: - From 0 to 150: Relative frequency = 0.193 - 0 = 0.193 - From 150 to 300: Relative frequency = 0.376 - 0.193 = 0.183 - From 300 to 450: Relative frequency = 0.627 - 0.376 = 0.251 - From 450 to 600: Relative frequency = 0.775 - 0.627 = 0.148 - From 600 to 750: Relative frequency = 0.872 - 0.775 = 0.097 - From 750 to 900: Relative frequency = 0.938 - 0.872 = 0.066 - From 900 to 1050: Relative frequency = 0.957 - 0.938 = 0.019 - From 1050 to 1200: Relative frequency = 0.986 - 0.957 = 0.029 - From 1200 to 1350: Relative frequency = 0.991 - 0.986 = 0.005 - From 1350 to 1500: Relative frequency = 0.995 - 0.991 = 0.004 - From 1500 to 1650: Relative frequency = 0.996 - 0.995 = 0.001 - From 1650 to 1800: Relative frequency = 0.998 - 0.996 = 0.002 - From 1800 to 1950: Relative frequency = 1.000 - 0.998 = 0.002.
03

Construct the Histogram

Plot a histogram using the relative frequencies calculated. Each bar of the histogram will represent one of the fire load ranges with height corresponding to its relative frequency. This histogram will visually show the distribution of fire loads in the office rooms.
04

Comment on Histogram Features

Observe the histogram to note that the fire load distribution is skewed to the right with most rooms having lower fire loads (up to 450). There is a smaller proportion of rooms with very high fire loads (above 1200).
05

Calculate Proportions

To find the required proportions:- \(\text{Proportion less than 600:} = \text{Cumulative percentage at 600} = 0.775 = 77.5\%\)- \(\text{Proportion at least 1200:} = 1 - \text{Cumulative percentage just before 1200} = 1 - 0.986 = 0.014 = 1.4\%\)- To find the proportion between 600 and 1200, calculate the difference in cumulative percentages: \(= \text{Cumulative percentage at 1200} - \text{Cumulative percentage at 600} = 0.986 - 0.775 = 0.211 = 21.1\%\).

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.

Relative Frequency Histogram
A relative frequency histogram is a powerful visual tool that helps us understand how data is spread across different intervals. This is especially useful in analyzing fire load patterns in office buildings, allowing for insightful conclusions regarding safety and structural assessments.
To create a relative frequency histogram, you begin by dividing your data into distinct intervals, or "bins." In this case, fire load values are divided into intervals such as 0 to 150, 150 to 300, and so on.
Next, compute the relative frequency for each interval. You do this by taking the difference between the cumulative percentages of neighboring intervals. For instance, for the interval 0 to 150, the relative frequency is 0.193, calculated by subtracting the cumulative percentage for 0 from that for 150.
  • 0 to 150: 0.193
  • 150 to 300: 0.183
  • 300 to 450: 0.251
  • 450 to 600: 0.148
Once you have determined all the relative frequencies, plot them in a bar graph. Each bar's height represents the relative frequency of its corresponding interval.
The histogram provides a clear visual representation of the distribution of fire loads across different ranges. This makes it easier to observe where most data points (rooms with specific fire loads) concentrate, offering valuable insights into how fire loads are distributed in different rooms.
Cumulative Percentages
Cumulative percentages are a handy way of understanding data in broader terms. They help us grasp the whole picture by showing the proportion of data points that fall below certain values.
Imagine cumulative percentages as a running total. They keep adding percentages as you move through your data intervals. For example, the cumulative percentage for a fire load of 300 MJ/m² is 37.6%, indicating that nearly 38% of rooms have fire loads below this level.
Here's a quick overview of how cumulative percentages work:
  • Start with the first value and its cumulative percentage.
  • Move to the next value, adding its relative frequency to the previous cumulative percentage.
  • Repeat this process for all intervals.
These percentages are derived from a cumulative frequency distribution, which has vital uses in statistical analysis. With cumulative percentages, we can easily calculate proportions for any interval of interest, such as determining what proportion of fire loads are less than 600 MJ/m² or at least 1200 MJ/m².
Using cumulative percentages, it's possible to see the progression and distribution of fire loads over the entire range of data, helping to assess safety risks and design appropriate preventive measures.
Skewed Distribution
The concept of a skewed distribution helps in recognizing patterns in your data, which is essential in making data-informed decisions.
In the context of a fire load analysis, it describes how data is distributed around the mean. A distribution is skewed when it is not symmetric, meaning more data points lean to one side. This skewness may be indicated by the shape of the histogram or other descriptive data measures.
When analyzing the fire load histogram:
  • A distribution skewed to the right, or positively skewed, means there are many rooms with lower fire loads but a few with much higher loads.
  • A left-skewed distribution would indicate the opposite, which is not the case here.
This "right skew" pattern indicates that while most rooms have fire loads lower than 450 MJ/m², a few rooms have significantly higher fire loads, extending far to the right of the histogram. Such patterns suggest critical considerations for safety and risk assessment in fire management.
Understanding skewness aids in predicting future data behavior. For example, knowing that only a small proportion of rooms have high fire loads can help prioritize safety measures where they are most needed.

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

Consider a sample \(x_{1}, \ldots, x_{n}\) with \(n\) even. Let \(\bar{x}_{L}\) and \(\bar{x}_{U}\) denote the average of the smallest \(n / 2\) and the largest \(n / 2\) observations, respectively. Show that the mean absolute deviation from the median for this sample satisfies $$ \Sigma\left|x_{i}-\tilde{x}\right| / n=\left(\bar{x}_{U}-\bar{x}_{L}\right) / 2 $$ Then show that if \(n\) is odd and the two averages are calculated after excluding the median from each half, replacing \(n\) on the left with \(n-1\) gives the correct result. [Hint: Break the sum into two parts, the first involving observations less than or equal to the median and the second involving observations greater than or equal to the median.]

Every corporation has a governing board of directors. The number of individuals on a board varies from one corporation to another. One of the authors of the article "'Does Optimal Corporate Board Size Exist? An Empirical Analysis" (J. of Applied Finance, 2010: 57-69) provided the accompanying data on the number of directors on each board in a random sample of 204 corporations. \(\begin{array}{lrrrrrr}\text { No. directors: } & 4 & 5 & 6 & 7 & 8 & 9 \\\ \text { Frequency: } & 3 & 12 & 13 & 25 & 24 & 42 \\ \text { No. directors: } & 10 & 11 & 12 & 13 & 14 & 15 \\ \text { Frequency: } & 23 & 19 & 16 & 11 & 5 & 4\end{array}\) \(\begin{array}{lrrrrr}\text { No. directors: } & 16 & 17 & 21 & 24 & 32 \\ \text { Frequency: } & 1 & 3 & 1 & 1 & 1\end{array}\) a. Construct a histogram of the data based on relative frequencies and comment on any interesting features. b. Construct a frequency distribution in which the last row includes all boards with at least 18 directors. If this distribution had appeared in the cited article, would you be able to draw a histogram? Explain. c. What proportion of these corporations have at most 10 directors? d. What proportion of these corporations have more than 15 directors?

The accompanying data came from a study of collusion in bidding within the construction industry ("Detection of Collusive Behavior," J. of Construction Engr. and \(M g m n t, 2012: 1251-1258)\). \(\begin{array}{cc}\text { No. Bidders } & \text { No. Contracts } \\ 2 & 7 \\\ 3 & 20 \\ 4 & 26 \\ 5 & 16 \\ 6 & 11 \\ 7 & 9 \\ 8 & 6 \\ 9 & 8 \\\ 10 & 3 \\ 11 & 2\end{array}\) a. What proportion of the contracts involved at most five bidders? At least five bidders? b. What proportion of the contracts involved between five and 10 bidders, inclusive? Strictly between five and 10 bidders? c. Construct a histogram and comment on interesting features.

Automated electron backscattered diffraction is now being used in the study of fracture phenomena. The following information on misorientation angle (degrees) was extracted from the article "Hbservations on the Faceted Initiation Site in the Dwell-Fatigue Tested Ti-6242 Alloy: Crystallographic Orientation and Size Effects" (Metallurgical and Materials Trans., 2006: 1507-1518). \(\begin{array}{lcccc}\text { Class: } & 0-<5 & 5-<10 & 10-<15 & 15-<20 \\\ \text { Rel freq: } & .177 & .166 & .175 & .136 \\ \text { Class: } & 20-<30 & 30-<40 & 40-<60 & 60-<90 \\ \text { Rel freq: } & .194 & .078 & .044 & .030\end{array}\) a. Is it true that more than \(50 \%\) of the sampled angles are smaller than \(15^{\circ}\), as asserted in the paper? b. What proportion of the sampled angles are at least \(30^{\circ} ?\) c. Roughly what proportion of angles are between \(10^{\circ}\) and \(25^{\circ} ?\) d. Construct a histogram and comment on any interesting features.

Allowable mechanical properties for structural design of metallic aerospace vehicles requires an approved method for statistically analyzing empirical test data. The article "Establishing Mechanical Property Allowables for Metals" (J. of Testing and Evaluation, 1998: 293-299) used the accompanying data on tensile ultimate strength (ksi) as a basis for addressing the difficulties in developing such a method. \(\begin{array}{lllllllll}122.2 & 124.2 & 124.3 & 125.6 & 126.3 & 126.5 & 126.5 & 127.2 & 127.3 \\ 127.5 & 127.9 & 128.6 & 128.8 & 129.0 & 129.2 & 129.4 & 129.6 & 130.2 \\ 130.4 & 130.8 & 131.3 & 131.4 & 131.4 & 131.5 & 131.6 & 131.6 & 131.8 \\ 131.8 & 132.3 & 132.4 & 132.4 & 132.5 & 132.5 & 132.5 & 132.5 & 132.6 \\ 132.7 & 132.9 & 133.0 & 133.1 & 133.1 & 133.1 & 133.1 & 133.2 & 133.2 \\ 133.2 & 133.3 & 133.3 & 133.5 & 133.5 & 133.5 & 133.8 & 133.9 & 134.0 \\ 134.0 & 134.0 & 134.0 & 134.1 & 134.2 & 134.3 & 134.4 & 134.4 & 134.6 \\ 134.7 & 134.7 & 134.7 & 134.8 & 134.8 & 134.8 & 134.9 & 134.9 & 135.2 \\ 135.2 & 135.2 & 135.3 & 135.3 & 135.4 & 135.5 & 135.5 & 135.6 & 135.6 \\ 135.7 & 135.8 & 135.8 & 135.8 & 135.8 & 135.8 & 135.9 & 135.9 & 135.9 \\ 135.9 & 136.0 & 136.0 & 136.1 & 136.2 & 136.2 & 136.3 & 136.4 & 136.4 \\ 136.6 & 136.8 & 136.9 & 136.9 & 137.0 & 137.1 & 137.2 & 137.6 & 137.6 \\ 137.8 & 137.8 & 137.8 & 137.9 & 137.9 & 138.2 & 138.2 & 138.3 & 138.3 \\ 138.4 & 138.4 & 138.4 & 138.5 & 138.5 & 138.6 & 138.7 & 138.7 & 139.0 \\ 139.1 & 139.5 & 139.6 & 139.8 & 139.8 & 140.0 & 140.0 & 140.7 & 140.7 \\ 140.9 & 140.9 & 141.2 & 141.4 & 141.5 & 141.6 & 142.9 & 143.4 & 143.5 \\ 143.6 & 143.8 & 143.8 & 143.9 & 144.1 & 144.5 & 144.5 & 147.7 & 147.7\end{array}\) a. Construct a stem-and-leaf display of the data by first deleting (truncating) the tenths digit and then repeating each stem value five times (once for leaves 1 and 2 , a second time for leaves 3 and 4 , etc.). Why is it relatively easy to identify a representative strength value? b. Construct a histogram using equal-width classes with the first class having a lower limit of 122 and an upper limit of 124 . Then comment on any interesting features of the histogram.

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.