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91Ó°ÊÓ

The accompanying frequency distribution of fracture strength (MPa) observations for ceramic bars fired in a particular kiln appeared in the article "Evaluating Tunnel Kiln Performance" (Amer: Ceramic Soc. Bull., Aug. 1997: 59-63). \(\begin{array}{lccccc}\text { Class } & 81-<83 & 83-<85 & 85-<87 & 87-<89 & 89-<91 \\ \text { Frequency } & 6 & 7 & 17 & 30 & 43 \\ \text { Class } & 91-<93 & 93-<95 & 95-<97 & 97-<99 \\ \text { Frequency } & 28 & 22 & 13 & 3\end{array}\) a. Construct a histogram based on relative frequencies, and comment on any interesting features. b. What proportion of the strength observations are at least 85? Less than 95 ? c. Roughly what proportion of the observations are less than 90 ?

Short Answer

Expert verified
a. The histogram peaks at 89-<91, skewed left. b. At least 85: ~92.3%, less than 95: ~90.5%. c. Less than 90: ~35.5%.

Step by step solution

01

Calculate Total Frequency

Add up all the frequencies to get the total number of observations. We have frequencies: 6, 7, 17, 30, 43, 28, 22, 13, and 3. Total frequency = 6 + 7 + 17 + 30 + 43 + 28 + 22 + 13 + 3 = 169.
02

Calculate Relative Frequencies

Divide each class frequency by the total frequency to find the relative frequency for each class.For example, for the class 81-<83, the relative frequency is \( \frac{6}{169} \approx 0.036 \).Do this for all classes:- 81-<83: \( \frac{6}{169} \approx 0.036 \)- 83-<85: \( \frac{7}{169} \approx 0.041 \)- 85-<87: \( \frac{17}{169} \approx 0.101 \)- 87-<89: \( \frac{30}{169} \approx 0.178 \)- 89-<91: \( \frac{43}{169} \approx 0.254 \)- 91-<93: \( \frac{28}{169} \approx 0.166 \)- 93-<95: \( \frac{22}{169} \approx 0.130 \)- 95-<97: \( \frac{13}{169} \approx 0.077 \)- 97-<99: \( \frac{3}{169} \approx 0.018 \).
03

Construct the Histogram

Draw the histogram with classes on the horizontal axis and relative frequencies on the vertical axis. Each bar will represent one class, and its height will correspond to the relative frequency. - 81-<83 with height 0.036 - 83-<85 with height 0.041 - 85-<87 with height 0.101 - 87-<89 with height 0.178 - 89-<91 with height 0.254 - 91-<93 with height 0.166 - 93-<95 with height 0.130 - 95-<97 with height 0.077 - 97-<99 with height 0.018.
04

Comment on the Histogram

The histogram shows that the majority of fracture strengths fall in the 89-<91 category, indicating a peak at this range. The distribution is slightly skewed to the left with a gradual decrease in frequency as we move towards higher strength values.
05

Calculate Proportion at Least 85

Add all frequencies from the class 85-<87 onwards and divide by the total frequency to find the proportion of observations at least 85.Proportion = \( \frac{17 + 30 + 43 + 28 + 22 + 13 + 3}{169} = \frac{156}{169} \approx 0.923 \).
06

Calculate Proportion Less Than 95

Add frequencies from the beginning up to but not including the class 95-<97 and divide by the total frequency to determine the proportion of observations less than 95.Proportion = \( \frac{6 + 7 + 17 + 30 + 43 + 28 + 22}{169} = \frac{153}{169} \approx 0.905 \).
07

Calculate Proportion Less Than 90

Add all frequencies from the beginning to class 89-<91 and divide by the total frequency to find the proportion of observations less than 90.Proportion = \( \frac{6 + 7 + 17 + 30}{169} = \frac{60}{169} \approx 0.355 \).

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Key Concepts

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

Relative Frequency
Relative frequency is the count of a certain class of data divided by the total number of observations. It provides insight into how common or rare a particular class is within a data distribution. To calculate it, you take the frequency of a class and divide it by the total number of data points.

For example, in the exercise, if you have a class of fracture strength observations, such as 81-<83, with a frequency of 6, and the total frequency (or total observations) is 169, then the relative frequency is calculated as follows:

\[\text{Relative Frequency for 81-<83} = \frac{6}{169} \approx 0.036\]

Repeat this calculation for each class to see how each compares to the entire dataset. This method transforms raw data into a form that is easy to interpret and compare across different categories.
Proportion Calculation
Proportion calculation is a way of understanding what fraction of your dataset meets certain criteria. It is crucial for comparing different parts of a dataset or analyzing specific conditions.

For instance, if you want to find the proportion of observation strengths that are at least 85 MPa, you sum the frequencies of the classes 85-<87 and above, then divide by the total number of observations.

This is how it's done for this problem:
  • Observations "at least 85" start from the class 85-<87. Add the frequencies for these classes: 17, 30, 43, 28, 22, 13, and 3.
  • Total: 156.
  • Proportion of observations at least 85: \( \frac{156}{169} \approx 0.923 \).

Similarly, you would add the frequencies of classes under 95 to find the proportion less than 95 MPa. This technique allows you to focus on sections of the data that meet specific thresholds, facilitating more targeted analysis.
Data Distribution
Data distribution involves understanding how data points are spread across different classes or intervals. This is typically visualized using a histogram, which provides a graphical representation of frequency distribution.

A histogram helps identify patterns, such as:

  • Where most data is concentrated (shown by taller bars).
  • Patterns of data skewness (left or right).
  • Gaps or dips in the distribution.

In the example exercise, a histogram drawn with relative frequencies on the vertical axis demonstrates that the most common observations lie within the 89-<91 range. It indicates that more ceramic bars have a fracture strength within this class. The histogram also lets us see that as strengths increase above 91, the frequency gradually decreases, showing a slight skew to the left. This visualization is essential for understanding the data's overall shape and any anomalies it may possess.
Frequency Distribution Analysis
Frequency distribution analysis examines how often data points occur within specified ranges. It is particularly useful for deriving insights from large datasets and allows for easier identification of significant patterns and trends.

Consider our example. By analyzing the frequency of data points across the ceramic bars' fracture strengths, specific observations are made:

  • The class with the highest frequency (43) is 89-<91, indicating a concentration of data in this strength range.
  • The distribution exhibits a left-skewed pattern, meaning most data points are concentrated towards the higher strength values but quickly drop off.
  • Lower and higher classes like 81-<83 and 97-<99 have notably fewer observations, represented by much lower frequencies.

This kind of analysis helps in making informed decisions based on data patterns. Such evaluations become foundational in predicting behaviors, making strategic adjustments, or even justifying changes in processes.

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Most popular questions from this chapter

The number of contaminating particles on a silicon wafer prior to a certain rinsing process was determined for each wafer in a sample of size 100 , resulting in the following frequencies: \(\begin{array}{lrrrrrrrr}\text { Number of particles } & 0 & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\ \text { Frequency } & 1 & 2 & 3 & 12 & 11 & 15 & 18 & 10 \\ \text { Number of particles } & 8 & 9 & 10 & 11 & 12 & 13 & 14 & \\ \text { Frequency } & 12 & 4 & 5 & 3 & 1 & 2 & 1 & \end{array}\) a. What proportion of the sampled wafers had at least one particle? At least five particles? b. What proportion of the sampled wafers had between five and ten particles, inclusive? Strictly between five and ten particles? c. Draw a histogram using relative frequency on the vertical axis. How would you describe the shape of the histogram?

Every score in the following batch of exam scores is in the \(60 \mathrm{~s}, 70 \mathrm{~s}, 80 \mathrm{~s}\), or \(90 \mathrm{~s}\). A stem-and-leaf display with only the four stems \(6,7,8\), and 9 would not give a very detailed description of the distribution of scores. In such situations, it is desirable to use repeated stems. Here we could repeat the stem 6 twice, using \(6 \mathrm{~L}\) for scores in the low 60 s (leaves \(0,1,2,3\), and 4 ) and \(6 \mathrm{H}\) for scores in the high 60 s (leaves \(5,6,7,8\), and 9 ). Similarly, the other stems can be repeated twice to obtain a display consisting of eight rows. Construct such a display for the given scores. What feature of the data is highlighted by this display? \(\begin{array}{lllllllllllll}74 & 89 & 80 & 93 & 64 & 67 & 72 & 70 & 66 & 85 & 89 & 81 & 81 \\ 71 & 74 & 82 & 85 & 63 & 72 & 81 & 81 & 95 & 84 & 81 & 80 & 70 \\ 69 & 66 & 60 & 83 & 85 & 98 & 84 & 68 & 90 & 82 & 69 & 72 & 87 \\ 88 & & & & & & & & & & & & \end{array}\)

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 "Observations 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 Evahuation, 1998: 293-299) used the accompanying data on tensile ultimate strength (ksi) as a basis for addressing the difficulties in developing such a method. \(\begin{array}{lllllllll}122.2 & 124.2 & 124.3 & 125.6 & 126.3 & 126.5 & 126.5 & 127.2 & 127.3 \\ 127.5 & 127.9 & 128.6 & 128.8 & 129.0 & 129.2 & 129.4 & 129.6 & 130.2 \\ 130.4 & 130.8 & 131.3 & 131.4 & 131.4 & 131.5 & 131.6 & 131.6 & 131.8 \\ 131.8 & 132.3 & 132.4 & 132.4 & 132.5 & 132.5 & 132.5 & 132.5 & 132.6\end{array}\) \(\begin{array}{lllllllll}132.7 & 132.9 & 133.0 & 133.1 & 133.1 & 133.1 & 133.1 & 133.2 & 133.2 \\ 133.2 & 133.3 & 133.3 & 133.5 & 133.5 & 133.5 & 133.8 & 133.9 & 134.0 \\ 134.0 & 134.0 & 134.0 & 134.1 & 134.2 & 134.3 & 134.4 & 134.4 & 134.6 \\ 134.7 & 134.7 & 134.7 & 134.8 & 134.8 & 134.8 & 134.9 & 134.9 & 135.2 \\ 135.2 & 135.2 & 135.3 & 135.3 & 135.4 & 135.5 & 135.5 & 135.6 & 135.6 \\ 135.7 & 135.8 & 135.8 & 135.8 & 135.8 & 135.8 & 135.9 & 135.9 & 135.9 \\ 135.9 & 136.0 & 136.0 & 136.1 & 136.2 & 136.2 & 136.3 & 136.4 & 136.4 \\ 136.6 & 136.8 & 136.9 & 136.9 & 137.0 & 137.1 & 137.2 & 137.6 & 137.6 \\ 137.8 & 137.8 & 137.8 & 137.9 & 137.9 & 138.2 & 138.2 & 138.3 & 138.3 \\ 138.4 & 138.4 & 138.4 & 138.5 & 138.5 & 138.6 & 138.7 & 138.7 & 139.0 \\ 139.1 & 139.5 & 139.6 & 139.8 & 139.8 & 140.0 & 140.0 & 140.7 & 140.7 \\ 140.9 & 140.9 & 141.2 & 141.4 & 141.5 & 141.6 & 142.9 & 143.4 & 143.5 \\ 143.6 & 143.8 & 143.8 & 143.9 & 144.1 & 144.5 & 144.5 & 147.7 & 147.7\end{array}\) 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.

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