/*! 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 65 The accompanying frequency distr... [FREE SOLUTION] | 91Ó°ÊÓ

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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) Histogram shows left skew; peak at 89-<91. b) 92.3% at least 85; 90.5% less than 95. c) Roughly 35.5% are less than 90.

Step by step solution

01

Calculate Total Number of Observations

Add up all the frequencies given in the table. This sum is the total number of observations.\[ 6 + 7 + 17 + 30 + 43 + 28 + 22 + 13 + 3 = 169 \]So, there are 169 observations in total.
02

Calculate Relative Frequencies

For each class interval, calculate the relative frequency by dividing the frequency of each class by the total number of observations.\[ \text{Relative Frequency} = \frac{\text{Class Frequency}}{\text{Total Observations}} \]Applying this:- For 81-<83: \( \frac{6}{169} \approx 0.036 \)- For 83-<85: \( \frac{7}{169} \approx 0.041 \)- For 85-<87: \( \frac{17}{169} \approx 0.101 \)- For 87-<89: \( \frac{30}{169} \approx 0.178 \)- For 89-<91: \( \frac{43}{169} \approx 0.254 \)- For 91-<93: \( \frac{28}{169} \approx 0.166 \)- For 93-<95: \( \frac{22}{169} \approx 0.130 \)- For 95-<97: \( \frac{13}{169} \approx 0.077 \)- For 97-<99: \( \frac{3}{169} \approx 0.018 \).
03

Construct the Histogram

Use the relative frequencies obtained in Step 2 to draw the histogram. The x-axis represents the class intervals, and the y-axis represents the relative frequencies. Each bar's height corresponds to the relative frequency of the class interval it represents. This histogram visualizes the distribution of fracture strengths.
04

Analyze Interesting Features of the Histogram

From the histogram, observe that the distribution has a peak in the 89-<91 interval. The shape indicates a unimodal distribution that is somewhat skewed to the left, as the majority of the data is on the higher end (89-<93).
05

Calculate Proportion of Observations at Least 85

Sum the frequencies of classes starting from 85-<87 to the last interval and divide by the total number of observations.\[ = \frac{17 + 30 + 43 + 28 + 22 + 13 + 3}{169} = \frac{156}{169} \approx 0.923 \] So, approximately 92.3% of the observations are at least 85.
06

Calculate Proportion of Observations Less Than 95

Sum the frequencies of classes from the start up to the 93-<95 interval and divide by the total number of observations.\[ = \frac{6 + 7 + 17 + 30 + 43 + 28 + 22}{169} = \frac{153}{169} \approx 0.905 \]Thus, approximately 90.5% of the observations are less than 95.
07

Calculate Proportion of Observations Less Than 90

Sum the frequencies of classes from the start up to the 87-<89 interval and divide by the total number of observations.\[ = \frac{6 + 7 + 17 + 30}{169} = \frac{60}{169} \approx 0.355 \]Therefore, roughly 35.5% of the observations are less than 90.

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Key Concepts

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

Histogram
A histogram is a powerful tool utilized in data visualization to showcase the distribution of numerical data. It consists of adjoining bars, each representing frequency data within specified intervals or bins. In the context of our exercise, the histogram for fracture strength observations serves as a graphical depiction of the frequency distribution data of ceramic bars.
To create a histogram, the class intervals (for example, 81-<83, 83-<85) are plotted along the x-axis, while the relative frequency of occurrences is plotted on the y-axis. This allows for a straightforward interpretation of data, highlighting how often data points fall within each range. Each bar's height corresponds to its class interval's relative frequency.
When observing the histogram, certain patterns may emerge. For instance, a peak in one of the intervals indicates a higher concentration of data points, giving insights into dominant outcomes or potential skewness. Such visual cues can reveal that the data is unimodal, skewed to the left or right, or more uniformly distributed.
Understanding these features provides meaningful insights into the nature of the data and how the fracture strengths are spread across the ceramic bars in question.
Relative Frequency
Relative frequency is a critical concept in statistics that helps in understanding how often a specific event occurs within a data set relative to the total number of events. Unlike absolute frequency, which simply counts occurrences, relative frequency expresses these occurrences as a proportion of the whole dataset, making it easier to compare different data sets.
To calculate the relative frequency of each class in our fracture strength exercise, you divide the frequency of a particular class by the total number of observations. For example, if 6 observations fall into the 81-<83 MPa category and there are 169 total observations, the relative frequency is calculated as \( \frac{6}{169} \approx 0.036 \).
This process is repeated for all class intervals, providing a clearer understanding of the distribution proportions. Relative frequency is vital for constructing a histogram, as it defines the y-axis values that determine the height of the bars. Analyzing these values gives insight into the proportion of data falling within certain intervals, essential for a meaningful interpretation of how the data is distributed within the sample.
Data Analysis
Data analysis involves examining, cleaning, transforming, and modeling data to discover useful information and support decision-making processes. In the context of the exercise, analyzing fracture strength data involves interpreting the histogram and calculating the proportion of observations within certain class intervals.
Some of the essential steps in data analysis include calculating cumulative frequencies, determining percentages, and understanding the data's spread. For example, when asked how many observations are at least 85 MPa, we examine all intervals beginning with 85-<87. Via these calculations, we find that approximately 92.3% of the observations meet this criterion.
Similarly, if examining the proportion of data less than 95 MPa, we sum the frequencies in intervals up to 93-<95, showing that around 90.5% of observations fall within this range. Such analyses help contextualize data and allow extrapolation of potential trends. They add layers of understanding necessary for interpreting key statistics and making factual conclusions about the quality or characteristics of the ceramic bars tested.

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Most popular questions from this chapter

Temperature transducers of a certain type are shipped in batches of 50 . A sample of 60 batches was selected, and the number of transducers in each batch not conforming to design specifications was determined, resulting in the following data: \(\begin{array}{llllllllllllllllllll}2 & 1 & 2 & 4 & 0 & 1 & 3 & 2 & 0 & 5 & 3 & 3 & 1 & 3 & 2 & 4 & 7 & 0 & 2 & 3 \\ 0 & 4 & 2 & 1 & 3 & 1 & 1 & 3 & 4 & 1 & 2 & 3 & 2 & 2 & 8 & 4 & 5 & 1 & 3 & 1 \\ 5 & 0 & 2 & 3 & 2 & 1 & 0 & 6 & 4 & 2 & 1 & 6 & 0 & 3 & 3 & 3 & 6 & 1 & 2 & 3\end{array}\) a. Determine frequencies and relative frequencies for the observed values of \(x=\) number of nonconforming transducers in a batch. b. What proportion of batches in the sample have at most five nonconforming transducers? What proportion have fewer than five? What proportion have at least five nonconforming units? c. Draw a histogram of the data using relative frequency on the vertical scale, and comment on its features.

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