/*! 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 27 The paper "Study on the Life Dis... [FREE SOLUTION] | 91Ó°ÊÓ

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The paper "Study on the Life Distribution of Microdrills" (J. of Engr. Manufacture, 2002: 301-305) reported the following observations, listed in increasing order, on drill lifetime (number of holes that a drill machines before it breaks) when holes were drilled in a certain brass alloy. \(\begin{array}{rrrrrrrrrr}11 & 14 & 20 & 23 & 31 & 36 & 39 & 44 & 47 & 50 \\\ 59 & 61 & 65 & 67 & 68 & 71 & 74 & 76 & 78 & 79 \\ 81 & 84 & 85 & 89 & 91 & 93 & 96 & 99 & 101 & 104 \\ 105 & 105 & 112 & 118 & 123 & 136 & 139 & 141 & 148 & 158 \\ 161 & 168 & 184 & 206 & 248 & 263 & 289 & 322 & 388 & 513\end{array}\) a. Why can a frequency distribution not be based on the class intervals \(0-50,50-100,100-150\), and so on? b. Construct a frequency distribution and histogram of the data using class boundaries \(0,50,100, \ldots\), and then comment on interesting characteristics. c. Construct a frequency distribution and histogram of the natural logarithms of the lifetime observations, and comment on interesting characteristics. d. What proportion of the lifetime observations in this sample are less than 100? What proportion of the observations are at least 200 ?

Short Answer

Expert verified
A frequency distribution cannot use the given class intervals because 50 falls on a boundary. Construct distributions and histograms with adjusted intervals. Proportion under 100: 54%, at least 200: 14%.

Step by step solution

01

Analyze Class Intervals for Part (a)

The class intervals are \(0-50\), \(50-100\), \(100-150\), and so on. These intervals have equal widths of 50. A frequency distribution cannot be based on these intervals because one of the data values (50) falls exactly on the boundary between two intervals. Standard practice requires class intervals to either be non-overlapping and mutually exclusive or definen classes in a way that removes ambiguity.
02

Frequency Distribution for Part (b)

To construct a frequency distribution with class boundaries at \(0, 50, 100, \ldots\), we count the number of observations in each interval:\- \(0-50\): 10 observations\- \(50-100\): 17 observations\- \(100-150\): 13 observations\- \(150-200\): 4 observations\- \(200-250\): 1 observation\- \(250-300\): 2 observations\- \(300-350\): 1 observation\- \(350-400\): 1 observation\- \(400-450\): 0 observations\- \(450-500\): 0 observations\- \(500-550\): 1 observation.
03

Histogram Creation for Part (b)

Plot a histogram using the intervals from Step 2 on the x-axis, and the frequency of observations on the y-axis. This histogram will show a skewed distribution with a long right tail, indicating that most drills have a relatively low lifetime before breaking.
04

Frequency Distribution with Logarithms for Part (c)

To construct a frequency distribution of the natural logarithms of lifetime observations, calculate the logarithm of each value. Use class intervals such as \((\ln(10), \ln(20))\), \((\ln(20), \ln(30))\), and similarly divide the logs into suitable intervals. Count the frequency of logarithmic values in each interval.
05

Histogram Creation with Logarithms for Part (c)

Plot the histogram using the intervals from Step 4, which may reveal a more symmetric distribution, possibly normal, indicating that the logarithms normalize the distribution skewness apparent in the original data.
06

Calculate Proportions for Part (d)

Count the number of observations below 100 and divide by the total to find the proportion: \( \frac{27}{50} = 0.54 \). For observations at least 200, count values from 200 upwards, yielding \( \frac{7}{50} = 0.14 \).

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Key Concepts

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

Frequency Distribution
A frequency distribution is a statistical tool used to show how often each value in a set of data occurs. To create one, you divide the range of your data into intervals, which are called "class intervals". Each interval is tallied up to see how many data points fall within it. This gives a clear picture of how values are distributed across your dataset.

In the context of the microdrill study, the original intervals like \(0-50\) and \(50-100\) were problematic because they placed a data value (50) on the boundary of two intervals. This overlap can create ambiguity, as it's not clear which interval the value should belong to. It is crucial when designing class intervals to ensure they are non-overlapping and mutually exclusive.

The corrected frequency distribution used intervals starting from 0 to 50, then 50 to 100, and so on, which are clear and do not overlap.
Histogram Construction
A histogram is a type of bar graph that represents the frequency distribution of a dataset. It gives a visual impression of the distribution of data, showing the number of data points in each interval as a bar.

In our exercise, the histogram constructed with intervals \(0, 50, 100, \ldots\) displayed a skewed distribution with a long right tail. This suggests that while only a few drills can machine many holes before breaking, most have a considerably lower lifetime. This type of visualization helps identify the central tendency and variability of the data.
  • Skewed Distribution: The right tail indicates a significant skewness, showing more lower value observations.
  • Long Right Tail: Implies potential outliers or a small number of high values in the dataset.
By observing the peaks and spread, one can better understand the lifetime distribution of the drills.
Data Transformation
Data transformation involves changing the data using a function, like taking the logarithm, to simplify its structure or handle outliers. This is a useful technique when dealing with skewed data.

For instance, in the microdrill exercise, transforming the drill lifetime data using natural logarithms aimed to adjust the distribution's skewness. When the raw data has a long tail or is not normally distributed, working with the logarithmic values can often produce a more symmetric and, sometimes, normally distributed data set.

In this exercise, the histogram of the logarithmic values helped in revealing a potentially normal distribution. This result implies that transformation can be valuable in making data easier to interpret and analyze.
  • Normalization: The distribution appeared more normal, facilitating further statistical analysis.
  • Symmetry: Log transformation can make skewed data appear more symmetrical.
This technique is commonly used in statistical data analysis to improve model performance and understanding.

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Most popular questions from this chapter

The article "Determination of Most Representative Subdivision" (J. of Energy Engr., 1993: 43-55) gave data on various characteristics of subdivisions that could be used in deciding whether to provide electrical power using overhead lines or underground lines. Here are the values of the variable \(x=\) total length of streets within a subdivision: $$ \begin{array}{rrrrrrr} 1280 & 5320 & 4390 & 2100 & 1240 & 3060 & 4770 \\ 1050 & 360 & 3330 & 3380 & 340 & 1000 & 960 \\ 1320 & 530 & 3350 & 540 & 3870 & 1250 & 2400 \\ 960 & 1120 & 2120 & 450 & 2250 & 2320 & 2400 \\ 3150 & 5700 & 5220 & 500 & 1850 & 2460 & 5850 \\ 2700 & 2730 & 1670 & 100 & 5770 & 3150 & 1890 \\ 510 & 240 & 396 & 1419 & 2109 & & \end{array} $$ a. Construct a stem-and-leaf display using the thousands digit as the stem and the hundreds digit as the leaf, and comment on the various features of the display. b. Construct a histogram using class boundaries 0,1000 , \(2000,3000,4000,5000\), and 6000 . What proportion of subdivisions have total length less than 2000 ? Between 2000 and 4000 ? How would you describe the shape of the histogram?

A Consumer Reports article on peanut butter (Sept. 1990) reported the following scores for various brands: \(\begin{array}{lllllllllll}\text { Creamy } & 56 & 44 & 62 & 36 & 39 & 53 & 50 & 65 & 45 & 40 \\ & 56 & 68 & 41 & 30 & 40 & 50 & 56 & 30 & 22 & \\ \text { Crunchy } & 62 & 53 & 75 & 42 & 47 & 40 & 34 & 62 & 52 & \\ & 50 & 34 & 42 & 36 & 75 & 80 & 47 & 56 & 62 & \end{array}\) Construct a comparative stem-and-leaf display by listing stems in the middle of your page and then displaying the creamy leaves out to the right and the crunchy leaves out to the left. Describe similarities and differences for the two types.

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Consider numerical observations \(x_{1}, \ldots, x_{n^{*}}\) It is frequently of interest to know whether the \(x_{i} \mathrm{~s}\) are (at least approximately) symmetrically distributed about some value. If \(n\) is at least moderately large, the extent of symmetry can be assessed from a stem-and-leaf display or histogram. However, if \(n\) is not very large, such pictures are not particularly informative. Consider the following alternative. Let \(y_{1}\) denote the smallest \(x_{i}, y_{2}\) the second smallest \(x_{i}\), and so on. Then plot the following pairs as points on a two-dimensional coordinate system: \(\left(y_{n}-\tilde{x}, \tilde{x}-y_{1}\right),\left(y_{n-1}-\tilde{x}, \tilde{x}-y_{2}\right),\left(y_{n-2}-\tilde{x}\right.\), \(\left.\tilde{x}-y_{3}\right), \ldots\) There are \(n / 2\) points when \(n\) is even and \((n-1) / 2\) when \(n\) is odd. a. What does this plot look like when there is perfect symmetry in the data? What does it look like when observations stretch out more above the median than below it (a long upper tail)? b. The accompanying data on rainfall (acre-feet) from 26 seeded clouds is taken from the article "A Bayesian Analysis of a Multiplicative Treatment Effect in Weather Modification" (Technometrics, 1975: 161-166). Construct the plot and comment on the extent of symmetry or nature of departure from symmetry. \(\begin{array}{rrrrrrr}4.1 & 7.7 & 17.5 & 31.4 & 32.7 & 40.6 & 92.4 \\ 115.3 & 118.3 & 119.0 & 129.6 & 198.6 & 200.7 & 242.5 \\ 255.0 & 274.7 & 274.7 & 302.8 & 334.1 & 430.0 & 489.1 \\ 703.4 & 978.0 & 1656.0 & 1697.8 & 2745.6 & & \end{array}\)

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