/*! 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 64 A sample of 77 individuals worki... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

A sample of 77 individuals working at a particular office was selected and the noise level (dBA) experienced by each one was determined, yielding the following data ("Acceptable Noise Levels for Construction Site Offices, Build. Serv. Engr. Res. Technol., 2009: 87-94). \(\begin{array}{lllllllll}55.3 & 55.3 & 55.3 & 55.9 & 55.9 & 55.9 & 55.9 & 56.1 & 56.1 \\ 56.1 & 56.1 & 56.1 & 56.1 & 56.8 & 56.8 & 57.0 & 57.0 & 57.0 \\\ 57.8 & 57.8 & 57.8 & 57.9 & 57.9 & 57.9 & 58.8 & 58.8 & 58.8 \\ 59.8 & 59.8 & 59.8 & 62.2 & 62.2 & 63.8 & 63.8 & 63.8 & 63.9 \\ 53.9 & 63.9 & 64.7 & 64.7 & 64.7 & 65.1 & 65.1 & 65.1 & 65.3 \\ 65.3 & 65.3 & 65.3 & 67.4 & 67.4 & 67.4 & 67.4 & 68.7 & 68.7 \\ 58.7 & 68.7 & 69.0 & 70.4 & 70.4 & 71.2 & 71.2 & 71.2 & 73.0 \\ 73.0 & 73.1 & 73.1 & 74.6 & 74.6 & 74.6 & 74.6 & 79.3 & 79.3 \\ 79.3 & 79.3 & 83.0 & 83.0 & 83.0 & & & & \end{array}\) Use various techniques discussed in this chapter to organize, summarize, and describe the data.

Short Answer

Expert verified
The mean noise level is 65.73 dBA, median is 63.8 dBA, and modes are 56.1 and 63.9 dBA.

Step by step solution

01

Organize the Data into a Frequency Distribution

First, we will create a frequency distribution table by counting how many times each noise level appears in the data set. For example, 55.3 appears 3 times, 55.9 appears 4 times, and so on.
02

Construct a Histogram

Using the frequency distribution table from Step 1, construct a histogram. Plot the noise levels on the x-axis and the frequencies on the y-axis. This visual representation will help us see the distribution of noise levels.
03

Calculate the Mean Noise Level

To find the mean noise level, add up all the noise levels and divide by the number of observations (77). Use the formula: \[\text{Mean} = \frac{55.3\times3 + 55.9\times4 + \cdots + 83.0\times3}{77} \approx 65.73\]
04

Determine the Median Noise Level

To find the median, sort the data in ascending order and find the middle value. Since there are 77 data points, the median is the 39th value, which is 63.8 dBA.
05

Identify the Mode(s) of the Data

The mode is the noise level that appears most frequently. From the frequency distribution table in Step 1, it is clear that certain noise levels appear more often than others. Ensure identifying those repeat values is clear: noise levels 56.1 and 63.9 occur the most frequently, both appearing 5 times.
06

Calculate the Range of Noise Levels

The range is the difference between the maximum and minimum values in the data set. Here, it is \[\text{Range} = 83.0 - 55.3 = 27.7\] dBA.
07

Find the Standard Deviation

To calculate the standard deviation, first find the variance by computing the average of the squared differences from the mean. Then take the square root of this value. Using the formula: \[s = \sqrt{\frac{1}{77} \sum_{i=1}^{77} (x_i - \bar{x})^2} \] where \(\bar{x}\) is the mean computed in Step 3.

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.

Frequency Distribution
Frequency distribution is a method to organize data in a way that shows how often each value occurs. It's like sorting a pile of Lego bricks by color before you start building. In this exercise, the noise levels measured in decibels (dBA) are sorted. First, we list each unique noise level and then count the occurrence or frequency of each. For example, if 55.3 dBA appears 3 times, we note it down as:
  • 55.3: 3 times
  • 55.9: 4 times
And we continue this process for each noise level. This approach gives us a clear picture of how the noise levels are distributed across the office environment. Understanding frequency distributions is fundamental in descriptive statistics because it simplifies data, making it easier to understand and analyze.
Histogram Construction
A histogram is a visual representation of the frequency distribution we just created. Imagine drawing a bar to represent each unique noise level, where the height of the bar shows how often the noise level occurs. This is like making a bar graph.

To construct a histogram:
  • Place noise levels on the x-axis.
  • Frequencies on the y-axis.
  • Draw bars for each noise level based on their frequency.
The result is a visual summary that makes it easy to see patterns, such as which noise levels are common or rare. For example, higher bars indicate noise levels that occur often. Histograms are excellent for identifying the distribution shape, whether it's symmetric, skewed, or has multiple peaks.
Calculation of Mean
The mean, often referred to as the average, provides a single number that represents the center of the data. To calculate the mean noise level, we sum up all the noise measurements and then divide by the total number of observations, which is 77 in this case. Mathematically, this can be expressed as:
\[ \text{Mean} = \frac{\sum \text{(all noise levels)}}{77} \]
In this exercise, calculating the mean involved adding up the noise levels using their frequencies to get a total sum, and then dividing by 77, resulting in approximately 65.73 dBA. The mean offers insights into the general tendency of the dataset, helpful for understanding the overall noise environment in the office.
Median and Mode
The median and mode are other measures of central tendency that provide different views of the data. The median is the middle value when the data is sorted in ascending order. It is very useful because it isn't affected by extremely high or low values. For 77 numbers, the median is the 39th one, which we found to be 63.8 dBA. The mode is the most frequently occurring value in the data.

In our example:
  • The median gives us a central point, showing the noise level around which the other data points cluster.
  • The modes here are 56.1 and 63.9, each occurring 5 times.
Modes indicate the common noise levels, providing insight into the settings that employees most frequently experience in their environment.
Range and Standard Deviation
The range and standard deviation are measures of variability, showing how spread out the data is. The range is simple: it's the difference between the highest and lowest values. In this case, it's 83.0 minus 55.3 dBA, which equals 27.7 dBA. It gives a quick overview of the spread but doesn't indicate how data is distributed.

The standard deviation goes deeper. It measures the average distance of each data point from the mean. A small standard deviation means data points are close to the mean, while a large one indicates they are spread out. Calculating it involves a few steps:
  • Find each point's deviation from the mean.
  • Square those deviations.
  • Average the squared deviations.
  • Take the square root of that average.
This gives us an idea of the noise level's consistency in the office environment. Understanding both range and standard deviation helps in grasping the variability within any data set.

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

Let \(\bar{x}_{n}\) and \(s_{n}^{2}\) denote the sample mean and variance for the sample \(x_{1}, \ldots, x_{n}\) and let \(\bar{x}_{n+1}\) and \(s_{n+1}^{2}\) denote these quantities when an additional observation \(x_{n+1}\) is added to the sample. a. Show how \(\bar{x}_{n+1}\) can be computed from \(\bar{x}_{n}\) and \(x_{n+1}\). b. Show that $$ n s_{n+1}^{2}=(n-1) s_{n}^{2}+\frac{n}{n+1}\left(x_{n+1}-\bar{x}_{n}\right)^{2} $$ so that \(s_{n+1}^{2}\) can be computed from \(x_{n+1}, \bar{x}_{n}\), and \(s_{n^{-}}^{2}\). c. Suppose that a sample of 15 strands of drapery yarn has resulted in a sample mean thread elongation of \(12.58 \mathrm{~mm}\) and a sample standard deviation of \(.512 \mathrm{~mm}\). A 16 th strand results in an elongation value of \(11.8\). What are the values of the sample mean and sample standard deviation for all 16 elongation observations?

Consider the following data on type of health complaint \((J=\) joint swelling, \(F=\) fatigue, \(B=\) back pain, \(\mathrm{M}=\) muscle weakness, \(\mathrm{C}=\) coughing, \(\mathrm{N}=\) nose running/irritation, \(\mathbf{O}\) - other) made by tree planters. Obtain frequencies and relative frequencies for the various categories, and draw a histogram. (The data is consistent with percentages given in the article "Physiological Effects of Work Stress and Pesticide Exposure in Tree Planting by British Columbia Silviculture Workers," Ergonomics, 1993: 951-961.) \(\begin{array}{llllllllllllll}\mathrm{O} & \mathrm{O} & \mathrm{N} & \mathrm{J} & \mathrm{C} & \mathrm{F} & \mathrm{B} & \mathrm{B} & \mathrm{F} & \mathrm{O} & \mathrm{J} & \mathrm{O} & \mathrm{O} & \mathrm{M} \\ \mathrm{O} & \mathrm{F} & \mathrm{F} & \mathrm{O} & \mathrm{O} & \mathrm{N} & \mathrm{O} & \mathrm{N} & \mathrm{J} & \mathrm{F} & \mathrm{J} & \mathrm{B} & \mathrm{O} & \mathrm{C} \\ \mathrm{J} & \mathrm{O} & \mathrm{J} & \mathrm{J} & \mathrm{F} & \mathrm{N} & \mathrm{O} & \mathrm{B} & \mathrm{M} & \mathrm{O} & \mathrm{J} & \mathrm{M} & \mathrm{O} & \mathrm{B} \\ \mathrm{O} & \mathrm{F} & \mathrm{J} & \mathrm{O} & \mathrm{O} & \mathrm{B} & \mathrm{N} & \mathrm{C} & \mathrm{O} & \mathrm{O} & \mathrm{O} & \mathrm{M} & \mathrm{B} & \mathrm{F} \\\ \mathrm{J} & \mathrm{O} & \mathrm{F} & \mathrm{N} & & & & & & & & & & \end{array}\)

Do running times of American movies differ somehow from times of French movies? The authors investigated this question by randomly selecting 25 recent movies of each type, resulting in the following running times: \(\begin{array}{lrrrrrrr}\text { Am: } & 94 & 90 & 95 & 93 & 128 & 95 & 125 \\\ & 91 & 104 & 116 & 162 & 102 & 90 & 110 \\ & 92 & 113 & 116 & 90 & 97 & 103 & 95 \\ & 120 & 109 & 91 & 138 & & & \\ \text { Fr: } & 123 & 116 & 90 & 158 & 122 & 119 & 125 \\ & 90 & 96 & 94 & 137 & 102 & 105 & 106 \\ & 95 & 125 & 122 & 103 & 96 & 111 & 81 \\ & 113 & 128 & 93 & 92 & & & \end{array}\) Construct a comparative stem-and-leaf display by listing stems in the middle of your paper and then placing the Am leaves out to the left and the Fr leaves out to the right. Then comment on interesting features of the display.

The article "A Thin-Fim Oxygen Uptake Test for the Evaluation of Automotive Crankcase Lubricants" (Lubric. Engrg., 1984: 75-83) reported the following data on oxidation-induction time (min) for various commercial oils: \(\begin{array}{lllllllllll}87 & 103 & 130 & 160 & 180 & 195 & 132 & 145 & 211 & 105 & 145 \\ 153 & 152 & 138 & 87 & 99 & 93 & 119 & 129 & & & \end{array}\) a. Calculate the sample variance and standard deviation. b. If the observations were reexpressed in hours, what would be the resulting values of the sample variance and sample standard deviation? Answer without actually performing the reexpression.

The amount of flow through a solenoid valve in an automobile's pollution- control system is an important characteristic. An experiment was carried out to study how flow rate depended on three factors: armature length, spring load, and bobbin depth. Two different levels (low and high) of each factor were chosen, and a single observation on flow was made for each combination of levels. a. The resulting data set consisted of how many observations? b. Does this study involve sampling an existing population or a conceptual population?

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.