/*! 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 9 The January 1990 issue of Arizon... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

The January 1990 issue of Arizona Trend contains a supplement describing the 12 "best" golf courses in the state. The yardages (lengths) of these courses are as follows: 6981 , 7099,6930,6992,7518,7100,6935,7518,7013,6800,7041 and \(6890 .\) Calculate the sample mean and sample standard deviation. Construct a dot diagram of the data.

Short Answer

Expert verified
Sample mean is approximately 7059.75 and standard deviation is approximately 208.48.

Step by step solution

01

Calculate the Sample Mean

To calculate the sample mean, sum up all the yardages and divide by the number of golf courses. The yardages are: 6981, 7099, 6930, 6992, 7518, 7100, 6935, 7518, 7013, 6800, 7041, and 6890. First, sum these values:\[ 6981 + 7099 + 6930 + 6992 + 7518 + 7100 + 6935 + 7518 + 7013 + 6800 + 7041 + 6890 = 84717 \]Then, divide by 12 (the number of courses):\[ \text{Sample mean} = \frac{84717}{12} \approx 7059.75 \]
02

Calculate the Sample Standard Deviation

First, find the deviation of each yardage from the mean, square these deviations, and sum them up:\[ (6981 - 7059.75)^2 + (7099 - 7059.75)^2 + \ldots + (6890 - 7059.75)^2 \]This results in:\[ 6176.0625 + 1538.0625 + 1669.0625 + 451.5625 + 210119.0625 + 1617.5625 + 1561.5625 + 210119.0625 + 69.0625 + 6760.0625 + 334.5625 + 28714.5625 = 478130 \]Now divide the sum by (n-1) to get the variance, where n is the number of courses:\[ \text{Variance} = \frac{478130}{11} \approx 43466.36 \]Take the square root of the variance to get the standard deviation:\[ \text{Sample standard deviation} = \sqrt{43466.36} \approx 208.48 \]
03

Construct a Dot Diagram

To create a dot diagram, place a number line that covers the range of the yardages. Then, above each yardage value, place a dot for each corresponding data point. For example, the value 6981 will have one dot and the value 7518 will have two dots. This visually represents the distribution of yardage lengths.

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.

Sample Mean Calculation
The sample mean is an important concept in statistics, providing an average that represents the dataset as a whole. To calculate the sample mean of a group of numbers, you simply sum all the numbers and divide by how many numbers there are.
In this exercise, to find the sample mean of the yardages of the 12 best golf courses in Arizona, start by adding up all the yardage values:
  • Add the yardages: 6981, 7099, 6930, 6992, 7518, 7100, 6935, 7518, 7013, 6800, 7041, and 6890. The sum comes out to be 84717.
  • The formula for the sample mean is: \( \bar{x} = \frac{\sum x}{n} \), where \( \sum x \) is the total sum of values and \( n \) is the number of values.
  • Divide the total sum, 84717, by the number of courses (12): \( \frac{84717}{12} \approx 7059.75 \).
The sample mean yardage of these golf courses is approximately 7059.75 yards. This average provides a central value around which the other data points cluster.
Sample Standard Deviation Calculation
The sample standard deviation is a measure of how spread out the numbers in your dataset are around the mean. It gives insight into the variability within your data set. A smaller standard deviation means the values are closely clustered around the mean, while a larger one indicates more spread.To find the sample standard deviation of the yardages:
  • First, calculate each yardage's deviation from the mean and square these differences.
  • Using the data, compute the squared deviations: \((6981 - 7059.75)^2, (7099 - 7059.75)^2, \ldots\)
  • The sum of these squared deviations is 478130.
  • Calculate the variance by dividing by (n-1), where \(n = 12\): \( \frac{478130}{11} \approx 43466.36 \).
  • The sample standard deviation is the square root of this variance: \(\sqrt{43466.36} \approx 208.48\).
This sample standard deviation of approximately 208.48 yards tells us there is some variability in the golf course lengths, but they do generally cluster around the mean of 7059.75 yards.
Dot Diagram Construction
A dot diagram, also known as a dot plot, is a simple way to visualize the frequency of data points along a number line. It's a helpful tool for displaying small sets of data to quickly observe patterns or clusters. To construct a dot diagram for the golf course yardages:
  • Draw a horizontal number line that encompasses the range of the yardages from 6800 to 7518.
  • Above each yardage on the line, place a dot for each occurrence of that value in the data set. For example, since the value 7518 appears twice, it gets two dots stacked vertically.
  • Continue placing dots over each yardage value from your list: 6981, 7099, 6930, 6992, 7518, etc.
This visual representation immediately shows the distribution and frequency of the golf yardages, making it easier to spot which yardages are most common and detect any outliers or clusters.

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

Preventing fatigue crack propagation in aircraft structures is an important element of aircraft safety. An engineering study to investigate fatigue crack in \(n=9\) cyclically loaded wing boxes reported the following crack lengths (in \(\mathrm{mm}): 2.13,2.96,3.02\) \(1.82,1.15,1.37,2.04,2.47,2.60 .\) Calculate the sample mean and sample standard deviation. Prepare a dot diagram of the data.

An article in the Journal of Physiology ["Response of Rat Muscle to Acute Resistance Exercise Defined by Transcriptional and Translational Profiling" (2002, Vol. 545, pp. \(27-41\) ) ] studied gene expression as a function of resistance exercise. Expression data (measures of gene activity) from one gene are shown in the following table. One group of rats was exercised for six hours while the other received no exercise. Compute the sample mean and standard deviation of the exercise and no-exercise groups separately. Construct a dot diagram for the exercise and no-exercise groups separately. Comment on any differences for the groups.

The following data are the joint temperatures of the O-rings \(\left({ }^{\circ} \mathrm{F}\right)\) for each test firing or actual launch of the space shuttle rocket motor (from Presidential Commission on the Space Shuttle Challenger Accident, Vol. \(1,\) pp. \(129-131\) ): 84,49 61,40,83,67,45,66,70,69,80,58,68,60,67,72,73,70,57 63,70,78,52,67,53,67,75,61,70,81,76,79,75,76,58,31 (a) Compute the sample mean and sample standard deviation and construct a dot diagram of the temperature data. (b) Set aside the smallest observation \(\left(31^{\circ} \mathrm{F}\right)\) and recompute the quantities in part (a). Comment on your findings. How "different" are the other temperatures from this last value?

The following data represent the yield on 90 consecutive batches of ceramic substrate to which a metal coating has been applied by a vapor-deposition process. Construct a stem-andleaf display for these data. Calculate the median and quartiles of these data. $$\begin{array}{lllllllll}94.1 & 86.1 & 95.3 & 84.9 & 88.8 & 84.6 & 94.4 & 84.1 \\ 93.2 & 90.4 & 94.1 & 78.3 & 86.4 & 83.6 & 96.1 & 83.7 \\ 90.6 & 89.1 & 97.8 & 89.6 & 85.1 & 85.4 & 98.0 & 82.9 \\ 91.4 & 87.3 & 93.1 & 90.3 & 84.0 & 89.7 & 85.4 & 87.3 \\ 88.2 & 84.1 & 86.4 & 93.1 & 93.7 & 87.6 & 86.6 & 86.4 \\ 86.1 & 90.1 & 87.6 & 94.6 & 87.7 & 85.1 & 91.7 & 84.5 \\ 95.1 & 95.2 & 94.1 & 96.3 & 90.6 & 89.6 & 87.5 & \\ 90.0 & 86.1 & 92.1 & 94.7 & 89.4 & 90.0 & 84.2 & \\ 92.4 & 94.3 & 96.4 & 91.1 & 88.6 & 90.1 & 85.1 & \\ 87.3 & 93.2 & 88.2 & 92.4 & 84.1 & 94.3 & 90.5 & \\ 86.6 & 86.7 & 86.4 & 90.6 & 82.6 & 97.3 & 95.6 & \\ 91.2 & 83.0 & 85.0 & 89.1 & 83.1 & 96.8 & 88.3 &\end{array}$$

The following data are direct solar intensity measurements (watts/m \(^{2}\) ) on different days at a location in southern Spain: 562,869,708,775,775,704,809,856,655,806,878 909,918,558,768,870,918,940,946,661,820,898,935 \(952,957,693,835,905,939,955,960,498,653,730,\) and 753\. Calculate the sample mean and sample standard deviation. Prepare a dot diagram of these data. Indicate where the sample mean falls on this diagram. Give a practical interpretation of the sample mean.

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.