On at most 25% of the weekday mornings last year the number of vehicles passing through the intersection from 8:00 a.m.On at most 12.5% of the weekday mornings last year the number of vehicles passing through the intersection from 8:00 a.m.was either less than 675 or greater than 775. On at least 189 weekday mornings last year the number of vehicles passing through the intersection from 8:00 a.m.On at least 75% of the weekday mornings last year the number of vehicles passing through the intersection from 8:00 a.m.On approximately 95% of the weekday mornings last year the number of vehicles passing through the intersection from 8:00 a.m.Identify which of the following statements must be true. The sample mean is x - = 725 and the sample standard deviation is s = 25. was observed and recorded on every weekday morning of the last year. The number of vehicles passing through a busy intersection between 8:00 a.m. These tallies are not coincidences, but are in agreement with the following result that has been found to be widely applicable. All of the measurements are within three standard deviations of the mean, that is, between 69.92 − 3 ( 1.70 ) = 64.822 and 69.92 + 3 ( 1.70 ) = 75.02 inches. If we count the number of observations that are within two standard deviations of the mean, that is, that are between 69.92 − 2 ( 1.70 ) = 66.52 and 69.92 + 2 ( 1.70 ) = 73.32 inches, there are 95 of them. If we go through the data and count the number of observations that are within one standard deviation of the mean, that is, that are between 69.92 − 1.70 = 68.22 and 69.92 + 1.70 = 71.62 inches, there are 69 of them. The mean and standard deviation of the data are, rounded to two decimal places, x - = 69.92 and s = 1.70. A relative frequency histogram for the data is shown in Figure 2.15 "Heights of Adult Men". Table 2.2 "Heights of Men" shows the heights in inches of 100 randomly selected adult men. We start by examining a specific set of data. zip file containing this book to use offline, simply click here. You can browse or download additional books there. More information is available on this project's attribution page.įor more information on the source of this book, or why it is available for free, please see the project's home page. Additionally, per the publisher's request, their name has been removed in some passages. However, the publisher has asked for the customary Creative Commons attribution to the original publisher, authors, title, and book URI to be removed. Normally, the author and publisher would be credited here. This content was accessible as of December 29, 2012, and it was downloaded then by Andy Schmitz in an effort to preserve the availability of this book. See the license for more details, but that basically means you can share this book as long as you credit the author (but see below), don't make money from it, and do make it available to everyone else under the same terms. We will not reference the standard normal tables in these notes, and instead rely on software to generate probabilities for us.This book is licensed under a Creative Commons by-nc-sa 3.0 license. We will use this information extensively as we progress into hypothesis testing in future modules. However in the immediate application, a simple application for Susie illustrates this point. If Susie knows that a hospital has a z score of +4.00, then she can use software or consult a Standard Normal Table to convert this to a probability. For example with a mean of 50 heart attacks and a standard deviation of 10, Susie can calculate the probability of a hospital reporting more than 100 heart attacks is very low (because 100 heart attacks is in the 99th percentile, for a hospital to report more than that would be very UNLIKELY!) We can also use these properties to make statements about probabilities about relationships among observations. \(Z = \dfrac\)Ī more frequent application of standard normal curves are expressions of percentiles or probabilities. Because the value of a z score can be aligned with a specific position within a standard normal distribution, it is also possible to find the equivalent percentile for that observations.įor example, an observation that has a z score of zero would be at the 50th percentile of the data. To find the z-score for a particular observation we apply the following formula: The Z-value (or sometimes referred to as Z-score or simply Z) represents the number of standard deviations an observation is from the mean for a set of data.
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