. Elements of statistics. Statistics. 96 BINOMIAL AND NORMAL DISTRIBUTIONS Ch. 4 distribution function. The mathematical procedures needed in this process are beyond the level of this course; but the reader can under- stand that the curve of Figure plays the same general role in the analysis of normal data that the curves did in Chapter 2. Figure Relative cumulative frequency distribution for the standard nor- mal frequency distribution described by formula The following problems will illustrate the uses to which Figure can be put. Problem Determ
. Elements of statistics. Statistics. 96 BINOMIAL AND NORMAL DISTRIBUTIONS Ch. 4 distribution function. The mathematical procedures needed in this process are beyond the level of this course; but the reader can under- stand that the curve of Figure plays the same general role in the analysis of normal data that the curves did in Chapter 2. Figure Relative cumulative frequency distribution for the standard nor- mal frequency distribution described by formula The following problems will illustrate the uses to which Figure can be put. Problem Determine the limits on X for the third quartile of a standard normal population of measurements. The limits required are obviously the median and Qs, respectively. If we read horizontally from .50 on the vertical scale over to the normal curve and then downward to the horizontal scale, we find that A = 0, as is to be expected. Doing likewise for .75 on the vertical scale, we find that A = ; therefore, the limits on the third quartile are A = 0 to A = Since these limits apply to any standard normal distribution, the limits of the third quartile for any particular normal distribution in terms of a measurement, .Y, can be obtained from the relation: A = (X — }*)/a. Problem 4^2. What is the probability that a measurement chosen at random from a normal population with /^ = 50 and o- = 5 will be found to lie between 50 and 52? Between 48 and 50? Between 60 and 65? To reduce this specific normal distribution to the standard normal distribution, substitute /a = 50 and o- = 5 into A = {X — /a)/o- so. Please note that these images are extracted from scanned page images that may have been digitally enhanced for readability - coloration and appearance of these illustrations may not perfectly resemble the original Fryer, Holly Claire. Wiley
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