Berenson-BusinessStat-Ch6-pp196-210

Berenson-BusinessStat-Ch6-pp196-210 - II-JnuL; into...

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Unformatted text preview: II-JnuL; into lYffllLKlHL MRI BE PROTECTED BY COPYRJGHT 196 CHAPTER SIX Th N alDistrib tion and Other Contin o D' tr'b tion 5 °”" “ “ “S ‘5 ‘“ ‘ LAW (TITLE 17 us CODE)" Using Statistics t is also imperative that the ign of the home page and . . . .I ‘ _ .'To check how fastthe OnC‘ampus! -- .- - - - - - - 7' artemofonCampusland . .. _ I ' .. . ‘pagetobefullydisplayed ‘ttjne is 7 seconds and the stan- _ [oilinload times are between en 3 and 11 seconds. In iii: _ ge. is distributed as a bell- flow can you use this infor- as what proportion of the I I fsecondsi'elapse before 10% ‘ .h'or‘ne page is redesigned to TRI-CITIES INTRODUCTION Numerical random variables are classified as either discrete or continuous. Chapter 5 discussed dis- crete random variables (i.e., those arising from a counting process), and in this chapter, continuous random variables are discussed. Continuous random variables arise from a measuring process where the response can take on any value within a continuum or interval. Examples include the weight of a product, time to download a Web page, advertising expenditures, sales revenues, cus- tomer servicing times, and the time between customers arriving at a bank. The mathematical expression that defines the distribution of the values for a continuous ran- dom variable is called a continuous probability density function. Figure 6.1 graphically displays the three continuous probability density functions introduced in this chapter. Panel A depicts a n0!- mal distribution. The normal distribution is symmetric and bell-shaped implying that most values tend to cluster around the mean, which, due to its symmetric shape, is equal to the median- Although the values in a normal distribution can range from negative infinity to positive infinity, the shape of the distribution makes it very unlikely that extremely large or extremely small values will occur. Panel B depicts a uniform distribution where the probability of occurrence of a value is equally likely to occur anywhere in the range between the smallest value a and the largest value it. Sometimes referred to as the rectangular distribution, the uniform distribution is symmetric and therefore the mean equals the median. An exponential distribution is illustrated in panel C. This dis‘ tribution is skewed to the right, making the mean larger than the median. The range for an exP‘" nential distribution is zero to positive infinity but note that its shape makes the occurrence 0f extremely large values unlikely. 6.1: THE NORMAL DISTRIBUTION __—___—____—__——_—_—___’_/ The most common continuous distribution used in statistics is the normal distribution Sometimes referred to as the Gaussian distribution, the normal distribution is represented by d?” classic bell-shape depicted in panel A of Figure 6.1. The probability that various values occur with1n certain ranges or intervals of the normal distribution can be calculated. However,-the exact plot?“ bility of a particular value from a continuous distribution is zero. This is what distinguishes CO“uflr uous variables, which are measured, from discrete variables, which are counted. As an example)” (in seconds) is measured, not counted. Thus, the probability that the download time for a home 6.1: The Normal Distribution 197 page on a Web browser is between 7 and 10 seconds can be computed. Narrowing this interval, the probability that the download time is between 8 and 9 seconds can be computed. The probability that the download time is between 7.99 and 8.01 seconds can also be computed. However, the prob- ability that the download time is exactly 8 seconds is zero. ' Obtaining probabilities or computing expected values and standard deviations for continuous variables involves mathematical expressions that require a knowledge of integral calculus and are Values of X beyond the scope of this book. Nevertheless, the normal distribution is considered so important for Panel A applications that special probability tables such as Tables E2 and EM have been devised to elimi- N0rmal D'SIT'bUW" nate the need for what would otherwise require laborious mathematical computations. The normal distribution is vitally important in statistics for three main reasons: 1. Numerous continuous variables common in the business world have distributions that closely resemble the normal distribution. 2. The normal distribution can be used to approximate various discrete probability distribu- tions (see section 6.5 on the CD-ROM). 3. The normal distribution provides the basis for classical statistical inference because of its Values of x relationship to the central limit theorem (which is discussed in section 7.1). Pan?“ B The normal distribution has several important theoretical properties, as illustrated in Exhibit 6.1. Uniform Distribution - EXHIBIT 6.I PROPERTIES OE THE NORhlAl DISTRIBUTION ' There are four key properties associated with the normal distribution: _ 1. It is bell—shaped (and thus symmetrical) in its appearance. _ 2. Its measures of central tendency_(mean, median, and mode) are all identical. Values of X . 3. Its “middle spread.”is equal to 1.33 standarddeviations. This means that the interquartile ' panel c . range is con , ed within aninterval of two—thirds of a standard deviation below the mean - Exponential Distribution to two-thirds of a standard deviation above the mean. ’ t ' ' ' “G u R E 6 _ 1 4. Its associated random variable. has an infinite range (-60 < X < oo). Three Continuous Distributions In practice, many variables have distributions that only approximate the theoretical properties of the normal distribution given in Exhibit 6.1. As a case in point, refer to Table 6.1. TA B L E 6 .1 - - _ . . Thickness of 10,000 brass “1mm” ""9131 “mmmw-EW “She's Under 0.0100 48/10,000=D.0048 0.0100<0.0182 122/10,000=0.0122 0.0182<0,0184 325/10,000=0.0325 0.0184<0.0186 595/10,ooo=00095 0.0186(00188 1,198/10,000=D.1198 _ 0.0100<0.0190 1,664/10,000=0.1664 0.0190<0.0192 1.896/10.000=0.1896 0.0192<o.0194 1,664/10.000=0.1664 0.0194<0.0196 1.198/10,000=0.1198 0.0r90<0.0190 695/10,000=0.0695 0.0190 <00200 325/10.0m=0.0025 0.0200<0.0202 122/10.0oo=0.0122 0.0202 orabove 40/1o,000=0.004s Total 1.0000 The data in Table 6.1 above represent the thickness (in inches) of 10,000 brass washers manufac- tured by a large company. The continuous variable of interest, thickness, can be approximated by the normal distribution. The measurements of the thickness of the 10,000 brass washers cluster in the inter— val 0.0190 < 0.0192 inch and distribute symmetrically around that grouping, forming a “bell-shaped” pattern. As demonstrated in this table, if the nonoverlapping (mutually exclusive) listing contains all pos- 198 CHAPTER SIX The Normal Distribution and Other Continuous Distributions F I G U R E 6 . 2 Relative frequency histogram and polygon of the thickness of 10,000 brass washers Source: Data are taken from ’Iizble 6.1 onpage197. sible class intervals (is collectively axhamtive), the probabilities will sum to 1. Such a probability distribu- tion can be considered as a relative frequency distribution, as described in section 2.2, where, extept for the two open-ended classes, the midpoint of every other class interval represents the data in that interval, Figure 6.2 depicts the relative frequency histogram and polygon for the distribution of the thickness of 10,000 brass washers. For these data, the first three theoretical properties of the normal distribution seem to be satisfied; however, the fourth does not hold. The random variable of interest, thickness, cannot possibly take on values of zero or below, and a washer cannot be so thick that it becomes unusable. From Table 6.1 on page 197 note that only 48 out of every 10,000 brass washers manufactured are expected to have a thickness of 0.0202 inch or more, whereas an equal number are expected to have a thickness under 0.0180 inch. Thus, the chance of randomly obtaining a washer so thin or so thick is 0.0043 + 0.0048 = 0.0096or less than 1 in 100. 3r. '5 g E. a .o 9. n. 'o 01 .9180 .0184 .0183 .0192 .0106 .0200 .0132 .0186 .0190 .0194 .0198 .0201 ' Thickness finches} _ ~ The mathematical model or expression representing a probability density function is denoted by the symbol f (X). For the normal distribution, the normal probability density function is given in Equation {6.1). [M] Note that because e and 11: are mathematical constants, the probabilities of the random variable X are dependent only on the two parameters of the normal distribution—the population mean 1' and the population standard deviation 0. Every time a particular combination of p and 0' is specifiedv a difierent normal probability distribution is generated. Figure 6.3 illustrates three different normal distributions. Distributions A and B have the same mean (u) but have different standard deviation?- On the other hand, distributions A and C have the same standard deviation (0') but have diffefem . means. Furthermore, distributions B and C depict two normal probability density functions 313‘ differ with respect to both it and o. FIG U R E 6 . 3 Three normal distributions having different parameters p and a Standardized Normal Distribution A standardized normal distribution is one whose random variable Z always hasameanp=0anda standard deviation 0‘ = 1. 1This text use: Table [5.2, the Cumulative Standardized Normal fable. To use the Standardized Normal table see Table 5.14 and 59511-431! 6.1a, Using the Standardized Normal Distribution table, on the CD~ROM. 6.1: The Normal Distribution 199 Unfortunately, the mathematical expression in Equation (6.1) is computationally tedious. To avoid such computations, a set of tables that provide the desired probabilities would be useful. However, because an infinite number of combinations of the parameters u and 6 exist, an infinite number of such tables would be required. By standardizing the data, only one table is needed (see Table 13.2 or E.l4). By the use of the transformation formula given in Equation (6.2), any normal random variable X can be converted to a standardized normal random variable Z. Although the original data for the random variable X had mean u and standard deviation 0, the standardized random variable Z will always have mean u = 0 and standard deviation 6 = 1. By substituting u = 0 and a = 1 in Equation (6.1), the probability density function of a stan— dardized normal variable 2' is given in Equation (6.3). Any set of normally distributed data can be converted to its standardized form, and then any desired probabilities can be determined from a table of the cumulative standardized normal distribu- tiOn, such as Table E.2. To see how the transformation formula is applied and the results used to find probabilities from Table E2,1 the Using Statistics scenario on page 196 can be examined. Recall that past data indicate that the time to download the Web page is normally distributed with a mean u = 7 seconds and a standard deviation 6 = 2 seconds. From Figure 6.4, observe that every measurement X has a corresponding standardized measure- ment Z obtained from the transformation formula [Equation (6.2) above]. Therefore a download time of 9 seconds is equivalent to l standardized unit (i.e., 1 standard deviation above the mean) because 200 CHAPTER Six The Normal Distribution and Other Continuous Distributions and that a download time of 1 second is equivalent to 3 standardized units (3 standard deviations) below the mean because 1 — 7 z = —— = —3 2 Thus, the standard deviation is the unit of measurement. In other words, a time of 9 seconds is 2 seconds (i.e., 1 standard deviation) higher, or slower, than the average time of 7 seconds, and a time of 1 second is 6 seconds (i.e., 3 standard deviations) lower, or faster, than the average time, On Campus! Home Page Download Time 13 XScale (u: 7. 6=2l FIGURE 6.4 +3 ,ZScale(p=0.o=1l Transformation of scales In order to further illustrate the transformation formula, suppose that the home page of another Web site had a download time that was normally distributed with a mean u of 4 seconds and a standard deviation 0' of 1 second. This distribution is depicted in Figure 6.5. Other Home Page Download Time _—b_J—|—l—I—l—|____. 1'2 3 4 5 6 7 XScale(p.=4,o=1) F I G U R E 6 . 5 A different transformation ‘3 '2 ‘1 0 +1 +2 *3 23”” (P = 0' U“ 1’ of scales l____ L—~—.——.uw—._ r A B I. E 6 . 2 Obtaining a cumulative area under the normal curve no u R E 6 .6 Detennining the area up to Z 0'" a cumulative standardized nonnal distribution 6.1: The Normal Distribution 201 Comparing these results with those of the On Campus! Web site, observe that a download time of 5 seconds is 1 standard deviation above the average download time because _s—4 1 and a time of 1 second is also 3 standard deviations below the average download time because Z =+1 Z= =-3 l — 4 l The two bell—shaped curves in Figures 6.4 and 6.5 depict the relative frequency polygons of the normal distributions representing the download time (in seconds) for the two Web sites. Because the download times represent the entire population, the probabilities or proportion of area under the entire curve must add to 1. Suppose you wanted to find the probability that the download time for the On Campus! site is less than 9 seconds. Because 9 seconds is one standard deviation above the mean, you need to find the probability that the download time is below +1 standard deviation. Table E.2 represents the cUmulative probabilities or areas under the normal curve calculated less than (i.e., to the left of) the particular values of interest X. Using Equation (6.2) on page 199, this corresponds to the probabili- ties or areas under the standardized normal curve less than the transformed values of interest Z. To use Table E.2, note that Z values need to be recorded to two decimal places. Thus, the particular Z value of interest is recorded as +1.00. To read the probability or area under the curve less than Z = +1.00, scan down the Z column from Table E.2 until you locate the Z value of interest (in 10ths) in the Z row for 1.0. Next read across this row until you intersect the column that contains the 100ths place of the Z value. Therefore, in the body of the table, the tabulated probability for Z = 1.00 corre- sponds to the intersection of the row Z = 1.0 with the column Z = .00 as shown in Table 6.2 below, which is extracted from Table E.2. This probability is 0.8413. As illustrated in Figure 6.6, there is an 84.13% chance that the download time will be less than 9 seconds. Source: Extracted from Table E.2. On Campus! Home Page Download Time Area 0.8413 X Scale 1 3 5 7 9 11 13 -3.00 —-2.00 —1.00 0 +1.00 +2.00 +3.00 Z Scale 202 CHAPTER SIX The Normal Distribution and Other Continuous Distributions On the other hand, observe from Figure 6.5 on page 200 that for the other home page, a time of 5 seconds is l standardized unit above the mean time of 4 seconds. Thus, the likelihood that the download time will be less than 5 seconds is also 0.8413. Figure 6.7 shows that regardless of the value of the mean u and standard deviation 0 of a normally distributed variable, a transformation to a standardized scale can always be made from Equation (6.2) on page 199 and, by using Table E.2, any probability or portion of area under the curve can be obtained. From Figure 6.7 observe that the probability or area under the curve less than 5 seconds for the second home page is identical to the probability or area under the curve less than 9 seconds for the On Campus! home page. Other Home Page On Campus! Home Page F I G U R E 6 . 7 Demonstrating a transformation of scales for corresponding portions under two cumulative normal curves Now that you have learned to use Table E.2 in conjunction with Equation (6.2), many different types of probability questions pertaining to the normal distribution can be resolved. Examples 6.1—6.7 illustrate this technique for the On Campus! home page. ” What is the probability that the download time will be more than 9 seconds? FINDING P”) 9) SOLUTION Because the probability that the download time will be less than 9 seconds was already determined from Figure 6.6 on page 201, the desired probability must be its complement, 1 ' 0.8413 = 0.1587. This is depicted in Figure 6.8. On Campus! Home Page Download Time Area 0.1587 1 F I G U R E 6 . 8 —3.00 —2.00 -1.00 +1.00 +2.00 +3.00 ZScaIe Finding P(X> 9) 6.1: The Normal Distribution 203 ” What is the probability that the download time will be between 7 and 9 seconds? FINDING H7 < X < 9) FIGURE 6.9 Finding PI7<X< 9) FINDING P(X < 7 0R X > 9) FleuRE 6.10 finding P(X <7orX > 9) SOLUTION From Figure 6.6 on page 201 you already determined that the probability that a down- load time will be less than 9 seconds is 0.8413. To obtain the desired results you now must determine the probability that the download time will be under 7 seconds and subtract this from the probabil- ity that the download time is under 9 seconds. This is depicted in Figure 6.9. On Campus! Home Page Download Time Area 0.5000 1 3 ' 13 4.00 —2.00 4.00 o X Scale +1.00 +2.00 +3.00 Z Scale Because the mean and median are theoretically the same for a normally distributed variable, ' 50% of the download times will be under 7 seconds. To show this, using Equation (6.2) on page 199 = 7—‘Z = 0.00 2 Using Table E.2, observe the area under the normal curve less than the mean of Z = 0.00 is 0.5000. Hence, the area under the curve between Z = 0.00 and Z = 1.00 must be 0.8413 — 0.5000 = 0.3413. What is the probability that the download time is under 7 seconds or over 9 seconds? SOLUTION Because the probability is 0.3413 that the download time between 7 and 9 seconds has already been determined, from Figure 6.9 observe that the probability must be its complement, l — 0.3413 = 0.6587. Another way to view this problem, however, is to separately obtain both the probability of a dOWnload time of under 7 seconds and the probability of a download time of over 9 seconds and to then add these two probabilities together to obtain the desired result. This result is depicted in Figure 6.10. Because the mean and median are theoretically the same for normally distributed data, it follows that 50% of download times are under 7 seconds. From Example 6.1, the probability of a download time of over 9 seconds is 0.1587. Hence, the probability that a download time is under 7 or over 9 seconds, P(X < 7 or X> 9), is 0.5000 + 0.1587 = 0.6587. On Campus! Home Page Download Time 0' Area 0.3413 since 2= x" " =>+1.oo Area 0.1587 0 XScaIe -3.00 —2.00 —1.00 0 +1.00 +2.00 +3.00 ZScaIe 204 CHAPTER SIX The Normal Distribution and Other Continuous Distributions What is the probability that the download time will be between 5 and 9 seconds, that is, P(S < X < 9)? HNDmG "5 < x< 9) SOLUTION Note from Figure 6.11 that one of the values of interest is above the mean download time of 7 seconds and the other value is below the mean. Because Table 13.2 permits you only to find probabilities less than a particular value of interest, the desired probability can be found in three steps: 1. Determine the probability of less than 9 seconds. 2. Determine the probability of less than 5 seconds. 3. Subtract the smaller result from the larger. For this example, you have already completed step I; the area under the normal curve less than 9 seconds is 0.8413. To find the area under the normal curve less than 5 seconds (step 2), Z = §——7 = —1.00 2 Using Table 13.2, look up the value Z = - 1.00 and find the probability to be 0.1587. Hence, from step 3, the probability that the download time will be between 5 and 9 seconds is 0.8413 — 0.1587 = 0.6826, as displayed in Figure 6.11. W. Area‘s 0.1587 since Cumulatiw Area = 0.8413 because 2: “1;” =4.00 2=—x—;-"—=+1.oo Area shaded blue is 03413 — 0.158? = 0.6826 1 3 5 7' 9 11 13 X Scale F I G U R E 6 . 'l 'l «3.00 -2.00 ~11!) 0 +1.00 +2.00 +3.00 ZScale Finding P(S < X < 9) ' 2The empirical rule presented The result of Example 6.4 is rather important and allows you to generalize the findings.2 For any 0'} Page {03 1'5 based 0" the normal normal distribution there is a 0.6826 chance that a randomly selected item will fall within i1 stan- d’ijzm “$5952: 3"“ 3:” dard deviation of the mean. From Figure 6.12, observe that slightly more than 95% of the items will {2m $663222: ejpiric'zl me fall within i2 standard deviations of the mean. Thus, 95.44% of the download times can be expectfd [fa dam 5e, ,adimu}, dwmfmm a to be between 3 and 11 seconds. From Figure 6.13, observe that 99.73% of the items will fall withln normal distribution, the percent- i3 standard deviations above or below the mean. Thus, 99.73% of the download times can be “805 given by the empirical rule expected to be between 1 and 13 seconds. Therefore, it is unlikely (0.0027, or only 27 in 10,000) tllat Wm beg’m’y i"““”’“" a download time will be so fast or so slow that it will take under 1 second or more than 13 seconds. This is why 66 (i.e., 3 standard deviations above the mean to 3 standard deviations below the mean) is often used as a practical approximation of the range for normally distributed data. Area below is 0.0228 because __-_r'- Area below is 0.9772 because " " = +2.00 I 1 13 XScale F I G U R E 6 . 'I 2 -3.00 -2.00 -1.00 0 +1.00 +2.00 +3.00 ZScale Finding P(3<X< H) 6.1: The Normal Distribution 205 Area below is 0.00135 because ' Area below is 0.99865 because 2: x'" =+3.oo J11 hi3 XScale F | G U R E 6 . ll 3 -3.00 —2.00 -1.00 0 +1.00 +2.00 +3.00 ZScale Finding P“ < X < 13) FINDING P” < 3.5l What is the probability that a download time will be under 3.5 seconds? SOLUTION To obtain the probability that a download time will be under 3.5 seconds, you need to examine the shaded lower left-tail region of Figure 6.14. Area is 0.0401 because 2: 1:“ a 4.75 9 11 13 'X5cale FIG U R E 6 .1 4 +1.00 +2.00 +3.00 ZSCaIe Finding P1X < 3.5) - To determine the area under the curve below 3.5 seconds, first calculate X—u 3.5—7 0 2 = —l.75 Look up the Z value of —l.75 in Table 15.2 by matching the appropriate Z row (— 1.70) with the appropriate Z column (.05) as shown in Table 6.3 (which is extracted from Table E2). The resulting probability or area under the curve less than —l.75 standard deviation below the mean is 0.0401. T A _B I. E 6 . 3 Obtaining a cumulative area under the normal curve :- =:: :-:- .0302 .0084 .0075 .0357 —1.50 .0548 .0537 .0525 .0515 .0505 .0495 .0485 .0475 .0455 .0455 Source: Extracted from Table E.2. EXPLORING THE NORMAL DISTRIBUTION Use the Visual Explorations in Statistics Normal Distribution procedure to observe the effects of changes in the arithmetic mean and standard deviation on the area under the curve. To use this procedure, open the Visual Ex lorations in Statistics Excel workbook (Visual Explorationsxla) and select VisualExplorations Normal Distribution from the Microsoft Excel menu bar. The procedure produces a normal curve for the Using Statistics home page download example and a floating control panel that allows you to adjust the shape of the curve and the shaded area under the curve (see illustration on page 206). 206 CHAPTER SIX The Normal Distribution and Other Continuous Distributions Nun’le Ui Arrhu'II-II . but u:- ” man vb lut-dm-h-ADNH nfllm ~15. r I llifla! x . an ctr! L-Q-A" 9¢v9°o¢++saesooooooeeeaoovéué99 . ' -: Hun-m: I_.Iifi!ril.1utis.n'| mm: fideuuum: [7?'"§i xvi»: Mofx<- thstdvek .._.—l Jed In Examples 6.1—6.5 the probabilities associated with various measured values were computed- Examples 6.6 and 6.7 illustrate how to determine particular numerical values of the variable Of interest that correspond to known probabilities. How much time (in seconds) will elapse before 10% of the downloads are complete? HNDING THE X VALUE FOR A SOLUTION Because 10% of the home pages are expected to download in under X seconds, the are“ CUMULATWE PROBABILITY under the normal curve less than this Z value must be 0.1000. Using the body of Table E2; You 0F 0.10 search for the area or probability of 0.1000. The closest result is 0.1003, as shown in Table 6.4 (which is extracted from Table 13.2). F 4 1' A B L E 6 . Obtaining a Z value b corresponding 10 a particular cumulative area (0.10) under the normal curve FIG U R E 6 . 'l 5 Finding Z to determineX 6.1: The Normal Distribution 207 l—1.5 .0088 .0855 -—1.4 .0808 .0793 .0778 :0764 :0749 10735 .0721 :0708 —4.3 .0988 .0951 .0881 .0823 .0985 Source: Extracted from Table 5.2. Working from this area to the margins of the table, observe that the Z value corresponding to the particular Z row ( — 1.2) and Z column (.08) is — 1.28 (see Figure 6.15). Amaunsmm Area is 0.1000 Once Z is obtained, the transformation formula Equation (6.2) on page 199 is used to deter- mine the value of interest, X. Let z=x_u 6 then Z6=X—p and p+Z6=X or X=p+ZO Substituting p = 7, 6 = 2, and Z = — 1.28, X: 7 + (—1.28)(2) = 4.44 seconds Thus, you can expect that 10% of the download times will be 4.44 seconds or less. From Example 6.6, the X value is defined in Equation (6.4), 208 CHAPTER SIX The Normal Distribution and Other Continuous Distributions To find a particular value associated with a known probability, follow the steps as displayed in Exhibit 6.2. Emmi-6.2 "Home Airmncuuii’vnu: ASSOCIATED wirH-A KNOWNPROBABILITY -. - _ - . To and a particular value assodafid'with a probability,- 1. _2. '. USiflfingajblé 1:112. germane appropriate Z v‘al'ue Corresponding to the area uhderthe nor-' _m_d®flé'lé$‘mm the-desiredx: 7 ‘ ’- ' ‘ - ' - US$18 Equafiofl' (5:31); .5551“ f0"? ‘ Sketch the normal curve, and then place the values for- themean's on the respective X and Z ,Find theirrumula'tive'aréa lees than undesired-X, _ . " Shade'the‘iarea ofiJ'ttelreSt, _, ,... ' X=nj+azd _‘ What are the lower and upper values of X, located symmetrically around the mean, that will include 95% of the download times? FINDING THE A' VALUES THAT INCIUDE 95% OF THE DOWNLOAD TIMES . SOLUTION First, you need to find the lower value of X (called X,). Then you can find the upper value of X (called XU). Since 95% of the values will be between X L and XU, and XL and XU are equal distance from the mean, 2.5% of the values will be below X L (see Figure 6.16). F I G U R E 6 . ‘l 6 Finding Z to determine X, Area is 0.9750 Although X L is not known, the corresponding standardized value Z can be obtained because the area under the normal curve less than this Z is 0.0250. Using the body of Table 6.5, search for the probability 0.0250. 1' A B L E 6 . 5 Obtaining a Zvalue corresponding to a cumulative area of 0.025 under the normal curve '—20 —1,8 .0228 .0222 .0192 .0i88 .0133 . : _ : . - . .. _ . . .0244 .0239 .0233 .0359 .0351 .0344 .0336 .0329 .0232 .0314 .0307 .0301 .0294 M Source: Extracted from Table 5.2. 6.1: The Normal Distribution 209 Working from the body of the table to the margins of the table, observe that the Z value corre- sponding to the particular 2' row (— 1.9) and Z column (.06) is —1.96. Once Z is obtained, the final step is to use Equation (6.4) on page 207 as follows, x = n + 250 = 7 + (—1.96)(2) = 7 —- 3.92 = 3.08 seconds To find XU. note that only 2.5% of the Web page downloads take longer than XU seconds, and 97.5% of the Web page download takes less than XU seconds. From the symmetry of the normal distribution, the desired Z value as shown in Figure 6.17 is +1.96 (because Z lies to the right of the standardized mean of 0). However, this Z value can also be extracted from Table 6.6, since 0.975 of the area under the normal curve is less than the standardized 2 value of +1.96. Area is 0.9750 Area is 0.0250 F I G U R E 6 . 'l 7 Finding Z to determine XU TA B l. E 6 . 6 Obtaining a Z value corresponding to a cumulative area of 0.975 under the normal curve .9893 .9899 .9706 . z . .9758 .9781 .9787 .9793 .9798 .9803 .9808 .9812 .9817 +2.0 .9772 .9778 :9783 .9788 Source: Extracted from Table 8.2. Therefore, using Equation (6.4) on page 207, X = u + 20 = 7 + (+1.96)(2) = 7 + 3.92 = 10.92 seconds Therefore, 95% of the download times will be between 3.08 and 10.92 seconds. Normal probabilities can also be obtained from Microsoft Excel or Minitab. Figure 6.18 illus— trates how Microsoft Excel can be used for Examples 6.1, 6.2, 6.5, and 6.6, and Figure 6.19 illustrates Minitab output for Examples 6.1 and 6.6. at... 210 CHAPTER 51x The Normal Distribution and Other Continuous Distributions F I G U R E 6 . 'I 8 Obtaining normal probabilities using PHStat and Microsoft Excel Cumulatlve Distribution Function Norma]. with mean - 7.00000 and standard deviation - 2.00000 x P: x <- a) 5.0000 0.8413 Inverse Cumulative Distribution Function Normal with mean = 7.00000 and standard deviation - 2.00000 F I G U R E 6 . 'l 9 Obtaining normal probabilities using Minitab Pt X <- x) x 0.1000 4.4369 Learning the Basics 6.1 Given a standardized normal distribution (with a mean of 0 and a standard deviation of 1, as in Table E.2), what is the probability that Z is less than 1.57? Z is greater than 1.84? Z is between 1.57 and 1.84? Z is less than 1.57 or greater than 1.84? Z is between — LS7 and 1.84? Z is less than — 1.57 or greater than 1.84? What is the value of Z if 50.0% of all possible Z values are larger? . What is the value of Z if only 2.5% of all possible Z values are larger? i. Between what two values of Z (symmetrically distributed around the mean) will 68.26% of all possible Z values be contained? =‘ carbon-99‘.» 6.2 Given a standardized normal distribution (with a mean of 0 and a standard deviation of 1 as in Table E.2), determine the following probabilities: a. P(Z> +1.34) d.P(Z< —1.17) b. P(Z< +1.17) e.P(-1.17<Z< +1.34) 6. P(0<Z< +1.17) f. P(—l.17<Z<—0.50) 6.3 Given a standardized normal distribution (with a mean of0 and a standard deviation of 1 as in Table E.2), what is the probability that Z is less than +1.08? Z is greater than -0.21? Z is between the mean and +1.08? Z is less than the mean or greater than +1.08? Z is between -0.21 and the mean? Z is less than -0.21 or greater than the mean? Z is between -0.21 and +1.08? Z is less than -0.21 or greater than +1.08? rm ran {in 9‘s ...
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Berenson-BusinessStat-Ch6-pp196-210 - II-JnuL; into...

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