ࡱ> C` bjbj ~GpAR &&&&---4.h">|. ƙ((t)))$hUt-e1)e1e1&&tJxŨcCcCcCe1&<^-tcCe1cCcCJH\b#| -t 3/k42BDۨ0 ɭ1:ɭɭ- )n,,cC-dL/)))C^))) e1e1e1e1...d`vD...v...&&&&&& Chapter 6 Continuous Random Variables Aims In this chapter we deal with: Continuous random variables Probability density functions Normal distributions Working in standard units (i.e., z-scores) By the end of this chapter you should: know the properties of a probability density function know the properties of a Normal probability density function in particular, the 689599.7 rule be able to use software and computer output to solve Normal probability problems Inverse Normal problems understand and be able to apply z-scores  Problem: Diagnosing Spina Bifida A screening test for spina bifida in a foetus involves measuring the concentration of alpha fetoprotein in the mothers urine. For mothers carrying healthy foetuses, the mean is 15.7 micromoles/litre ((M/L) and the standard deviation is 0.7 (M/L. For mothers carrying foetuses with spina bifida, the mean is 23.1 (M/L and the standard deviation is 4.1 (M/L. To operate this screening test for spina bifida, medical professionals must set a threshold concentration of alpha fetoprotein, T, say. If the alpha fetoprotein concentration is below T, the foetus is diagnosed as not having spina bifida, whereas if it is above T further testing is required. Let: X be the concentration of alpha fetoprotein for a mother carrying a healthy foetus Y be the concentration of alpha fetoprotein for a mother carrying a foetus with spina bifida.  Suppose T was set at 17.8 (M/L: What is the probability that a foetus with spina bifida is correctly diagnosed? What is the probability that a foetus without spina bifida is correctly diagnosed? To ensure that 99% of foetuses with spina bifida are correctly diagnosed, at what value should T be set? Properties of a Probability Density Function (p.d.f.) The p.d.f. curve is always above or on the x-axis. Probabilities are represented by under the p.d.f. curve. pr(a d" X d" b) = under the p.d.f. curve The total area under a p.d.f. curve = Endpoints of Intervals For continuous random variables: pr(a d" X d" b ) = pr(a < X d" b ) = pr(a d" X < b ) = pr(a < X < b ) In calculations involving continuous random variables, we do not have to worry about whether interval endpoints are The Normal Distribution Heights of Male Students In a recent survey of 544 male Introductory Statistics students the mean height was 177.1cm and the standard deviation (sd) was 7.7cm. In this survey: mean 1 sd = 177.1cm 7.7cm = 169.4cm mean + 1 sd = 177.1cm + 7.7cm = 184.8cm have a height within 1 standard deviation of the mean. mean 2 sd = 177.1cm 2 7.7cm = 161.7cm mean + 2 sd = 177.1cm + 2 7.7cm = 192.5cm have a height within 2 standard deviations of the mean. mean 3 sd = 177.1cm 3 7.7cm = 154.0cm mean + 3 sd = 177.1cm + 3 7.7cm = 200.2cm 542 / 544 = 99.6% have a height within 3 standard deviations of the mean.  Percentage Body Fat for Competitive Cyclists The distribution of percentage body fat for competitive cyclists is modelled by a Normal distribution with a mean of 9 and a standard deviation of 3. Let X be the percentage body fat of a competitive cyclist. So X ~ Normal( = 9,  = 3) Based on this model: a. approximately 95% of competitive cyclists have percentage body fat somewhere between    b. for a randomly chosen competitive cyclist, there is a probability of 0.68 that the cyclist has percentage body fat somewhere between    Annual Compound Share Returns for Large Companies The distribution of annual compound share returns for large companies in the United States (as a percentage) is modelled by a Normal distribution with a mean of 11.0 and a standard deviation of 20.3. Let X be the annual compound share return for a large company in the United States (as a percentage). So X ~ Normal( = 11.0,  = 20.3) Using the computer output given, find the probability that a randomly chosen large US company has an annual compound share return of: a. less than 0% (i.e., a negative return).    b. more than 50%.    c. between 10% and 30%.  Computer Output Normal with mean = 11.0 and standard deviation = 20.3 x P(X <= x) 0 0.2940 10 0.4804 30 0.8254 50 0.9726 Spina Bifida We will model the distributions of alpha fetoprotein with Normal distributions. Let: X be the concentration of alpha fetoprotein for a mother carrying a healthy foetus Y be the concentration of alpha fetoprotein for a mother carrying a foetus with spina bifida So X ~ Normal(X = 15.7, X = 0.7) Y ~ Normal(Y = 23.1, Y = 4.1) Note: The threshold is set at 17.8 (M/L, i.e., if the concentration of alpha fetoprotein is below 17.8 (M/L the foetus would be diagnosed as not having spina bifida. a. For a foetus without spina bifida, what is the probability that it is correctly diagnosed?   b. What is the probability that a foetus with spina bifida is correctly diagnosed?    Computer Output Normal with mean = 15.7 and Normal with mean = 23.1 and standard deviation = 0.7 standard deviation = 4.1 x P(X <= x) x P(X <= x) 17.8 0.9987 17.8 0.0981 Inverse Normal Problems IQ Scores The distribution of IQ scores is modelled by a Normal distribution with a mean of 100 and a standard deviation of 15. Let X be the IQ of a person. So X ~ Normal( = 100,  = 15) a. Find the IQ score that the bottom 20% of the population fall below (i.e., the 20th percentile).    b. What IQ score is exceeded by only the top 10% of the population?    c. Find the interquartile range for IQ scores.      Computer Output Normal with mean = 100 and standard deviation = 15 P(X <= x) x 0.10 80.78 0.20 87.38 0.25 89.88 0.75 110.12 0.80 112.62 0.90 119.22 Human Gestation Period The distribution of the natural human gestation period (in weeks) is modelled by a Normal distribution with a mean of 40 and a standard deviation of 2.3. Let X be the natural human gestation period in weeks. So X ~ Normal( = 40,  = 2.3) a. Before how many weeks do 10% of births occur?   b. What is the range of gestation period for the central 50% of births?      Range is to weeks c. 95% of births occur after how many weeks?    Computer Output Normal with mean = 40 and standard deviation = 2.3 P(X <= x) x 0.05 36.22 0.10 37.05 0.25 38.45 0.75 41.55 0.90 42.95 0.95 43.78 Working in Standard Units Test and Examination Results In a recent semester a student obtained the following marks for the Introductory Statistics mid-semester test and exam. Test: 19/25 Exam: 40/50 In which of these assessments did this student achieve better results in terms of ranking with students in the same course? The distribution of test marks is modelled by a Normal distribution with mean 13.5 and standard deviation 4.6 and the distribution of exam marks is modelled by a Normal distribution with mean 30.1 and standard deviation 9.4. Let T be the test mark E be the exam mark So T ~ Normal(T = 13.5, T = 4.6) E ~ Normal(E = 30.1, E = 9.4) Test: z-score for 19 = Exam: z-score for 40 = This student did better in the in terms of ranking with students in the same course. Percentage Body Fat for Competitive Swimmers The distribution of percentage body fat for competitive swimmers is modelled by a Normal distribution with a mean of 10 and a standard deviation of 4. Let X be the percentage body fat of a competitive swimmer. So X ~ Normal( = 10,  = 4) Find the z-score for these percentage body fat values for competitive swimmers: a. x = 18 z = (i.e., 18 is sd the mean of 10) b. x = 6 z = (i.e., 6 is sd the mean of 10) c. x = 12.6 z = (i.e., 12.6 is sd the mean of 10)  Find the z-score for these percentage body fat values for competitive swimmers: d. x = 15.7 z = e. x = 8.9 z = Calculate the percentage body fat values (i.e., x-scores) for competitive swimmers with the following z-scores: f. z = 2.3  x is 2.3 the mean so  x = g. z = 1.8  x is 1.8 the mean so  x = Continuous Random Variables If a random variable, X, can take any value in some interval it is called a continuous random variable. There are no gaps between the values a continuous random variable can take. Continuous random variables measure characteristics of items chosen from a: clearly defined finite population, or random process producing observations Examples: time, weight, concentration of alpha fetoprotein, annual share return for a company. Properties of a Probability Density Function (p.d.f.) 1. The p.d.f. curve is always above or on the x-axis. 2. Probabilities are represented by areas under the p.d.f. curve. pr(a d" X d" b) = area under the p.d.f. curve between x = a and x = b. 3. The total area under a p.d.f. curve = 1 Mean (Expected Value) and Standard Deviation The expected value of a random variable X, E(X ), is: the long-run average for X-values also called the population mean, X (often shortened to ) where the probability density curve balances    The standard deviation of a random variable X, sd(X ), is: a measure of the variability (spread) of X-values roughly, the average distance of the X-values from the population mean also called the population standard deviation, X (often shortened to ) The Normal Distribution If the distribution of a random variable, X, has a Normal distribution with mean and standard deviation  we write: X ~ Normal (, ) and  are called the parameters of the distribution. Features of the Normal density curve: Symmetric and bell-shaped Centred at  determines the spread (and hence the height)  Obtaining Normal Probabilities Use statistical software, e.g. Excel, SPSS. When obtaining Normal probabilities we give the software an x-value and it returns a probability. Most software calculates the value of pr(X d" x), the cumulative or lower-tail probability. Method for obtaining Normal probabilities: Sketch a Normal curve, marking on the mean and value(s) of interest. Shade the area under the curve corresponding to the required probabilities. Obtain the desired probabilities from the lower-tail probabilities provided by the software.         Inverse Normal Problems Also use statistical software, e.g. Excel, SPSS. When solving inverse Normal problems we give the software a probability and it returns an x-value. Most software requires the value of pr(X d" x), the cumulative or lower-tail probability, to be given. Method for solving inverse Normal problems: Sketch a Normal curve, marking on the mean. Shade the area under the curve corresponding to the given probability. Obtain the desired x-value from the lower-tail probability given to the software.       Standard Units (z-scores) The z-score for an observation, x, is the number of standard deviations x is from the mean, . If x is an observation from a Normal distribution with mean and standard deviation  then the z-score is  EMBED Equation.3  If x is above the mean then the z-score is positive. If x is below the mean then the z-score is negative. Standard Normal Distribution If X ~ Normal(, ) and  EMBED Equation.3  then Z ~ Normal(0, 1). Z ~ Normal(0, 1) is called the standard Normal distribution. Sample Exam / Cecil Test Questions Questions 1 to 8 refer to the following information. For mothers carrying foetuses with spina bifida, the distribution of the concentration of alpha fetoprotein ((M/L) in a mothers urine is modelled by a Normal distribution with mean 23.1 and standard deviation 4.1. Use the following computer output to answer Questions 1 to 6. Normal with mean = 23.1 and standard deviation = 4.1 x P(X <= x) P(X <= x) x 17 0.0684 0.0500 16.356 18 0.1068 0.1000 17.846 19 0.1587 0.2000 19.649 20 0.2248 0.2248 20.000 21 0.3043 0.2400 20.204 22 0.3942 0.2500 20.335 23 0.4903 0.2600 20.462 31 0.9730 0.6785 25.000 32 0.9850 0.7500 25.865 33 0.9921 0.8000 26.551 0.9000 28.354 0.9500 29.844 For a randomly selected mother carrying a foetus with spina bifida, the probability that the concentration of alpha fetoprotein in her urine is less than 20.0 (M/L is approximately: 0.1587 0.6957 0.7752 0.2248 0.3043 For mothers carrying foetuses with spina bifida, the proportion who have an alpha fetoprotein concentration greater than 18.0 (M/L in their urine is approximately: 0.8413 0.1068 0.8932 0.1587 0.9316 For mothers carrying foetuses with spina bifida, the proportion who have an alpha fetoprotein concentration between 22.0 (M/L and 32.0 (M/L in their urine is approximately: 0.4827 0.4947 0.6878 0.5788 0.5908 For mothers carrying foetuses with spina bifida, the concentration of alpha fetoprotein in urine for which 25% lie below is approximately: 20.335 (M/L 25.865 (M/L 20.204 (M/L 0.679 (M/L 20.462 (M/L For mothers carrying foetuses with spina bifida, the concentration of alpha fetoprotein in urine for which 20% lie above is approximately: 19.649 (M/L 0.775 (M/L 28.354 (M/L 26.551 (M/L 0.225 (M/L For mothers carrying foetuses with spina bifida, the range of the central 90% of concentrations of alpha fetoprotein in urine is approximately between: 19.6 (M/L and 26.6 (M/L 16.4 (M/L and 29.8 (M/L 17.8 (M/L and 29.8 (M/L 16.4 (M/L and 28.4 (M/L 17.8 (M/L and 28.4 (M/L Which one of the following statements about mothers carrying foetuses with spina bifida is false? Approximately 68% have alpha fetoprotein concentrations between 19.0 (M/L and 27.2 (M/L. Approximately 95% have alpha fetoprotein concentrations between 14.9 (M/L and 31.3 (M/L. Approximately 16% have alpha fetoprotein concentrations greater than 27.2 (M/L. More than 5% have alpha fetoprotein concentrations less than 14.9 (M/L. Almost all have alpha fetoprotein concentrations between 10.8 (M/L and 35.4 (M/L. A mother who is carrying a foetus with spina bifida has an alpha fetoprotein concentration of 16.8 (M/L. 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Hypholoma Capnoides is a pleasant tasting mushroom which looks very much like the generally taller, poisonous fungus Sulphur Tuft. Let H be the height (in centimetres) of a Hypholoma Capnoides mushroom and let S be the height (in centimetres) of a Sulphur Tuft fungus. The distribution of H is modelled by a Normal distribution with mean 6.5 and standard deviation 1.76 and the distribution of S is modelled by a Normal distribution with mean 9.5 and standard deviation 1.25. In summary, H ~ Normal(H = 6.5, H = 1.76) and S ~ Normal(S = 9.5, S = 1.25). Which one of the following statements is false? The proportion of Hypholoma Capnoides mushrooms that are: taller than 11.0 cm is less than the proportion of Sulphur Tuft fungi that are taller than 12.5 cm. taller than 5.0 cm is less than the proportion of Sulphur Tuft fungi that are taller than 8.0 cm. taller than 9.5 cm is greater than the proportion of Sulphur Tuft fungi that are shorter than 6.5 cm. shorter than 8.0 cm is less than the proportion of Sulphur Tuft fungi that are taller than 8.5 cm. taller than 4.0 cm is greater than the proportion of Sulphur Tuft fungi that are shorter than 11.0 cm. Which one of the following statements is false? Ӱֱ 2.5% of Sulphur Tuft fungi are taller than 12.0 cm. The proportion of Hypholoma Capnoides mushrooms that are taller than 10.5 cm is greater than the proportion of Sulphur Tuft fungi that are taller than 12.5 cm. Ӱֱ 16% of Hypholoma Capnoides mushrooms are shorter than 4.74 cm. Hypholoma Capnoides mushrooms are generally shorter than and more variable in height than Sulphur Tuft fungi. The proportion of Hypholoma Capnoides mushrooms that are shorter than 2.5 cm is less than the proportion of Sulphur Tuft fungi that are shorter than 6.0 cm.      Page  = PAGE 20 / 210 Chapter 6 Chapter 6 Notes Chapter 6 pr(X < 17.8) = 0.9987 40 T pr(X < a) = 0.25 a = 38.5 pr(X < b) = 0.75 b = 41.5 0.1 pr(X > x) = 0.95 pr(X < x) = 0.05 x = 36 weeks x 1 above 2 9 2 -1  EMBED Equation.3  = 1.425 515/544 = 95%  EMBED Equation.3  = -0.275 pr(X < a) = 0.25 a = 89.88 pr(X < b) = 0.75 b = 110.12 IQR = 110.12 89.88 = 20 pr(X > x) = 0.1 pr(X < x) = 0.9 x = 119 pr(X < x) = 0.2 x = 87 100 0.2 x pr(Y > 17.8) = 1 pr(Y < 17.8) = 1 0.0981 = 0.9019 100 10 1.8 4 = 2.8 pr(10 < X < 30) = pr (X < 30) pr(X < 10) = 0.8254 0.4804 = 0.3450 Concentration ((M/L) 23.1 6% and 12% pr(X < x) = 0.1 x = 37 weeks 17.8 23.1  EMBED Equation.3  3% and 15% 10 + 2.3 4 = 19.2 sd below 689599.7 Rule In a Normal distribution, approximately: 68% of observations are within 1 standard deviation of the mean 95% of observations are within 2 standard deviations of the mean 99.7% of observations are within 3 standard deviations of the mean 30 pr(X < 0) = 0.2940 ? Software gives x = 34.21 10 The z-score for an observation x: z =  EMBED Equation.3  375/544 = 69% 15.7 11.0 50 11.0 0 11.0 pr(X > 50) = 1 pr(X < 50) = 1 0.9726 = 0.0274 0.68 included (e" or d") or excluded (> or <). 1 area Y sd above areas X between x = a and x = b. pr(a d" X d" b) b Z ~ Normal (0, 1) 3  3 0 a 1.05 1.20 Z ~ Normal (0, 1) 3  3 0 Software gives x = 27.38 above ? 0.05 0.9 0.75 0.25 a b 100 17.8 15.7 0.95 0.25 0.25 0.5 x 40 b a 40 0.1 x x 30 30 x Area = 0.3 Area = 0.8 Find x, where pr(X d" x) = 0.3 Find x, where pr(X d" x) = 0.8 X024DFHbdxz|ʵʵ{jTjTjGhFh=CJOJQJ*hOh=5CJOJQJ^JmHnHu!h=CJOJQJ^JmHnHu(hOh=CJOJQJ^JaJmH sH %h=6CJOJQJ^JaJmH sH "h=CJOJQJ^JaJmH sH (h=6CJH*OJQJ^JaJmH sH !h=6CJOJQJ^JmH sH h=CJOJQJ^JmH sH 'hFh=6CJOJQJ^JmH sH (Nvx    / 0 R ^ z { ~               ƶ}q}жq}q[*jh=CJOJQJU^JmHnHuh=CJOJQJ^J'hOh=6CJOJQJ^JmH sH hyCJOJQJ^JmH sH (hUyh=CJ OJQJ^JaJ mH sH h=CJOJQJ^JmH sH hFCJOJQJhFCJOJQJ^JmH sH 'hOhF6CJOJQJ^JmH sH hFCJOJQJ^J# !)!H!I!L!M!\!]!!!!!!!!!!!!!!!""1"D"h"i"v""""""""𻪔}l\hFCJOJQJ^JmH sH  h6)h=CJ OJQJ^JaJ h=6CJOJQJ^Jh=CJOJQJ*h6)h=5CJOJQJ^JmHnHu!h=CJOJQJ^JmHnHuh=CJOJQJ^J(hUyh=CJ OJQJ^JaJ mH sH 'hOh=6CJOJQJ^JmH sH h=CJOJQJ^JmH sH $!!!""#r#$$$$$$$$$$$ $=$>$$a$gd%`$a$gd=$ & Fdh<<a$gd q$ & Fdh<<gd= & F dh<<gd= dh<<gd="""###^#j#q#r#s####### $ $$$$$$$$$$$$$ $%$|x|x|x|xc(h%`hw3CJOJQJ^JaJmH sH hi]ljhi]lUh=CJOJQJaJmH sH (hUyh=CJ OJQJ^JaJ mH sH *jh=CJOJQJU^JmHnHu'hOh=6CJOJQJ^JmH sH  h6)h=CJ OJQJ^JaJ h=CJOJQJ^Jh=CJOJQJ^JmH sH  %$&$($)$/$0$2$3$7$8$:$;$<$=$>$G$H$I$X$Y$Z$c$d$e$f$q$իxjTPjTPLhi]lh+hbDKh[Q6CJOJQJ^JaJmH sH h[QOJQJ^JmH sH h}}hw3CJaJ+hbDKhw36CJOJQJ^JaJmH sH hw3OJQJ^JmH sH hw3(h%`hw3CJOJQJ^JaJmH sH )h*0JCJOJQJ^JaJmHnHu$h%`hw30JCJOJQJ^JaJ-jh%`hw30JCJOJQJU^JaJ>$H$I$Y$Z$d$e$f$g$h$i$j$k$l$m$n$o$p$q$$$$$$$$$$$a$gd~N !q$t$u$$$$$$$$$$$$$$$$$$$$$$$$$$$$$²ϠόvvbvbvvbvbQ!hM5CJ OJQJaJ mH sH 'hMhM5CJ OJQJaJ mH sH *hMhw356CJ OJQJaJ mH sH 'hMhw35CJ OJQJaJ mH sH #hghg6CJOJQJ^JaJ8h~Nh~N5CJOJQJaJh~N5CJOJQJaJhw3h(hw35CJ OJQJaJ hw356CJ OJQJaJ hw35CJ OJQJaJ $$$$$$$$$$$$$$$$$$$ % %!%#%$%&%$a$gd$gy gd2$a$gdQ= gd$a$gdg$$$$$$$$$$$$%%% %%%% %!%"%#%$%%%ʴʴʣʴʴʣm]Ph$gy5CJ OJQJaJ hQ=hQ=5CJOJQJaJ"hQ=hQ=56CJOJQJaJ$hw356CJ OJQJaJ mH sH !hw35CJ OJQJaJ mH sH !hm5CJ OJQJaJ mH sH *h2hw356CJ OJQJaJ mH sH 'h2hw35CJ OJQJaJ mH sH hw3hQ=hQ=5CJ OJQJaJ hQ=5CJ OJQJaJ %%&%'%,%-%.%/%0%1%3%4%5%6%7%9%:%;%<%O%P%Q%R%Z%[%\%i%j%k%l%%멥p`XTh$gyjh$gyUh59hw35CJ OJQJaJ hKyh| 5CJ OJQJaJ hKy5CJ OJQJaJ jh| hKyEHUjF hKyUVh| jh| Uhzhw35CJ OJQJaJ hw35CJ OJQJaJ hIHUhkt5CJOJQJaJh$gy5CJ OJQJaJ hw3h$gyh$gy5CJ OJQJaJ &%'%-%.%0%1%3%4%6%7%:%;%[%\%j%k%%%%%%%%% gdiU% J gdEgd$gygd&?&B&ְ}k[Nhw35CJ OJQJaJ hQ=hQ=5CJOJQJaJ"hQ=hQ=56CJOJQJaJh?hQ=5CJ OJQJaJ h?hQ=5CJOJQJaJ$hw356CJ OJQJaJ mH sH !hw35CJ OJQJaJ mH sH hw3!h_E$5CJ OJQJaJ mH sH 'hiU%hw35CJ OJQJaJ mH sH *hiU%hw356CJ OJQJaJ mH sH %&&!&1&2&6&7&;&<&>&?&_&m&w&x&|&}&~&&&&&&&& Ogd$a$gdQ= gdiU%B&C&U&V&v&w&x&|&~&&&&&&&&&&&&&&&&&&&&&&&&&г䖉yhRh* jmhh]hCJOJQJ^JmH sH  hghgCJOJQJ^JaJ0hdhw35CJ OJQJaJ haP5CJ OJQJaJ ho5CJ OJQJaJ hWzhWz5CJ OJQJaJ hWz5CJ OJQJaJ h?hQ=5CJOJQJaJhw3hhw35CJ OJQJaJ hw35CJ OJQJaJ hw356CJ OJQJaJ &&&&&'''''('-'.'3'4'L'M'N'O'Z'['\'p'q'z'{'|'$a$gd($a$gdu- gd=Z$a$gdg&&&&''''' ''''&'''(','-'.'2'3'4'5'H'I'޼޼޼uucVKjF hG/UVhG/5CJ OJQJaJ "jhG/5CJ OJQJUaJ hu-hu-5CJOJQJaJhu-5CJOJQJaJh=Zhw35CJ OJQJaJ hM5CJ OJQJaJ h~N5CJ OJQJaJ "h=Zhw356CJ OJQJaJ hQhw35CJ OJQJaJ hw35CJ OJQJaJ hw3 hghgCJOJQJ^JaJ0I'J'K'L'O'Y'Z'\'o'p'q'y'z''''''öÙÙvcM8(hbFphw3CJOJQJ^JaJmH sH +hbFphw35CJOJQJ^JaJmH sH %hw35CJOJQJ^JaJmH sH %h_5CJOJQJ^JaJmH sH hWzhWz5CJ OJQJaJ hWz5CJ OJQJaJ hASnhw35CJ OJQJaJ hw35CJ OJQJaJ hw3hG/hG/5CJ OJQJaJ "jhG/5CJ OJQJUaJ ,jR hG/hG/5CJ EHOJQJUaJ |'}'~''''|(}(((((((((((((((gdz$a$gd6$a$gdVg$a$gdo$dh<<a$gd0" dh<<gdbFp $dh<a$gdbFp''''''''(((8(9(>(Z([(\({(|(}(((((((((((ԾuhXhhw35CJ OJQJaJ ho5CJ OJQJaJ hw356CJ OJQJaJ hw35CJ OJQJaJ hoho5CJ OJQJaJ ho5CJ OJQJaJ hoho5CJOJQJaJhw3"h_CJOJQJ^JaJmH sH (hbFphw3CJOJQJ^JaJmH sH +h0"hw35CJOJQJ^JaJmH sH ((((((((((((((((((((ݺ뫙}k}]}K=hFCJOJQJ^JaJ#hE[@hw36CJOJQJ^JaJh.c[CJOJQJ^JaJ#hzhw36CJOJQJ^JaJhw3CJOJQJ^JaJh$gyCJOJQJ^JaJ#h6h65CJOJQJ^JaJh65CJOJQJ^JaJ h/6KhVgCJOJQJ^JaJ#h/6KhVg6CJOJQJ^JaJhVgCJOJQJ^JaJhw3hzhw35CJ OJQJaJ (((((((((((())) ))))){kZM=h;;h65CJOJQJaJh65CJOJQJaJ hghgCJOJQJ^JaJ0h59hw35CJ OJQJaJ hw35CJ OJQJaJ hw3 hzhw3CJOJQJ^JaJj h!Sh!SEHUjiG h!SUVh!Sjh!SUhw3CJOJQJ^JaJ#hzhw36CJOJQJ^JaJh$gyCJOJQJ^JaJ hFhw3CJOJQJ^JaJ((()) )))))))) )!)&)')C)Q)[)\)a)b)$a$gd 3gd+77$a$gd;;$a$gd=$a$gd6$a$gdggd59xgd!S)))))))) )!)%)&)')*)+)/);)<)@)C)P)Q)W)Z)[)\)`)a)*N*ޤ{k^NޤhIHUh5CJ OJQJaJ h5CJ OJQJaJ h+77hw35CJ OJQJaJ haP5CJ OJQJaJ ho5CJ OJQJaJ hw356CJ OJQJaJ hw35CJ OJQJaJ h;;h;;5CJOJQJaJh;;5CJOJQJaJh;;h=5CJOJQJaJhw3h=h=5CJOJQJaJh=5CJOJQJaJb)P*R*V*X*Z*\*^*`*j*l*p*r*t**************gdEg$a$gd($a$gdggddN*P*R*V*`*h*j*l*p*t********D+T+V+\+^+h+j+p+r+t+v+x+~+++ξ띍ξxbxbxbxbxQxbx h% hOJQJ^JmH sH +h! h6CJOJQJ^JaJmH sH (h! hCJOJQJ^JaJmH sH hWzhWz5CJ OJQJaJ hWz5CJ OJQJaJ &hghg56CJOJQJ^JaJ8hShd5CJ OJQJaJ hd5CJ OJQJaJ hShw35CJ OJQJaJ hw3hTahw35CJ OJQJaJ *****************************gdEg***************+++++ + +++++++++++ +"+$+&+(+*+,+.+0+2+4+6+8+:+<+>+@+B+D+v+x+++++$a$gd$ / dhx<a$gdi]l+++++++++++++++++++++++++++++++̻̬|̞k̞k̞k̻^N^NhMhM5CJ OJQJaJ hM5CJ OJQJaJ hhHMCJOJQJ^JaJ hMhHMCJOJQJ^JaJ hMhHMCJOJQJ^JaJhHMCJOJQJ^JaJhHM6CJOJQJ^JaJ hhCJOJQJ^JaJhw3h(h! hCJOJQJ^JaJmH sH +h! h6CJOJQJ^JaJmH sH +++++++++++++++,,, ,&,(,,,.,`,b,$a$gdVg$a$gd$a$gdM$a$gd$a$gdHM+++, ,,,,,, ,$,&,(,*,,,.,L,N,^,`,b,l,n,p,r,t,v,߽~m`P`Ph$gyhi5CJ OJQJaJ hi5CJ OJQJaJ h/6KhVgCJOJQJ^JaJ#h/6KhVg6CJOJQJ^JaJhVgCJOJQJ^JaJ hhCJOJQJ^JaJhCJOJQJ^JaJ hMhMCJOJQJ^JaJ hMhMCJOJQJ^JaJhMCJOJQJ^JaJhM6CJOJQJ^JaJhw3b,n,p,t,v,,,,,,,,,,,,,,,,,,,,$a$gd5%$a$gd$a$gd)$a$gd?$a$gdr$a$gdiv,~,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,-޲޲ޒޅuޅueeeeh?h5%5CJ OJQJaJ hu-h5CJOJQJaJh5CJOJQJaJh?h)5CJOJQJaJhQ=h)5CJOJQJaJh)56CJOJQJaJhQ=h?5CJ OJQJaJ h?5CJ OJQJaJ hw3hrhr5CJ OJQJaJ hr5CJ OJQJaJ ",,,,,,,--- -------"-$-&-.-0-4-6-:-<-$a$gdVg$a$gd~N$a$gdZc$a$gd5%---- ---------"-$-&-.-0-2-4-6-8-:-<-ͽڮڮڑځo_P>#h/6KhVg6CJOJQJ^JaJhVg6CJOJQJ^JaJhQ=h~N5CJOJQJaJ"hQ=h~N56CJOJQJaJh?h~N5CJ OJQJaJ h~N5CJOJQJaJh~Nh~N5CJOJQJaJh~N56CJOJQJaJh~NhZc5CJOJQJaJhZc5CJOJQJaJhw3hQ=hZc5CJOJQJaJ"hQ=hZc56CJOJQJaJ<-@-B-D-H-J-L-N-P-R-f-h-j-~---------------------.|||$|&|0|2|4|:|<|Z|ݽ݇t h$3hVgCJOJQJ^JaJU#h$3hVg6CJOJQJ^JaJ#h)hVg6CJOJQJ^JaJ#hDIhVg6CJOJQJ^JaJhVg6CJOJQJ^JaJ hDIhVgCJOJQJ^JaJhw3 h)hVgCJOJQJ^JaJhVgCJOJQJ^JaJ-<-B-D-J-L-P-R-h-j-------2|4|\|^|||||||||||gdVg$a$gdVg ~ Normal( = 30,  = 5) pr(X d" 35) = 0.0668 pr(X d" 70) = 0.9772 50 35 50 70 Software gives 0.0668 Software gives 0.9772 Find pr(X d" 35) Find pr(X d" 70) X ~ Normal( = 50,  = 10) 68 95 99.7 Rule In a Normal distribution with mean and standard deviation , approximately: 68% of observations are within 1 standard deviation of the mean, i.e., between  1 and + 1 (or 1) 95% of observations are within 2 standard deviations of the mean, i.e., between 2 99.7% of observations are witZ|\|^|d|f|||||||||||||||||||}}}}}"}$}&}6}8}D}F}H}J}^}`}n}p}|}~}}}}ݽݽݽݽݽݽݫݽݫݽݫݫݽ똂+hhh5CJOJQJ^JaJmH sH %hh5CJOJQJ^JaJmH sH #h$3hVg6CJOJQJ^JaJ h$3hVgCJOJQJ^JaJhVg6CJOJQJ^JaJhVgCJOJQJ^JaJhw3 hyhVgCJOJQJ^JaJ-|||}}$}&}F}H}~}}}<~xzgdh$a$gdh7dh<<^7gdh dh<<gdh $dh<a$gdh$a$gdVg $da$gdVg}}}}~~<~B~z~|~~~~~~~~~~~~  VX  jlntvxzٛٛٛ٭ٛ٭ٛ땇hhCJOJQJ^JaJhw3U"hhCJOJQJ^JaJmH sH +h 6hh5CJOJQJ^JaJmH sH +h:o'hh6CJOJQJ^JaJmH sH "hhCJOJQJ^JaJmH sH (hhhCJOJQJ^JaJmH sH 2hin 3 standard deviations of the mean, i.e., between 3 Mean 50% 50% Mean, 9 + 3 9 - 3 9 0.95 9 + 2 3 9  2 3 below test  EMBED Equation.3  41.5 38.5 ˆĈƈȈʈ̈Ј҈ڈعؚzؚzj]hIHU5CJ OJQJaJ hIHUh5CJOJQJaJhh5CJOJQJaJh(Th5CJOJQJaJh5CJOJQJaJ#h 6hh6CJOJQJ^JaJ h 6hhCJOJQJ^JaJhhCJOJQJ^JaJhw3 h"<hhCJOJQJ^JaJ#h"<hh6CJOJQJ^JaJʈ̈Ј҈܈ވ "$TVgdG/$a$gdG/$a$gd$gy$a$gdIHU$a$gd$a$gdhgdhڈ܈ވ  "$&L˾ˬ˚}p`Np"jhG/5CJ OJQJUaJ hG/hG/5CJ OJQJaJ hG/5CJ OJQJaJ h$gyh$gy5CJ OJQJaJ h$gy5CJ OJQJaJ #h3h35CJOJQJ^JaJ#h3hIHU5CJOJQJ^JaJhIHU5CJOJQJaJh3hIHU5CJOJQJaJhIHUhIHU5CJOJQJaJhw3hIHUhIHU5CJ OJQJaJ LNPRTV^`bdjl˻h=CJOJQJaJmH sH hOhMhM5CJ OJQJaJ hM5CJ OJQJaJ hw3hG/hG/5CJ OJQJaJ "jhG/5CJ OJQJUaJ ,jhG/hG/5CJ EHOJQJUaJ j䨹F hG/UVV`blnprtvxz|~$a$gd=$a$gdM2&P :pD . 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