Télécharger la présentation
## Chapter 7

- - - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - - -

**Chapter 7**Continuous Distributions**Continuous random variables**• Are numerical variables whose values fall within a range or interval • Are measurements • Can be described by density curves**Density curves**• Is always on or above the horizontal axis • Has an area exactly equal to one underneath it • Often describes an overall distribution • Describe what proportions of the observations fall within each range of values**Can be any shape**Are generic continuous distributions Probabilities are calculated by finding the area under the curve Unusual density curves**How do you find the area of a triangle?**P(X < 2) =**What is the area of a line segment?**P(X = 2) = 0 P(X < 2) = .25**In continuous distributions, P(X < 2) & P(X< 2) are**the same answer. Hmmmm… Is this different than discrete distributions?**Shape is a trapezoid –**How long are the bases? b1 = .5 b2 = .375 h = 1 P(X > 3) = P(1 < X < 3) = .5(.375+.5)(1)=.4375 .5(.125+.375)(2) =.5**P(X > 1) =**.75 .5(2)(.25) = .25 (2)(.25) = .5**P(0.5 < X < 1.5) =**.28125 .5(.25+.375)(.5) = .15625 (.5)(.25) = .125**Uniform Distribution**• Is a continuous distribution that is evenly (or uniformly) distributed • Has a density curve in the shape of a rectangle • Probabilities are calculated by finding the area under the curve How do you find the area of a rectangle? Where: a & b are the endpoints of the uniform distribution**1/.12**4.92 4.98 5.04 • The Citrus Sugar Company packs sugar in bags labeled 5 pounds. However, the packaging isn’t perfect and the actual weights are uniformly distributed with a mean of 4.98 pounds and a range of .12 pounds. • Construct the uniform distribution above. What shape does a uniform distribution have? What is the height of this rectangle? How long is thisrectangle?**1/.12**4.92 4.98 5.04 • What is the probability that a randomly selected bag will weigh more than 4.97 pounds? P(X > 4.97) = What is the length of the shaded region? .07(1/.12) = .5833**1/.12**4.92 4.98 5.04 • Find the probability that a randomly selected bag weighs between 4.93 and 5.03 pounds. What is the length of the shaded region? P(4.93<X<5.03) = .1(1/.12) = .8333**1/35**5 40 • The time it takes for students to drive to school is evenly distributed with a minimum of 5 minutes and a range of 35 minutes. • Draw the distribution What is the height of the rectangle? Where should the rectangle end?**1/35**5 40 b) What is the probability that it takes less than 20 minutes to drive to school? P(X < 20) = (15)(1/35) = .4286**c) What is the mean and standard deviation of this**distribution? m = (5 + 40)/2 = 22.5 s2 = (40 - 5)2/12 = 102.083 s = 10.104**Normal Distributions**• Symmetrical bell-shaped (unimodal) density curve • Above the horizontal axis • N(m, s) • The transition points occur at m+s • Probability is calculated by finding the area under the curve • As sincreases, the curve flattens & spreads out • As sdecreases, the curve gets taller and thinner How is this done mathematically?**Normal distributions occur frequently.**• Length of newborn child • Height • Weight • ACT or SAT scores • Intelligence • Number of typing errors • Chemical processes**s**s A B 6 Do these two normal curves have the same mean? If so, what is it? Which normal curve has a standard deviation of 3? Which normal curve has a standard deviation of 1? YES B A**Empirical Rule**• Approximately 68% of the observations fall within s of m • Approximately 95% of the observations fall within 2s of m • Approximately 99.7% of the observations fall within 3s of m**68%**71 Suppose that the height of male students at SHS is normally distributed with a mean of 71 inches and standard deviation of 2.5 inches. What is the probability that the height of a randomly selected male student is more than 73.5 inches? 1 - .68 = .32 P(X > 73.5) = 0.16**Standard Normal Density Curves**Always has m = 0 & s = 1 To standardize: Must have this memorized!**Standard Normal Distribution**• A normal distribution with mean 0 and standard deviation 1, is called the standard (or standardized) normal distribution.**Column labeled 0.06**Row labeled 0.4 P(z < 0.46) = 0.6772 Using the Normal Tables • Find P(z < 0.46)**Column labeled 0.04**P(z < -2.74) = 0.0031 Row labeled -2.7 Using the Normal Tables • Find P(z < -2.74)**P(z < 1.83)**• = 0.9664 (b) P(z > 1.83) = 1 – P(z < 1.83) = 1 – 0.9664 = 0.0336 Sample Calculations Using the Standard Normal Distribution Using the standard normal tables, find the proportion of observations (z values) from a standard normal distribution that satisfy each of the following:**(d) P(z > -1.83)**= 1 – P(z < -1.83) = 1 – 0.0336= 0.9664 c) P(z < -1.83) = 0.0336 Using the standard normal tables, find the proportion of observations (z values) from a standard normal distribution that satisfies each of the following:**Symmetry Property**• Notice from the preceding examples it becomes obvious that • P(z > z*) = P(z < -z*) P(z > -2.18) = P(z < 2.18) = 0.9854**P(Z<2.34)=0.9904**P(Z<-1.37)=0.0853 P(-1.37 < z < 2.34) = 0.9904 - 0.0853 = 0.9051 Using the standard normal tables, find the proportion of observations (z values) from a standard normal distribution that satisfies -1.37 < z < 2.34, that is find P(-1.37 < z < 2.34).**P(Z<1.61)=0.9463**P(Z<.54)=0.7054 P(0.54 < z < 1.61) = 0.9463 - 0.7054 = 0.2409 Using the standard normal tables, find the proportion of observations (z values) from a standard normal distribution that satisfies 0.54 < z < 1.61, that is find P(0.54 < z < 1.61).**Using the standard normal tables, find the proportion of**observations (z values) from a standard normal distribution that satisfy -1.42 < z < -0.93, that is find P(-1.42 < z < -0.93). P(Z<-0.93)=0.1762 P(Z<-1.42)=0.0778 P(-1.42 < z < -0.93) = 0.1762 - 0.0778 = 0.0984**Using the standard normal tables, in each of the following,**find the z values that satisfy : (a) The point z with 98% of the observations falling below it. The closest entry in the table to 0.9800 is 0.9798 corresponding to a z value of 2.05**Using the standard normal tables, in each of the following,**find the z values that satisfy : (b) The point z with 90% of the observations falling above it. The closest entry in the table to 0.1000 is 0.1003 corresponding to a z value of -1.28**If a variable X has a normal distribution with mean and**standard deviation , then the standardized variable has the normal distribution with mean 0 and standard deviation 1. Standard Normal Distribution Revisited This is called the standard normal distribution.**What is your z-score on a test that you scored an 86% on**that had a mean score of 80% with a standard deviation of 5?**Strategies for finding probabilities or proportions in**normal distributions • State the probability statement • Draw a picture • Calculate the z-score • Look up the probability (proportion) in the table**The lifetime of a certain type of battery is normally**distributed with a mean of 200 hours and a standard deviation of 15 hours. What proportion of these batteries can be expected to last less than 220 hours? Write the probability statement Draw & shade the curve P(X < 220) = .9082 Look up z-score in table Calculate z-score**The lifetime of a certain type of battery is normally**distributed with a mean of 200 hours and a standard deviation of 15 hours. What proportion of these batteries can be expected to last more than 220 hours? P(X>220) = 1 - .9082 = .0918**The lifetime of a certain type of battery is normally**distributed with a mean of 200 hours and a standard deviation of 15 hours. How long must a battery last to be in the top 5%? Look up in table 0.95 to find z- score P(X > ?) = .05 .95 .05 1.645**The heights of the female students at SHS are normally**distributed with a mean of 65 inches. What is the standard deviation of this distribution if 18.5% of the female students are shorter than 63 inches? What is the z-score for the 63? P(X < 63) = .185 -0.9 63**The heights of female teachers at SHS are normally**distributed with mean of 65.5 inches and standard deviation of 2.25 inches. The heights of male teachers are normally distributed with mean of 70 inches and standard deviation of 2.5 inches. • Describe the distribution of differences of heights (male – female) teachers. Normal distribution with m = 4.5 & s = 3.3634**4.5**• What is the probability that a randomly selected male teacher is shorter than a randomly selected female teacher? P(X<0) = .0901**Will my calculator do any of this normal stuff?**• Normalpdf – use for graphing ONLY • Normalcdf – will find probability of area from lower bound to upper bound • Invnorm (inverse normal) – will find z-score for probability**Ways to Assess Normality**Use graphs (dotplots, boxplots, or histograms) Normal probability (quantile) plot