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## Chapter The Normal Probability Distribution

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**3**7 Chapter The Normal Probability Distribution © 2010 Pearson Prentice Hall. All rights reserved**Section 7.1 Properties of the Normal Distribution**© 2010 Pearson Prentice Hall. All rights reserved**EXAMPLE Illustrating the Uniform Distribution**Suppose that United Parcel Service is supposed to deliver a package to your front door and the arrival time is somewhere between 10 am and 11 am. Let the random variable X represent the time from10 am when the delivery is supposed to take place. The delivery could be at 10 am (x = 0) or at 11 am (x = 60) with all 1-minute interval of times between x = 0 and x = 60 equally likely. That is to say your package is just as likely to arrive between 10:15 and 10:16 as it is to arrive between 10:40 and 10:41. The random variable X can be any value in the interval from 0 to 60, that is, 0 <X< 60. Because any two intervals of equal length between 0 and 60, inclusive, are equally likely, the random variable X is said to follow a uniform probability distribution.**The graph below illustrates the properties for the**“time” example. Notice the area of the rectangle is one and the graph is greater than or equal to zero for all x between 0 and 60, inclusive. Because the area of a rectangle is height times width, and the width of the rectangle is 60, the height must be 1/60.**Values of the random variable X less than 0 or greater than**60 are impossible, thus the equation must be zero for X less than 0 or greater than 60.**The area under the graph of the density function over an**interval represents the probability of observing a value of the random variable in that interval.**EXAMPLE Area as a Probability**The probability of choosing a time that is between 15 and 30 seconds after the minute is the area under the uniform density function. Area = P(15 <x< 30) = 15/60 = 0.25 15 30**A probability density function**(a) Shows the number of observations for a variable (b) Lists the probabilities for a discrete random variable (c) Shows how dense the mean and standard deviation are compared to the median (d) Is used to compute probabilities for continuous random variables (e) Not sure**True or False: The area under a probability density**function must equal 1.**Relative frequency histograms that are symmetric and**bell-shaped are said to have the shape of a normal curve.**If a continuous random variable is normally distributed, or**has a normal probability distribution, then a relative frequency histogram of the random variable has the shape of a normal curve (bell-shaped and symmetric).**The curve below is not a normal curve because**(a) It is skewed left (b) It is not continuous (c) It is skewed right (d) It has outliers (e) Not sure**What is the mean of the normal distribution shown?**(a) 120 (b) 20 (c) 140 (d) 100 (e) Not sure**Each graph represents a normal curve with mean μ = 100.**Which graph indicates the normal random variable X has more dispersion? (a) Blue graph (b) Red graph (c) Not sure**The normal density curve**(a) Is not symmetric (b) Has an area under the curve equal to one. (c) Always has a mean of 0 (d) Has positive and negative values (e) Not sure**EXAMPLE A Normal Random Variable**The data on the next slide represent the heights (in inches) of a random sample of 50 two-year old males. (a) Draw a histogram of the data using a lower class limit of the first class equal to 31.5 and a class width of 1. (b) Do you think that the variable “height of 2-year old males” is normally distributed?**36.0 36.2 34.8 36.0 34.6 38.4 35.4 36.8**34.7 33.4 37.4 38.2 31.5 37.7 36.9 34.0 34.4 35.7 37.9 39.3 34.0 36.9 35.1 37.0 33.2 36.1 35.2 35.6 33.0 36.8 33.5 35.0 35.1 35.2 34.4 36.7 36.0 36.0 35.7 35.7 38.3 33.6 39.8 37.0 37.2 34.8 35.7 38.9 37.2 39.3**In the next slide, we have a normal density curve drawn over**the histogram. How does the area of the rectangle corresponding to a height between 34.5 and 35.5 inches relate to the area under the curve between these two heights?**EXAMPLE Interpreting the Area Under a Normal Curve**• The weights of giraffes are approximately normally distributed with mean μ = 2200 pounds and standard deviation σ = 200 pounds. • Draw a normal curve with the parameters labeled. • Shade the area under the normal curve to the left of x = 2100 pounds. • Suppose that the area under the normal curve to the left of x = 2100 pounds is 0.3085. Provide two interpretations of this result. (a), (b) • (c) • The proportion of giraffes whose weight is less than 2100 pounds is 0.3085 • The probability that a randomly selected giraffe weighs less than 2100 pounds is 0.3085.**EXAMPLE Relation Between a Normal Random Variable and a**Standard Normal Random Variable The weights of giraffes are approximately normally distributed with mean μ = 2200 pounds and standard deviation σ = 200 pounds. Draw a graph that demonstrates the area under the normal curve between 2000 and 2300 pounds is equal to the area under the standard normal curve between the Z-scores of 2000 and 2300 pounds.**The area to the right of 0 under the standard normal curve**is equal to (a) 0.0 (b) 0.25 (c) 0.5 (d) 1.0 (e) Not sure**If the area to the right of 0.41 under the standard normal**curve is equal to 0.34, then the area to the left of 0.41 is equal to (a) 0.66 (b) 0.50 (c) 0.34 (d) 0.17 (e) Not sure**The table gives the area under the standard normal curve for**values to the left of a specified Z-score, zo, as shown in the figure.**EXAMPLE Finding the Area Under the Standard Normal Curve**Find the area under the standard normal curve to the left of z = -0.38. Area left of z = -0.38 is 0.3520.**Find the area under the standard normal curve to the left of**z = 1.54.**Area under the normal curve to the right of zo =1 – Area**to the left of zo**EXAMPLE Finding the Area Under the Standard Normal Curve**Find the area under the standard normal curve to the right of Z = 1.25. Area right of 1.25 = 1 – area left of 1.25 = 1 – 0.8944 = 0.1056**Find the area under the standard normal curve to the right**of z = -2.38.**EXAMPLE Finding the Area Under the Standard Normal Curve**Find the area under the standard normal curve between z = -1.02 and z = 2.94. • Area between -1.02 and 2.94 = (Area left of z = 2.94) – (area left of z = -1.02) • = 0.9984 – 0.1539 • = 0.8445