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Chapter 6 The Normal Probability Distribution CONTINUOUS RANDOM VARIABLES Continuous random variables can assume the infinitely many values corresponding to points on a line interval. This means that the range of the variable contains an interval of real numbers. Examples: Heights, weights length of life of a particular product experimental laboratory error Recall that the probability distribution p(x) of a discrete random variable is just the histogram of its values. We had the following examples in chapter 5: 1. Toss a fair coin three times and let x be the number of heads. Then x is binomial and x 0 1 2 3 p(x) 1/8 3/8 3/8 1/8 We can create what we call the probability polygon h(x) by starting from the midpoint of the left side of the first bar and connecting the midpoints of the upper sides of the subsequent bars by straight lines and finishing at the midpoint of the right side of the last bar. 2. A marksman hits a target 80% of the time. He fires five shots at the target. What is the probability that exactly 3 shots hit the target? If x is the number of shots that hit the target, then x is a binomial and k p= .80 0 .000 1 .007 2 .058 3 .263 4 .672 5 1.000 Again we can draw the probability polygon h(x) in the same fashion. Note that the total area under the probability histogram is 1 and that the area under the probability polygon is very close to 1 Now let x be a continuous random variable and consider a set of n measurements of x. we can construct the relative frequency histogram of these values and the corresponding probability polygon h(x) . As mentioned before, the area under h(x) is very close to 1. We can use this polygon approximate as the area under h(x) over the interval . rel freq 10 8 6 rel freq 4 2 0 0.835 0.865 0.895 0.925 0.955 0.985 1.015 1.045 1.075 More As (the number of measurements increases without bound) the relative frequency polygon h(x) approaches the graph of a smooth function whose area under the graph is one. This function is called the probability density function, pdf, of the random variable x. rel freq 10 8 h(x) 6 rel freq 4 2 0 0.835 0.865 0.895 0.925 0.955 0.985 1.015 1.045 1.075 More The pdf of x can be used, among other things, to find the exact value of which is the area under the graph of over the interval which is, from calculus, a b The probability density function for the random variable x describes the probability distribution of the continuous random variable x. This function is sometime referred to as the depth of probability. •In many situations the density function is given by a mathematical formula where as in many many situations finding a nice formula may not be possible. PROPERTIES OF CONTINUOUSPROBABILITY DISTRIBUTIONS 1. 2. The total area under the curve = 3. =area under the curve between a and b Exercise: Show that there is no probability attached to any single value of x. That is, P(x = a) = 0. There are many different types of continuous random variables. Here are some that occur often: 1. Uniform Random variable 2. Exponential Random Variable Some other random variable include The Gamma, Chi Square, and Bata random variables. The most important random variable is the Normal Random variable. To choose a model for real life situation, we try to pick a model that Fits the data well Allows us to make the best possible inferences using the data. As mentioned, one important continuous random variable is the normal random variable and we will discuss that in more detail. THE NORMAL DISTRIBUTION • The formula that generates the normal probability distribution is: 1 x 2 2 1 f ( x) e for x 2 e 2.7183 3.1416 and are the population mean and standard deviation. • The shape and location of the normal curve changes as the mean and standard deviation change. THE STANDARD NORMAL DISTRIBUTION To find P(a < x < b), we need to find the area under the appropriate normal curve. To simplify the tabulation of these areas, we standardize each value of x by expressing it as a z-score, the number of standard deviations it lies from the mean . z x THE STANDARD NORMAL (Z) DISTRIBUTION Mean = 0; Standard deviation = 1 When x = , z = 0 Symmetric about z = 0 Values of z to the left of center are negative Values of z to the right of center are positive Total area under the curve is 1. USING TABLE 3 The four digit probability in a particular row and column of Table 3 gives the area under the z curve to the left that particular value of z. Area for z = 1.36 EXAMPLE Use Table 3 to calculate these probabilities: P(z 1.36) = .9131 P(z >1.36) = 1 - .9131 = .0869 P(-1.20 z 1.36) = .9131 - .1151 = .7980 mpirical Rule: % of the within 2 ns of the mean. USING TABLE 3 To find an area to the left of a z-value, find the area directly from the table. To find an area to the right of a z-value, find the area in Table 3 and subtract from Remember the 1. Empirical Rule: 99.7% of the of z, find the two To find the areaApproximately between two values within 3 areas in Table 3, measurements and subtractlieone from the other. standard deviations of the mean. P(-3 z z3) 1.96) P(-1.96 = .9987 .9750 - .0250 = .9500 .0013=.9974 WORKING BACKWARDS Find the value of z that has area .25 to its left. 1. Look for the four digit area closest to .2500 in Table 3. 2. What row and column does this value correspond to? 3. z = -.67 4. What percentile does this value represent? 25th percentile, or 1st quartile (Q1) Find the value of z that has area .05 to its right. 1. The area to its left will be 1 - .05 = .95 2. Look for the four digit area closest to .9500 in Table 3. 3. Since the value .9500 is halfway between .9495 and .9505, we choose z halfway between 1.64 and 1.65. 4. z = 1.645 FINDING PROBABILITIES FOR THE GENERAL NORMAL RANDOM VARIABLE To find an area for a normal random variable x with mean and standard deviation , standardize or rescale the interval in terms of z. Find the appropriate area using Table 3. Example: x has a normal distribution with = 5 and = 2. Find P(x > 7). 75 P ( x 7) P ( z ) 2 P( z 1) 1 .8413 .1587 1 z EXAMPLE The weights of packages of ground beef are normally distributed with mean 1 pound and standard deviation .10. What is the probability that a randomly selected package weighs between 0.80 and 0.85 pounds? P(.80 x .85) P(2 z 1.5) .0668 .0228 .0440 What is the weight of a package such that only 1% of all packages exceed this weight? P( x ?) .01 ? 1 P( z ) .01 .1 ? 1 From Table 3, 2.33 .1 ? 2.33(.1) 1 1.233