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Normal Distributions Unit Plan A random variable is called discrete if we can represent the probabilities of its instances with a table. The distributions we studied in the last section, Geometric and Poisson, were still discrete, even though they were open ended, because we could use tables to find their probabilities. A random variable is called continuous if we cannot use tables to represent its probabilities. In Chapter 5 and beyond, the text discusses this kind of random variable. The most famous and widespread kind of continuous r.v.βs are the Normal Distributions. A random variable x is called normal if and only if: 1.) Its instances can be any real number. [This means we canβt represent a normal distribution with a table, even an open-ended one! Therefore, normal random variables are continuous] 2.) Instead, the probabilities in the normal distribution are represented by a shape with an area of 1. [Remember the frequency polygons in Chapter 2? This is similar.] The normal distribution a very special property worthy of noting: mean=median=mode (this means the distribution is symmetric) This shape is generated with a special equation. When graphed, it looks like a bell. Hence another famous name for the normal distribution: The βBell Curve.β Finding Probabilities on a Normal Distribution: A two-step process Step 1: Converting to the Standard Normal Just like the other random variables we have studied, the normal distribution is not a single distribution, but rather a family of similar distributions (see p. 217). Just like Poisson Distributions, Normals are parameterized by π, so we are given that ahead of time. Normals are parameterized by π as well, so we are also given that ahead of time. Probabilities on the normal distribution can only be found directly through integral calculus, but fortunately there are many resources that give us a table of approximate values [I will be printing out a table for you to use. There is one in the book, but mine is slightly more user-friendly.] But since Normals are a family of distributions, how can a single table give us these values? Fortunately, we know that the z-scores of any normal distribution are also normally distributed with a mean of 0 and a standard deviation of 1. This specific normal distribution is called the Standard Normal Distribution and this is the distribution for which the values are given in the table. Recall the formula for the z-score: π§π₯ = π₯βπ π Step 2: Using the table The table in the book assumes we are looking up a βless thanβ probability, so if we want a greater than or a between, we will have to use a conversion. Weβve seen the βgreater thanβ conversion before, but the between conversion is new. Here it is: π(π < π₯ < π) = π(π₯ < π) β π(π₯ < π) Note: Since the normal distribution can take on any real number, the difference between π(π₯ < π) and π(π₯ β€ π) is trivial. Therefore, the distinction between βless than or equal toβ and βstrictly less thanβ that was so important in Chapter 4 now matters not at all (whew!) To use the table, first note that the negative z-scores are on one side and the positive z-scores are on the other. Now, to find the probability, the whole number and the tenth digit are located on the left side of the table. The hundredths digit is located on the top of the table. To look up the βless thanβ probability associated with a particular z-score, go to the row with the proper whole number and tenth, and then over to the column with the proper hundredth. Note: Traditionally, z-scores are only calculated to two decimal places. If you need more accuracy, there are many computer programs that do the direct integral of the normal curve and can give you an arbitrary level of precision. p.229 Example 1: βLess thansβ π(π₯ < 2) = (2 β 2.4) π (π§ < )= 0.5 π(π§ < β0.80) = Now, use the table. Make sure you are on the βnegativeβ side. Go down to the row labeled -0.8. Since you are looking up -0.80, you will want the probability in the first column: 0.2119. What this means is the probability of a random variable having a z-score of less than -0.80 in any normal distribution is 0.2119. Therefore: π(π₯ < 2) = 0.2119 Now do the βTry it yourselfβ on the bottom of p.229, and p.232 #1,2 p. 230 Example 2-2a: βGreater thansβ π(π₯ > 39) = 1 β π(π₯ < 39) = (39 β 45) 1 β π (π§ < )= 12 1 β π(π§ < β0.50) = 1 β 0.3085 = 0.6915 Now do p.232 #3,4 p.230 Example 2-1: βBetweensβ π(24 < π₯ < 54) = π(π₯ < 54) β π(π₯ < 24) = (54 β 45) (24 β 45) π (π§ < ) β π (π§ < )= 12 12 π(π§ < 0.75) β π(π§ < β1.75) = 0.7734 β 0.0401 = 0.7333 Now do p.230 βTry it yourself #2β and p.232 #5,6 HW: p.232-5 #7-30 (or #1-30 if accel.) Finding Bounds on a Normal Distribution In some problems, we are given the probability, and we must find the bound (or bounds, in the case of a βbetweenβ) that generated the probability. To do these problems, we basically reverse the procedure in the previous section, with some complications. p.240 Example 4: Greater than Since theyβre only taking the top five percent, we must find the bound such that only 5% of the curve is greater than that bound. In other words: π(π₯ > π) = 0.05 Since this is a greater than, and the table assumes weβre looking up less thans, we need to do the conversion: π(π₯ < π) = 0.95 Which makes sense: If 5% of the curve is greater than our bound a, then 95% of the curve will be less than the bound. Go to the table and look up the P-value that is closest to 0.95, and see what z-value it corresponds to. You will find that the z-score of 1.64 yields P=0.9495 and the z-score of 1.65 yields P=0.9505. So, they are both 0.0005 away from the value we want. In this case, we say these βtieβ, and we will use their mean, z=1.645. So, now we know that since for any point on the x-axis, including our bound, π§= π₯βπ π we can re-write that equation as πβπ π§= π and we now already have a z, a π and a π, we can find the bound. You might find it useful to re-write this equation in terms of the bound, that way you donβt have to do the algebra each time: π = π + π§π or, more generally: πππ πππππ = π + ππππππ π Remind you of Chebyshevβs Theorem?? It should! So, now to solve example 4: π = 75 + 1.645(6.5) π = 85.69 So to pass the Civil Service Test, you need to score at least an 85.69. Now try p.240 βTry it yourselfβ #4, p.241 Ex. 5 and βTry it yourselfβ #5. These are βless thansβ so they should be a bit easier. For HW: p.244 #43-47 Example: p.244 #48: βBetweensβ: Letβs look at the middle question in this problem, βthe middle 40% receive a C.β What are the two grades between which a student would receive a C? In other words, π(π < π₯ < π) = 0.40 We know that since 40% of the curve is in between a and b, 60% is outside of these bounds. If weβre able to assume that a and b are symmetric about the mean (S.A.M), which we can here, since it says the βmiddleβ 40% get a C, then by the symmetry of the normal curve, we know that 30% of the curve is above b, and 30% is below a. Letβs examine the second part of that sentence: β30% of the curve is below a.β It turns out, this will give us all we need to solve the entire problem! We can write this as: π(π₯ < π) = 0.30 Now, by using the same procedure as in the previous example, we can find the corresponding z-score to be 0.52. Applying the equation πππ¦ πππ’ππ = π + π§πππ’ππ π, we can now find: π = 72 + β0.52(9) π = 67.32 Ok, but how does that help us find b? Well, by symmetry, we know that the z-score of b is simply the opposite of the z-score of a. In other words, π§π = 0.52. Applying the equation πππ¦ πππ’ππ = π + π§πππ’ππ π again, we can now find: π = 72 + 0.52(9) π = 76.68 In other words, to get a βCβ on this test, you have to score between a 67.32 and a 76.68 There is a formula shortcut we can use to get from a βbetweenβ probability to a βless thanβ probability: π β π·(π < π < π) π·(π < π) = π Given, of course, that a and b are symmetric about the mean. HW: Normal distributions worksheets