The Normal Distribution (Pt. 2)

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1 Chapter 5 The Normal Distribution (Pt 2) 51 Finding Normal Percentiles Recall that the Nth percentile of a distribution is the value that marks off the bottom N% of the distribution For review, remember that Q1 is the 25th percentile, the median is the 50th percentile, and Q3 is the 75th percentile A normal percentile is simply a percentile of a normally-distributed variable In the last lecture, we learned about the empirical rule s estimates for the normal distribution s central regions Let us extend that rule to make some estimates on normal percentiles Let X N(µ, σ) Then, by application of the empirical rule: 1 µ 2σ is approximately the 25th percentile 2 µ 1σ is approximately the 16th percentile 3 µ is approximately the 50th percentile (ie the median) 4 µ + 1σ is approximately the 84th percentile 5 µ + 2σ is approximately the 975th percentile Refer to the illustrations in the last lecture if you are puzzled as to how we got these numbers 1

2 Elem Business Statistics CHAPTER 5 THE NORMAL DISTRIBUTION (PT 2) Note that if x marks the Nth percentile, then it follows straightforwardly that x also marks off the top (100 N)% of the distribution The table of standard normal percentiles or, more colloquially, the z-table, percentile information on the standard normal distribution Thus, the percentiles in this case are actually z-scores The table ranges from 390 z 390 in increments of 001 and the percentiles are given to the fourth decimal place Given a z-score say z = 115 you can find the percentile to which it corresponds by first looking down the left margin and locating 11 You then move along the row next to 11 until you get to the entry in the column headed by 005 This entry is the desired percentile which, in this case, is z Figure 51: How to look up a z-score s percentile At this point, we must introduce an important piece of notation Since 8749% of the standard normal curve lies to the left of z = 115, it follows that if we were to randomly choose some z-score, 8749% of the time it would be less than 115 That is, the probability statement P(Z < 115) = holds true In general, for a particular z-score z, P(Z < z) = the z-table s entry for z (It is highly recommended that you get a handle on this notation ASAP) Also note that, for technical reasons, P(Z < z) = P(Z z) That is, including z does not make a difference Now, because any normal distribution (eg X N(µ, σ)) can be converted to the standard normal distribution (eg Z N(0, 1)), we can take any value x from 2

3 Elem Business Statistics CHAPTER 5 THE NORMAL DISTRIBUTION (PT 2) X, convert it to a z-score, and then use the z-table to find out what percentile x corresponds to in its original distribution Therein lies the usefulness of the standard normal distribution Example (Standard Normal Percentiles) For the following, we use the model Z N(0, 1) For each problem, you should first draw a bell curve and shade in the area of interest Determine the following: 1 The percentage of the standard normal curve below z = The percentage of the standard normal curve above z = P(Z < 102) 4 P(Z 102) 3

4 Elem Business Statistics CHAPTER 5 THE NORMAL DISTRIBUTION (PT 2) Example (Non-Standard Normal Percentiles) For the following, we use the distribution X N(1152 pounds, 84 pounds) describing the weights of young Angus steers in Problem 519 As before, draw a picture of the model and shade in the area of interest Also, note that the z-table only gives us z-values up to the second decimal place, so round accordingly Determine the following: 1 The percentage of Angus steers weighing less than 1250 pounds 2 The probability of randomly choosing an Angus steer weighing less than 1000 pounds 3 If a z-score is a measure of unusualness, then which would be more unusual: a steer weighing 1250 pounds or one weighing 1000 pounds? 4 If a particular Angus steer had a z-score of 135, how much did it weigh? 4

5 Elem Business Statistics CHAPTER 5 THE NORMAL DISTRIBUTION (PT 2) 52 Using Normal Percentiles to Find More General Areas of the Normal Curve The standard normal percentiles in the z-table give the proportion of the normal curve less than, or to the left of, a given z-value However, we might be interested in the areas of more general areas of the standard normal curve, like the proportion greater than a given z-score, or the region between two z-scores Here are the three basic rules for finding Normal curve areas using only percentiles We name them Rule L(ess Than), Rule G(reater Than), and Rule B(etween): Rule L: Let z be a z-value Then the proportion of the standard normal curve less than z is simply its entry in the z-table That is: P(Z < z) = Entry in the z-table dnorm(x) x 5

6 Elem Business Statistics CHAPTER 5 THE NORMAL DISTRIBUTION (PT 2) Rule G: Let z be as before The proportion of the standard normal curve greater than z is 1 minus z s entry in the table That is: P(Z > z) = 1 Entry in the z-table = 1 P(Z < z) dnorm(x) x dnorm(x) x 6

7 Elem Business Statistics CHAPTER 5 THE NORMAL DISTRIBUTION (PT 2) Rule B: Let z 1 < z 2 be two z-scores The proportion of the standard normal curve between z 1 and z 2 is the difference between the areas left of z 2 and z 1 That is: P(z 1 < Z < z 2 ) = Entry for z 2 Entry for z 1 = P(Z < z 2 ) P(Z < z 1 ) dnorm(x) dnorm(x) x x dnorm(x) x 7

8 Elem Business Statistics CHAPTER 5 THE NORMAL DISTRIBUTION (PT 2) Example (Areas of the Standard Normal Curve) For the following, use the standard normal model (Z N(0, 1)) As before, draw a picture of the bell curve, shading in the area of interest Using the 3 rules, determine the following: 1 P(Z > 065) 2 P( Z < 101) = P( 101 < Z < 101) 3 P( 025 < Z < 225) 4 P(Z = 112) 8

9 Elem Business Statistics CHAPTER 5 THE NORMAL DISTRIBUTION (PT 2) Example (Areas of Other Normal Distributions) For the following, we use the previous example of X N(1152 pounds, 84 pounds) describing the weights of Angus steers Determine the following: 1 The percentage of steers weighing more than 1200 pounds 2 The percentage of steers weighing between 1100 and 1300 pounds 53 Solving for Missing Parameters Recall the direct calculation of z: z = x µ σ This equation can be re-expressed in three other ways by isolating the other variables on the left hand side: x = zσ + µ σ = x µ z µ = x zσ In this way, if we have enough information (3 of the 4 variables), we can solve for the missing parameter by using one form of the equation 9

10 Elem Business Statistics CHAPTER 5 THE NORMAL DISTRIBUTION (PT 2) Example (Solving for Missing Parameters) 1 A student got a 580 on her verbal GRE, putting her at the lower 16th percentile The report said that the standard deviation of verbal GRE scores was 48 What was the mean of the scores? Use the Empirical Rule for this problem 2 In 2004, STAT 119 exam scores were normally distributed with a mean of 75 points Unfortunately, the standard deviation and all scores were lost, but one z-score of -215, which corresponded to a raw exam score of 60 What was the standard deviation of these exam scores? 54 Inverse Normal Probability Calculations In the previous sections, we were given a distribution and asked to find the proportion of the distribution to the left of, right of, or between certain values from the distribution However, sometimes we will be given a region of the normal curve, like the lower 30% and asked to find the value marking off that region Given a distribution X N(µ, σ), the general method for solving this type of problem is as follows: 10

11 Elem Business Statistics CHAPTER 5 THE NORMAL DISTRIBUTION (PT 2) 1 Draw a picture of the region 2 On the z-table, look for the z-score corresponding to the percentile closest to the one related to the area of interest (eg For the lower 30%, look in the middle of the table for the number closest to 3000 and choose its z-score) 3 Use the equation x = zσ + µ to unstandardize your z-score into a value from X Example (Inverse Probability Calculations) 1 In the standard normal model, what value(s) of z cut(s) off the region described? (a) The lowest 10% (b) The highest 30% (c) The middle 70% 11

12 Elem Business Statistics CHAPTER 5 THE NORMAL DISTRIBUTION (PT 2) 2 Polygraph machines rely in part on galvanic skin response, or the skin s natural ability to conduct electricity, to assess a person s emotional state Suppose that skin response scores are approximately normal with a mean of 4940 and a standard deviation of 300 (units unknown) Suspects with skin response scores in the upper 3% are thought to be lying Find the minimum score that indicates lying 3 Suppose (old) SAT verbal scores are normal with mean 500 and standard deviation 100 Find the two test scores representing the middle 60% 4 In MathXL, there are questions telling you µ = 20, 25% above 22, σ = Indicating that a distribution is centered at 22 with 25% of the distribution lying above 22 Find the missing parameter σ 12

13 Elem Business Statistics CHAPTER 5 THE NORMAL DISTRIBUTION (PT 2) 5 A machine is used to regulate the amount of dye dispensed for mixing shades of paint The amount of dye per can follows a normal distribution with σ = 04 If more than 6 ml of dye is discharged while making paint, the shade is unacceptable Determine the mean amount of paint discharged (µ) so that only 1% of the cans of paint will be unacceptable 55 Preparation for the Quiz The reading for this lecture is Chapter 6 Normal Probability Plots (pg ) will not be covered in class, but may be required knowledge for the quiz and exam Practice Problems Chapter 6: 32, 33, 39 50, 52, 55, 60 13

Section 7.1 Properties of the Normal Distribution

Section 7.1 Properties of the Normal Distribution In Chapter 6, talked about probability distributions. Coin flip problem: Difference of two spinners: The random variable x can only take on certain discrete