Introduction to the Practice of Statistics Fifth Edition Moore, McCabe

Introduction to the Practice of Statistics Fifth Edition Moore, McCabe Section 1.3 Homework Answers 1.80 If you ask a computer to generate "random numbers between 0 and 1, you uniform will get observations from a uniform distribution. Figure 1.35 graphs the distribution density curve for a uniform distribution. Use areas under this density curve to answer the following questions. Define the random variable X to be the value that is generated by the computer. 0 1 X Figure 1.35 The density curve of a uniform distribution, for exercise 1.80. (a) Why is the total area under this curve equal to 1? Since the figure is a defined as a density curve, then by definition it has a total area of 1 square unit. The area represents 100% of the population (b) What proportion of the observations lie above 0.75? To answer this question we need only to find the area above the curve corresponding to X > 0.75. P(X > 0.75) = Height of yellow rectangle(width of yellow rectangle) = (1)(1 0.75) = 0.25 Keep in mind that because the author chose a uniform 0 0.75 1 X distribution with endpoints 0 and 1, it is easy to see what the proportion should be without much thought. Make sure you learn the real lesson here, that in order to calculate proportions with density curves, the area underneath the curve is directly related to the corresponding proportion. (c) What proportion of the observations lie between 0.25 and 0.75? We need to calculate P(0.25 < X < 0.75). P(0.25 < X < 0.75) = 1(0.75 0.25) = 0.5 0 0.25 0.75 1 X

1.81 Many random number generators allow users to specify the range of the random numbers to be produced. Suppose that you specify that the outcomes are to be distributed uniformly between 0 and 2. Then the density curve of the outcomes has constant height between 0 and 2, and height 0 elsewhere. Let the random variable Y be the value generated by the computer. (a) What is the height of the density curve between 0 and 2? Draw a graph of the density curve. The height of the density curve is ½, 0.5. Why? Because, a density curve, must have an area equal to 1 square unit. If you look at the dimensions of the rectangle we get ½ (2) = 1 square unit. ½ 0 2 Y (b) Use your graph from (a) and the fact that areas under the curve are proportions of outcomes to find the proportion of outcomes that are less than 1. It is very easy to see that the area is one, but to be complete I will run through the calculation. ½ 0 1 P(Y < 1) = ½ (1 0) = 0.5 2 Y (c) Find the proportion of outcomes that lie between 0.5 and 1.3. 2 ½ 0 0.5 1.3 2 Y P(0.5 < Y < 1.3) = ½ (1.3 0.5) = 0.4

1. 82 What are the mean and the median of the uniform distribution from problem 1.80 (Figure 1.35)? What are the quartiles? Since this is a symmetric distribution, the median and the mean are the same value, the halfway point. Thus the mean is 0.5 as well as the median. To calculate the mean of any uniform distribution take the average of the two endpoints: (0 + 1)/2 = 0.5 Again, since the boundaries of the figure are 0 and 1, it is easy to see the position of the quartiles: Q 1 = 0.25 and Q 3 = 0.75. Now while it is easy to see the quartile values, it is also easy to confuse what it is I am looking at. It just happens that the value of X also corresponds to the area it represents when we consider the frequency to the left of the number. That is, 1 P(X < 0.25) = P(X < Q 1 ) = 0.25 (area not value of X) P(X < 0.75) = P(X < Q 3 ) = 0.75 (area not value of X) 0 0.25 0.75 1 X Q 1 Q 3 If you are unsure what the above notation means or how it is related to the picture on the left, see me quickly. 1.83 Figure 1.36 displays three density curves, each with three points marked on the axis. At which of these points on each curve do the mean and the median fall? A B C A B C A B C (a) (b) (c) In order to analyze these curves correctly, one needs to remember that for a density curve the median is the value that splits the area above exactly in half (the median is the point the cuts the ordered set of numbers in half); the mean is pulled by outliers. Thus for picture (a) The median appear to be B, which then makes the mean C. For picture (b), since we have a symmetric graph, the mean and median are represented by A. Lastly, for picture (c), the median appears to be B and thus, the mean is A, which is pulled by outliers.

1.84 The length of human pregnancies from conception to birth varies according to a distribution that is approximately normal with mean 266 and standard deviation 16 days. Draw a density curve for this distribution on which the mean and standard deviation are correctly related. Let the random variable X denote the length of human pregnancies. µ 3σ µ 2σ µ σ µ µ+σ µ+2σ µ+3σ µ + 218 234 250 266 282 298 314 X 1.89 The height of women aged 20 to 29 are approximately normal with mean 64 inches and standard deviation 2.7 inches. Men the same age have mean height 69.3 inches with standard deviation 2.8 inches. What are the z-scores for a woman 6 feet tall and a man 6 feet tall? What information do the z-scores give that the actual heights do not? Women: {µ = 64 inches, σ = 2.7 inches} Men:{µ = 69.3 inches, σ = 2.8 inches} 0.75 Man: z = 72-69.3 2.8 0.9643 Woman: z = 72-64.0 2.7 2.9630 I can see that the six-foot tall woman is, among her peers, very tall, an extremely unusual height. (z = 2.9630). While the man is at six feet is above average but not as far away from the norm as the woman.

1.93 Using either Table A or your calculator or software, find the proportion of observations from a standard normal distribution that satisfies each of the following statements. In each case, sketch a standard normal curve and shade the area under the curve that is the answer to the question. (a) Z -2 (this is a cumulative proportion) If I looked this value up on a table then, I need to realize that -2, implies that my accuracy will be -2.00. So I look up -2 on the column with the z-value and the first column gives you the rest of the accuracy 0.00. P(Z -2) = 0.0228. z 0.00 Standard Normal Probabilities 3.0 2.0 1.0 0.0 1.0 2.0 3.0 Z 4. 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09-3.4 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0002-3.3 0.0005 0.0005 0.0005 0.0004 0.0004 0.0004 0.0004 0.0004 0.0004 0.0003-3.2 0.0007 0.0007 0.0006 0.0006 0.0006 0.0006 0.0006 0.0005 0.0005 0.0005-3.1 0.0010 0.0009 0.0009 0.0009 0.0008 0.0008 0.0008 0.0008 0.0007 0.0007-3 0.0013 0.0013 0.0013 0.0012 0.0012 0.0011 0.0011 0.0011 0.0010 0.0010-2.9 0.0019 0.0018 0.0018 0.0017 0.0016 0.0016 0.0015 0.0015 0.0014 0.0014-2.8 0.0026 0.0025 0.0024 0.0023 0.0023 0.0022 0.0021 0.0021 0.0020 0.0019-2.7 0.0035 0.0034 0.0033 0.0032 0.0031 0.0030 0.0029 0.0028 0.0027 0.0026-2.6 0.0047 0.0045 0.0044 0.0043 0.0041 0.0040 0.0039 0.0038 0.0037 0.0036-2.5 0.0062 0.0060 0.0059 0.0057 0.0055 0.0054 0.0052 0.0051 0.0049 0.0048-2.4 0.0082 0.0080 0.0078 0.0075 0.0073 0.0071 0.0069 0.0068 0.0066 0.0064-2.3 0.0107 0.0104 0.0102 0.0099 0.0096 0.0094 0.0091 0.0089 0.0087 0.0084-2.2 0.0139 0.0136 0.0132 0.0129 0.0125 0.0122 0.0119 0.0116 0.0113 0.0110-2.1 0.0179 0.0174 0.0170 0.0166 0.0162 0.0158 0.0154 0.0150 0.0146 0.0143-2 0.0228 0.0222 0.0217 0.0212 0.0207 0.0202 0.0197 0.0192 0.0188 0.0183-1.9 0.0287 0.0281 0.0274 0.0268 0.0262 0.0256 0.0250 0.0244 0.0239 0.0233 If I used Excel, the command would be =Normsdist(-2); this of course provides more accuracy in the result than the table.

(b) Z -2 What you need to keep in mind when looking up values on the table, is what area the table provides versus what you want. I want P(Z -2). The area underneath the whole curve is Thus, P(Z -2) = 1 - P(Z -2) = 1-0.0228. = 0.09772 Notice if I looked up Z = 2 on the table this is the associated value. On Excel the command is =1 normsdist(-2) 3.0 2.0 1.0 0.0 1.0 2.0 3.0 Z 4. (c) Z > 1.67 P(Z > 1.67) = 1 P(Z < 1.67) = 1 0.9525 = 0.04750 On Excel the command would be = 1 normsdist(1.67) Standard Normal Probabilities 3.0 2.0 1.0 0.0 1.0 2.0 3.0 Z 4. z 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 0.0 0.5000 0.5040 0.5080 0.5120 0.5160 0.5199 0.5239 0.5279 0.5319 0.5359 0.1 0.5398 0.5438 0.5478 0.5517 0.5557 0.5596 0.5636 0.5675 0.5714 0.5753 0.2 0.5793 0.5832 0.5871 0.5910 0.5948 0.5987 0.6026 0.6064 0.6103 0.6141 0.3 0.6179 0.6217 0.6255 0.6293 0.6331 0.6368 0.6406 0.6443 0.6480 0.6517 0.4 0.6554 0.6591 0.6628 0.6664 0.6700 0.6736 0.6772 0.6808 0.6844 0.6879 0.5 0.6915 0.6950 0.6985 0.7019 0.7054 0.7088 0.7123 0.7157 0.7190 0.7224 0.6 0.7257 0.7291 0.7324 0.7357 0.7389 0.7422 0.7454 0.7486 0.7517 0.7549 0.7 0.7580 0.7611 0.7642 0.7673 0.7704 0.7734 0.7764 0.7794 0.7823 0.7852 0.8 0.7881 0.7910 0.7939 0.7967 0.7995 0.8023 0.8051 0.8078 0.8106 0.8133 0.9 0.8159 0.8186 0.8212 0.8238 0.8264 0.8289 0.8315 0.8340 0.8365 0.8389 1.0 0.8413 0.8438 0.8461 0.8485 0.8508 0.8531 0.8554 0.8577 0.8599 0.8621 1.1 0.8643 0.8665 0.8686 0.8708 0.8729 0.8749 0.8770 0.8790 0.8810 0.8830 1.2 0.8849 0.8869 0.8888 0.8907 0.8925 0.8944 0.8962 0.8980 0.8997 0.9015 1.3 0.9032 0.9049 0.9066 0.9082 0.9099 0.9115 0.9131 0.9147 0.9162 0.9177 1.4 0.9192 0.9207 0.9222 0.9236 0.9251 0.9265 0.9279 0.9292 0.9306 0.9319 1.5 0.9332 0.9345 0.9357 0.9370 0.9382 0.9394 0.9406 0.9418 0.9429 0.9441 1.6 0.9452 0.9463 0.9474 0.9484 0.9495 0.9505 0.9515 0.9525 0.9535 0.9545 1.7 0.9554 0.9564 0.9573 0.9582 0.9591 0.9599 0.9608 0.9616 0.9625 0.9633

(d) -2 < Z < 1.67 To get this result I will use the previous information. I could look it up on the tables but it would most likely be the information I already have. Here is one way. P(-2 < Z < 1.67) = 0.9525 0.0228 3.0 2.0 1.0 0.0 1.0 2.0 3.0Z 4. The 0.9525 I got from problem (c). I note that P(Z > -1.67) = P(Z < 1.67). Now I need to subtract that little portion to the leftof 2, mainly the area 0.0228. Another way. P(-2 < Z < 1.67) = 1 (0.0228 + (1 0.9525)) Here I am using the fact that the entire area is one. I then calculate the two missing end points either directly or by another calculation. Subtract from one and I have the area I want. Using Excel = normsdist(1.67) normsdist(-2). 1.94 Find the value of z of a standard normal variable Z that satisfies each of the following conditions. (If you use Table A, report the value of z that comes closest to satisfying the condition). In each case, sketch a standard normal curve with your value of z marked on the axis. (a) 20% of the observations fall below z. If I use table A, I find that P(Z < -0.84) =.2005 which is close to the 0.2000. Using software like Excel, I get z 0.84162 (=normsinv(0.2)). 0.4 0.3 0.2 0.1 3.0 2.0 1.0 1.0 2.0 3.0 Z 4. 0.1-0.84 desired (b) 30% of the observations fall above z. 0.4 If I look at the table I see that P(Z > 0.52) = 0.3015 and P(Z > 0.53) = 0.2981. The value I want is about halfway between the two. So a good approximation of z is the average of 0.52 and 0.53 which is 0.525. 0.3 0.2 0.1 3.0 2.0 1.0 1.0 2.0 3.0 4. 0.1 0.525 Z

Using software like Excel, I get z 0.5244; I entered =normsinv(0.7). 1.97 The Wechsler Adult Intelligence Scale (WAIS) is the most common IQ test. The scale of scores is set separately for each age group and is approximately normal mean with mean 100 and standard deviation 15. The organization MENSA which calls itself the high IQ society, requires a WAIS score of 130 or higher for membership. What percent of adults would qualify for membership? Let the random variable X denote the WAIS score. We want to calculate P(X > 130). I notice that the value 130 is 2 standard deviations from the mean; by the 68-95-99.7 rule then, P(X > 130) = 2.5%. So P(X > 130) = 1 0.97725 Notice if I use the tables or a computer by finding the z-score I will not get 2.5%. Z = 2 for X = 130. = 0.2275 less than 2.5% which is just an approximation. Using Excel, I type in =normsdist(2) and I get 0.97725, which is the area to the right. The TI-83 command is normalcdf(2,10) 1.99 Jacob scores 16 on the ACT. Emily scores 670 on the SAT. Assuming that both tests measure the same thing, who has the highest score? SAT: µ = 1026 σ = 209 ACT: µ = 20.8 σ = 4.8 Emily: z = 670 1026 209 Jacob: z = 16 20.8 4.8 = -1.70 = -1 The z-scores tells us how far away each value is away from their respective means. So Emily is 1.7 standard deviations below the mean, and Jacob is only one standard deviation below the mean. Since Emily is much further below the mean than Jacob, Jacob has the higher score.

1.102 Reports on a student s ACT or SAT usually give the percentile as well as the actual score. The percentile is just the cumulative proportion stated as percent: the percent of all scores that were lower than this one. Tonya scores 1318 on the SAT. What is the percentile? Let s see so far we have the words percentage, relative frequency, percentile, and soon to come probability. All are calculated exactly the same, but how we view it is slightly different, thus the name change. Basically I need to calculate the area to the left of 1318, for this normal distribution. µ = 1026 σ = 209 Let the random variable X denote an SAT score. P(X < 1318) The area above represents the frequency of the numbers found on the X-axis, (i.e. how often would I encounter a value less than 1318 for example). The z-score for 1318 is 0.9188 1318-1026 209 1.3971 If I use Excel I would enter = normsdist(1.3971) which results in 0.9188 P(X < 1318) = 0.9188 which ranks Tonya very high, almost at the 92 percentile. If I were to use the table then instead of interpolating(the correct thing to do) to make it easier I will round (which does not give me as good of an approximation as interpolating, whatever that means). My z-score is then z = 1.40 P(Z < 1.40) =0.9192 which essentially says the same thing as the other result, Tonya is almost at the 92 nd percentile. 1.103 Reports on a student s ACT or SAT usually give the percentile as well as the actual score. The percentile is just the cumulative proportion stated as percent: the percent of all scores that were lower than this one. Jacob scores 16 on the ACT. What is his percentile? Since I know the distribution is normal ( a very important fact) then I will turn the value in question to a z-score, so I can look up the frequency on Table A, or input it into software such as Excel and get the required frequency. Z = 16 20.8 4.8 = -1 Using the 68-95-99.7 rule I calculate that P(Z < -1) 16%. Using table A, P(Z < -1) = 15.87%.

1.111 Middle-aged men are more susceptible to high cholesterol than the young women of Exercise 1.110. The blood cholesterol levels of men aged 55 to 64 are approximately normal with mean 222 mg/dl and standard deviation 37 mg/dl. What percent of these men have high cholesterol (levels above 240 mg/dl)? What percent have borderline high cholesterol (between 200 and 240 mg/dl)? In order to do well in statistics, one needs to understand what information is available to them. This will be specially true in later chapters. So let us start good habits now. What is known? The distribution is normal (Knowing the distribution type is tremendously important). Do I know the parameters for the normal distribution? Yes, µ = 222 mg/dl and σ = 37 mg/dl. What do I want to know? How often I will see a value that is above 240 mg/dl, for one question and the other how often will I get a value between 200 mg/dl and 240 mg/dl. Since I know the distribution type and have all the necessary information I will find my z-scores so I can correlate the z-scores to the requested frequencies by looking on a table or using the computer. Let the random variable X denote the blood cholesterol levels. P(X > 240 mg/dl) = P Z > 240 222 37 P(Z > 0.49) 0.3121 We would expect a reading above 240mg/dl about 31% of the time. 240 222 200 222 P(200 < X < 240 ) = P Z < P Z < 37 37 P(Z < 0.49) P(Z < -0.59) 0.6879 0.2776 of the time. 0.4103 We would expect a reading between 200mg/dl and 240 mg/dl about 41%