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Statistics and, Probability, Quarter 3 – Module 3:, The Normal Distribution

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Statistics and, Probability, Quarter 3 – Module 3:, The Normal Distribution

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What I Need to Know, This module was designed and written to help you, understand the concept of the normal distribution. The, scope of this module helps you to appreciate and understand learning, situations that can be used in a day-to-day basis experience. The, language used in this learning kit recognized the diverse vocabulary level, of students for a higher understanding of the lesson. The lessons were, arranged to follow the standard sequence of the course. However, the, manner in which you read them can be possibly changed to correspond, with the textbook that you might be using now. Thus, this module is, divided into two lessons, respectively as follows, •, •, , Lesson 1 – The Normal Distribution and Its Properties, Lesson 2 – The Standard Normal Distribution, , After going through this module, you are expected to:, 1. illustrate a normal random variable and its characteristics, (M11/12SP-IIIc1);, 2. identify regions under the normal curve that correspond to, different standard normal values (M11/12SP-IIc-3);, 3. convert a normal random variable to a standard normal variable, and vice versa (M11/12SP-IIIc-4); and, 4. compute probabilities and percentiles using the standard normal, distribution (M11/12SP-IIIc-d-1).

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Lesson, , 1, , The Normal Distribution and, Its Properties, , The normal distribution is just one of the distributions to be discussed, in this course. It is also considered as the most important distribution in, Statistics because it fits many real-life situations. This lesson will bring, us a deeper understanding of the normal distribution and its, characteristics., , What is It, To give us a deeper understanding of the concept of the, normal distribution, let us learn more about its properties., The following are the properties that can be observed from the graph of, a normal distribution, also known as Gaussian distribution., 1. The graph is a continuous curve and has a domain -∞ < X < ∞., •, , This means that X may increase or decrease without, bound., 2. The graph is asymptotic to the x-axis. The value of the variable, gets closer and closer but will never be equal to 0., •, , As the x gets larger and larger in the positive direction, the, tail of the curve approaches but will never touch the, horizontal axis. The same thing when the x gets larger and, larger in the negative direction., , 3. The highest point on the curve occurs at x = µ (mean)., •, , The mean (µ) indicates the highest peak of the curve and is, found at the center., , •, , Take note that the mean is denoted by this symbol µ and, the standard deviation is denoted by this symbol ., • The median and mode of the distribution are also found at, the center of the graph. This indicates, that in a normal distribution, the, mean, median and mode are equal., 4. The curve is symmetrical about the mean., •, , This means that the curve will have, balanced proportions when cut in, halves and the area under the curve to, the right of mean (50%) is equal to the, area under the curve to the left of the mean (50%).

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5. The total area in the normal distribution under the curve is equal, to 1., •, , Since the mean divides the curve into halves, 50% of the, area is to the right and 50% to its left having a total of 100%, or 1., , 6. In general, the graph of a normal distribution is a bell-shaped, curve with two inflection points, one on the left and another on, the right. Inflection points are the points that mark the change in, the curve’s concavity., •, , Inflection point is the point at, which a change in the direction of, curve at mean minus standard, deviation, and mean plus, standard deviation., , •, , Note that each inflection point of, the normal curve is one standard deviation away from the, mean., , 7. Every normal curve corresponds to the “empirical rule” (also called, the 68 - 95 - 99.7% rule):, •, , about 68.3% of the area under, the curve falls within 1 standard, deviation of the mean, , •, , about 95.4% of the area under, the curve falls within 2 standard, deviations of the mean, , •, , about 99.7% of the area under, the curve falls within 3 standard, deviations of the mean.

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Consider the following examples:, 1. Suppose the mean is 60 and the standard deviation is 5, sketch a, normal curve for the distribution. This is how it would look like., , 2. A continuous random variable X is normally distributed with a, mean of 45 and standard deviation of 6. Illustrate a normal curve, and find the probability of the following:, a. P (39 < X < 51) = 68.3%, , *Since the area covered is 1 standard, of the deviation to the left and to the right., , c. P (X > 45) = 50%, , * Since the area covered is half, curve.

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Lesson, , 2, , The Standard Normal, Distribution, , The standard normal distribution, which is denoted by Z, is also a, normal distribution having a mean of 0 and a standard deviation of 1., Since the normal distribution can have different values for its mean and, standard deviation, it can be standardized by setting the µ = 0 and the =, 1., , What is It, As mentioned earlier, normal variable is standardized, by setting the mean to 0 and standard deviation to 1. This is for the, purpose of simplifying the process in approximating areas for normal, curves. As shown below is the formula used to manually compute the, approximate area., 1, , 𝑧2, , f(z) =, √2𝜋, , However, this formula is seldom used because a table was created to, summarize the approximate areas under the standard normal curve and, to further simplify the process. This table of probabilities is known as the, z- table., The Z - Table, Let us get a closer look at the z-table. The outermost column and row, represent the z-values. The first two digits of the z-value are found in the, leftmost column and the last digit (hundredth place) is found on the first, row., Suppose the z-score is equal to 1.85, locate the first two digits 1.8 in the, leftmost column and the last digit, .05, can be located at the first row., Then find their intersection which gives the corresponding area., Therefore, given z = 1.85, the area is equal to 0.9678., z, 0.00, 0.01, 0.02, 0.03, 0.04, 0.05, 0.06, 0.07, 0.08, 0.09, 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

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2.0, 0.9817, , 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 1.8, 0.9641, 0.9649 0.9656 0.9664 0.9671 0.9678 0.9686 0.9693 0.9699 0.9706 1.9, 0.9713 0.9719 0.9726 0.9732 0.9738 0.9744 0.9750 0.9756 0.9761, 0.9767, 0.9772 0.9778 0.9783 0.9788 0.9793 0.9798 0.9803 0.9808 0.9812, , Other examples are as follow:, 1. Find the area that corresponds to z = 2.67, z, 0.00, 0.01, 0.02, 0.03, 0.04, 0.09, 2.4 0.9918 0.9920 0.9922 0.9925 0.9927 0.9929, 2.5 0.9938 0.9940 0.9941 0.9943 0.9945 0.9946, 2.6 0.9953 0.9955 0.9956 0.9957 0.9959 0.9960, 2.7 0.9965 0.9966 0.9967 0.9968 0.9969 0.9970, 2.8, 0.9974 0.9975 0.9976 0.9977 0.9977, 0.9981, 2. Find the area that corresponds to z = 1.29, 3. Find the area that corresponds to z = 3, 4. Find the area that corresponds to z = - 0.64, , 0.05, 0.9931, 0.9948, 0.9961, 0.9971, 0.9978, , 0.06, , Answer: 0.9962, 0.07, 0.08, , 0.9932 0.9934 0.9936, 0.9949 0.9951 0.9952, 0.9962 0.9963 0.9964, 0.9972 0.9973 0.9974, 0.9979 0.9979 0.9980, Answer: 0.9015, , Answer: 0.9987, Answer: 0.2611 z 0.00, , 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09, -0.9, 0.1611, -0.8, 0.1867, -0.7, 0.2148, -0.6, 0.2451, -0.5, 0.2776, -0.4, 0.3121, , 0.1841 0.1814 0.1788 0.1762 0.1736 0.1711 0.1685 0.1660 0.1635, 0.2119 0.2090 0.2061 0.2033 0.2005 0.1977 0.1949 0.1922 0.1894, 0.2420 0.2389 0.2358 0.2327 0.2296 0.2266 0.2236 0.2206 0.2177, 0.2743 0.2709 0.2676 0.2643 0.2611 0.2578 0.2546 0.2514 0.2483, 0.3085 0.3050 0.3015 0.2981 0.2946 0.2912 0.2877 0.2843 0.2810, 0.3446 0.3409 0.3372 0.3336 0.3300 0.3264 0.3228 0.3192 0.3156, , 5. Find the area that corresponds to z = - 2.33, Answer: 0.0099, Note: The z-table used is the Cumulative Distribution Function (CDF) of the, Standard Normal Curve. Refer to the attachment on pages 19 and 20., Now that you already know how to use the z-table to find the, corresponding area for the z-score, let us identify the regions under the, nomal curve that corresponds to different standard normal values. In, order to find the regions, a probability notation is used.

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The probability notation P(a < Z < b) indicates that the z-value is between, a and b, P(Z > a) means z-value is above a and P(Z < a) means z-value is, below a. It would not matter whether we are considering P(Z < a) or P(Z ≤, a) or P(Z > a) or P(Z ≥ a). To illustrate, let us consider these examples:, 1. Find the proportion of the area between z = -1.25 and 2.19, this, can be expressed as P(-1.25 < Z < 2.19), read, as the probability that Z is greater than -1.25, but less than 2.19., Solution:, STEP 1: Draw a normal curve and locate the, scores and shade., , z -, , STEP 2: Locate the corresponding area of the, , z - scores in the z-, , table., z = -1.25 has a corresponding area, of 0.1056, , z, , =, , 2.19, , has, , a, , corresponding area of 0.9857, STEP 3: If you are looking for the area between two z - scores, simply, subtract the corresponding areas to arrive at the answer. Therefore,, 0.9857 0.1056 = 0.8801 and the P(-1.25 < Z < 2.19) = 0.8801 or 88.01%, 2. Compute the probability using the standard normal curve., a. P(Z < 1.67) = ______________, Solution:, STEP 1: Draw a normal curve and locate the, z - score and shade., STEP 2: Locate the corresponding area of the z - score in the z-table., z = 1.67 has a corresponding area of 0.9525, STEP 3: If you are looking for a less than area, the area in the table is, the answer, therefore the P(Z < 1.67) = 0.9525 or 95.25%., b. P(Z > -0.65) = ______________, Solution:, STEP 1: Draw a normal curve and locate the, score and shade., , z -, , STEP 2: Locate the corresponding area of the z - score in, the z-table., , z = -0.65 has a corresponding area of 0.2578, , STEP 3: If you are looking for a greater than area, the area in the, table is, subtracted from 1, therefore, 1.0000 – 0.2578 =, 0.7422 , and the, P(Z > -0.65) = 0.7422 or 74.22%

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The Z- Score, The z-score is an essential component in standard normal distribution., This allows us to describe a given set of data by finding the z-scores. This, leads us to a question of how z-scores are identified?, Given a normal random variable X with mean (µ) and standrad deviation, ( ), each value of x of the variable can be transformed into z-scores using, the formula,, 𝑥, −𝜇, , 𝑧=, 𝜎, , where z = z- score or standard score, x = observed value, µ = mean, = standard deviation, To illustrate how the value of x can be converted in z-score, here are, some examples., 1. A random variable X has a mean of 6 and a standard deviation of, 2. Find the corresponding z-score for x = 11., Given:, , x = 11, , µ=6, , =2, , 𝑥− 𝜇, , Solution: 𝑧 =, , Step 1: Write the formula., 𝜎, , Step 2: Substitute the given values., , =, , Step 3: Perform the operations., , =, z = 2.5, , 2., , Step 4: Write the corresponding z-score., , Given:, , x = 20, , µ = 12, , =3, , 𝑥− 𝜇, , Solution: 𝑧 =, , Step 2: Substitute the given values., , =, =, , Step 1: Write the formula., , 𝜎, , or 2.666666666667, , Step 3: Perform the, , operations., z = 2.67, , Step 4: Write the corresponding z-score.

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3., , Given:, , x = 18, , µ = 28, , =5, , 𝑥− 𝜇, , Solution: 𝑧 =, , Step 1: Write the formula., 𝜎, , =, , Step 2: Substitute the given values., , =, , Step 3: Perform the operations., , z = -2, , Step 4: Write the corresponding z-score., , 4. The scores in the summative test of 11- STEM B are normally, distributed with a mean of 65 and a standard deviation of 12. Find, the probability that some students got a score below 40., Solution:, STEP 1: Convert the normal value in z-score., Given: x = 40, , µ = 65, , = 12, , 𝑥− 𝜇, , Solution: 𝑧 =, 𝜎, , =, =, z = -2.08, Therefore, the P(X < 40) = P(Z < -2.08), STEP 2: Draw a normal curve and, locate the z - score and shade., STEP 3: Locate the corresponding area of the z - score in the z-table., z = -2.08 has a corresponding area of 0.0188, STEP 4: If you are looking for a less than area, the area in the table is, the answer, therefore, the P(Z < -2.08) = 0.0188 or 1.88%., 5. The height (in meters) of grade 11 students in section A follows a, normal distribution with the mean 1.6 and a standard deviation, of 0.3. Find the pobability that students chosen at random has a, height greater than 1.75., Solution:, STEP 1: Convert the normal value in z - score., Given: x = 1.75, µ = 1.6, = .3, 𝑥− 𝜇, , Solution: 𝑧 =

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𝜎, , 1, ., , =, ., , =, , z = 0.5, , Therefore, the P(X > 1.75) = P(Z > 0.5), STEP 2: Draw a normal curve and locate the z - score and shade., STEP 3: Locate the corresponding area of the z - score in, the z-table., , z = 0.5 has a corresponding area of 0.6915, , STEP 4: If you are looking for a greater than area, the area in the, table is, subtracted from 1, therefore, 1.0000 – 0.6915 =, 0.3085 , and the P(Z >, 0.5) = 0.3085 or 30.85%, The Percentile, A percentile is a measure used in statistics indicating the value below, which a given percentage of observations in a group of observations fall., Imagine you took a standardized test and you scored 91 at the 89th, percentile. This means that 89% of the examiners scored lower than 91, and 11% scored higher than 91. This explains that 89th percentile is, located where 89% of the total population lies below and 11% lies above, that point. To illustrate the 89th percentile of the normal curve here are, the steps:, 1. Express the given percentage as probability, remember 89% is the, same as, 0.8900., 2. Using the z-table (Cumulative Distribution Function (CDF) of the, Standard Normal Curve), locate the area of 0.8900., 3. There is no area corresponding exactly to 0.8900. It is between of, 0.8888 with a corresponding z - score of 1.22 and 0.8907 with a, corresponding z - score of 1.23. The nearest value to 0.8900 is, 0.8888 and therefore, the distribution lies below z = 1.22., 4. Construct a normal curve and shade the region to the left of 1.22.

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Cumulative Distribution Function (CDF) of the Standard, Normal Curve, 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, , 1.8, , 0.9641, , 0.9649, , 0.9656, , 0.9664, , 0.9671, , 0.9678, , 0.9686, , 0.9693, , 0.9699, , 0.9706, , 1.9, , 0.9713, , 0.9719, , 0.9726, , 0.9732, , 0.9738, , 0.9744, , 0.9750, , 0.9756, , 0.9761, , 0.9767, , 2.0, , 0.9772, , 0.9778, , 0.9783, , 0.9788, , 0.9793, , 0.9798, , 0.9803, , 0.9808, , 0.9812, , 0.9817, , 2.1, , 0.9821, , 0.9826, , 0.9830, , 0.9834, , 0.9838, , 0.9842, , 0.9846, , 0.9850, , 0.9854, , 0.9857, , 2.2, , 0.9861, , 0.9864, , 0.9868, , 0.9871, , 0.9875, , 0.9878, , 0.9881, , 0.9884, , 0.9887, , 0.9890, , 2.3, , 0.9893, , 0.9896, , 0.9898, , 0.9901, , 0.9904, , 0.9906, , 0.9909, , 0.9911, , 0.9913, , 0.9916, , 2.4, , 0.9918, , 0.9920, , 0.9922, , 0.9925, , 0.9927, , 0.9929, , 0.9931, , 0.9932, , 0.9934, , 0.9936, , 2.5, , 0.9938, , 0.9940, , 0.9941, , 0.9943, , 0.9945, , 0.9946, , 0.9948, , 0.9949, , 0.9951, , 0.9952, , 2.6, , 0.9953, , 0.9955, , 0.9956, , 0.9957, , 0.9959, , 0.9960, , 0.9961, , 0.9962, , 0.9963, , 0.9964

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2.7, , 0.9965, , 0.9966, , 0.9967, , 0.9968, , 0.9969, , 0.9970, , 0.9971, , 0.9972, , 0.9973, , 0.9974, , 2.8, , 0.9974, , 0.9975, , 0.9976, , 0.9977, , 0.9977, , 0.9978, , 0.9979, , 0.9979, , 0.9980, , 0.9981, , 2.9, , 0.9981, , 0.9982, , 0.9982, , 0.9983, , 0.9984, , 0.9984, , 0.9985, , 0.9985, , 0.9986, , 0.9986, , 3.0, , 0.9987, , 0.9987, , 0.9987, , 0.9988, , 0.9988, , 0.9989, , 0.9989, , 0.9989, , 0.9990, , 0.9990, , 3.1, , 0.9990, , 0.9991, , 0.9991, , 0.9991, , 0.9992, , 0.9992, , 0.9992, , 0.9992, , 0.9993, , 0.9993, , 3.2, , 0.9993, , 0.9993, , 0.9994, , 0.9994, , 0.9994, , 0.9994, , 0.9994, , 0.9995, , 0.9995, , 0.9995, , 3.3, , 0.9995, , 0.9995, , 0.9995, , 0.9996, , 0.9996, , 0.9996, , 0.9996, , 0.9996, , 0.9996, , 0.9997, , 3.4, , 0.9997, , 0.9997, , 0.9997, , 0.9997, , 0.9997, , 0.9997, , 0.9997, , 0.9997, , 0.9997, , 0.9998, , Cumulative Distribution Function (CDF) of the Standard, Normal Curve, z, , 0.00, , 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, , 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, , 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, , -1.8, , 0.0359, , 0.0351, , 0.0344, , 0.0336, , 0.0329, , 0.0322, , 0.0314, , 0.0307, , 0.0301, , 0.0294, , -1.7, , 0.0446, , 0.0436, , 0.0427, , 0.0418, , 0.0409, , 0.0401, , 0.0392, , 0.0384, , 0.0375, , 0.0367, , -1.6, , 0.0548, , 0.0537, , 0.0526, , 0.0516, , 0.0505, , 0.0495, , 0.0485, , 0.0475, , 0.0465, , 0.0455, , -1.5, , 0.0668, , 0.0655, , 0.0643, , 0.0630, , 0.0618, , 0.0606, , 0.0594, , 0.0582, , 0.0571, , 0.0559, , -1.4, , 0.0808, , 0.0793, , 0.0778, , 0.0764, , 0.0749, , 0.0735, , 0.0721, , 0.0708, , 0.0694, , 0.0681

Page 15 :
-1.3, , 0.0968, , 0.0951, , 0.0934, , 0.0918, , 0.0901, , 0.0885, , 0.0869, , 0.0853, , 0.0838, , 0.0823, , -1.2, , 0.1151, , 0.1131, , 0.1112, , 0.1093, , 0.1075, , 0.1056, , 0.1038, , 0.1020, , 0.1003, , 0.0985, , -1.1, , 0.1357, , 0.1335, , 0.1314, , 0.1292, , 0.1271, , 0.1251, , 0.1230, , 0.1210, , 0.1190, , 0.1170, , -1.0, , 0.1587, , 0.1562, , 0.1539, , 0.1515, , 0.1492, , 0.1469, , 0.1446, , 0.1423, , 0.1401, , 0.1379, , -0.9, , 0.1841, , 0.1814, , 0.1788, , 0.1762, , 0.1736, , 0.1711, , 0.1685, , 0.1660, , 0.1635, , 0.1611, , -0.8, , 0.2119, , 0.2090, , 0.2061, , 0.2033, , 0.2005, , 0.1977, , 0.1949, , 0.1922, , 0.1894, , 0.1867, , -0.7, , 0.2420, , 0.2389, , 0.2358, , 0.2327, , 0.2296, , 0.2266, , 0.2236, , 0.2206, , 0.2177, , 0.2148, , -0.6, , 0.2743, , 0.2709, , 0.2676, , 0.2643, , 0.2611, , 0.2578, , 0.2546, , 0.2514, , 0.2483, , 0.2451, , -0.5, , 0.3085, , 0.3050, , 0.3015, , 0.2981, , 0.2946, , 0.2912, , 0.2877, , 0.2843, , 0.2810, , 0.2776, , -0.4, , 0.3446, , 0.3409, , 0.3372, , 0.3336, , 0.3300, , 0.3264, , 0.3228, , 0.3192, , 0.3156, , 0.3121, , -0.3, , 0.3821, , 0.3783, , 0.3745, , 0.3707, , 0.3669, , 0.3632, , 0.3594, , 0.3557, , 0.3520, , 0.3483, , -0.2, , 0.4207, , 0.4168, , 0.4129, , 0.4090, , 0.4052, , 0.4013, , 0.3974, , 0.3936, , 0.3897, , 0.3859, , -0.1, , 0.4602, , 0.4562, , 0.4522, , 0.4483, , 0.4443, , 0.4404, , 0.4364, , 0.4325, , 0.4286, , 0.4247, , 0.0, , 0.5000, , 0.4960, , 0.4920, , 0.4880, , 0.4840, , 0.4801, , 0.4761, , 0.4721, , 0.4681, , 0.4641