Bell shaped curves

Learning Objectives for Section 11.5 Normal Distributions  The student will be able to identify what is meant by a normal distribution.  The student will be able to find the area under normal curves.  The student will be able to approximate the binomial distribution with a normal distribution.

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Normal Distributions We have seen that the histogram for a binomial distribution with n = 20 trials and p = 0.50 was shaped like a bell if we join the tops of the rectangles with a smooth curve. Real world data, such as IQ scores, weights of individuals, heights, test scores have histograms that have a symmetric bell shape. We call such distributions normal distributions. They will be the focus of this section. Barnett/Ziegler/Byleen Finite Mathematics 11e

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Mathematicians and the Normal Curve Three mathematicians contributed to the mathematical foundation for this curve. They are Abraham De Moivre, Pierre Laplace and Carl Friedrich Gauss.

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Abraham De Moivre De Moivre pioneered the development of analytic geometry and the theory of probability. He published The Doctrine of Chance in 1718. The definition of statistical independence appears in this book together with many problems with dice and other games. He also investigated mortality statistics and the foundation of the theory of annuities

De Moivre

http://www-gap.dcs.st-and.ac.uk/~history/Mathematicians/De_Moivre.html Barnett/Ziegler/Byleen Finite Mathematics 11e

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Pierre Laplace Laplace systematized and elaborated probability theory in "Essai Philosophique sur les Probabilités" (Philosophical Essay on Probability, 1814). He was the first to publish the value of the Gaussian integral. We will talk about Gauss later.

Laplace Barnett/Ziegler/Byleen Finite Mathematics 11e

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Bell-Shaped Curves  Many frequency distributions have a symmetric, bell shaped histogram.  Example 1: The frequency distribution of heights of males is symmetric about a mean of 69.5 inches.  Example 2: IQ scores are symmetrically distributed about a mean of 100, with a standard deviation of 15 or 16. The frequency distribution of IQ scores is bell shaped.  Example 3: SAT test scores have a bell shaped, symmetric distribution.

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Properties of Normal Curves  Normal curves are bell-shaped and are symmetrical with respect to the vertical line x =  (the mean).  The curve approaches, but does not touch, the horizontal axis as x gets very large (or x gets very small)  The shape of a normal curve is completely determined by its mean and standard deviation - a small standard deviation indicates a tight clustering about the mean and thus a tall, narrow curve; a large standard deviation indicates a large deviation from the mean and thus a broad, flat curve.

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Graphs of Normal Curves

1.4 1.2 1 0.8 0.6 0.4 0.2 0 -0.2

-10

-5

0

5

10

Several normal curves Barnett/Ziegler/Byleen Finite Mathematics 11e

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Probability and Area under the Normal Curve  Key fact: For a normally distributed variable, the percentage of observations that lie within a specified range equals the corresponding area under its associated normal curve.  This is approximately true for a variable that is approximately normally distributed.  p(a < x < b) = probability that the random variable X is between a and b = area under the normal curve between x = a and x = b.  The total area under a normal curve is 1. Barnett/Ziegler/Byleen Finite Mathematics 11e

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Finding Areas Under a Normal Curve Finding the area under a normal curve between x = a and x = b requires calculus. We can circumvent this problem by looking up the values in a table. However, the shape of each normal curve is determined by the standard deviation; the greater the standard deviation, the “flatter” and more spread out the normal curve will be. We would need infinitely many tables. The solution is to standardize a normally distributed variable, and to use the table for the standard normal curve.

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Standard Normal Distribution The standard normal distribution has a mean of 0 and a standard deviation of 1. Values on the horizontal axis are called z values. Values on the y axis are probabilities and are decimal numbers between 0 and 1, inclusive.

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Areas under the Standard Normal Curve For the following examples, 1. Draw a diagram 2. Shade the appropriate area 3. Use a table or a TI 83 to find the probability (A) Find p(0 < z < 1.2)

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Areas under the Standard Normal Curve For the following examples, 1. Draw a diagram 2. Shade the appropriate area 3. Use a table or a TI 83 to find the probability (A) Find p(0 < z < 1.2) Look up the z value of 1.2 Answer: 0.3849 Barnett/Ziegler/Byleen Finite Mathematics 11e

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Areas (continued) (B) Find p(-1.3 < z < 0)

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Areas (continued) (B) Find p(-1.3 < z < 0) We can ignore the sign of z since the graph is symmetrical. Look up the z value of 1.3. Answer: 0.432.

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Areas (continued) (C) Find p(-1.25 < z < 0.89)

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Areas (continued) (C) Find p(-1.25 < z < 0.89) Use table to find two different areas, and add: area left of y axis = 0.3944, area right of y axis = 0.3133 Answer: .7077

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Areas (continued) (D) Find p(z > .75)

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Areas (continued) (D) Find p(z > .75) Use table to find p(0 < z < 0.75) = 0.2744 Subtract this area from 0.5000 Answer: 0.2266

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Standardizing a Normally Distributed Variable To find p(a < x < b) for a normal curve with mean  and standard deviation , we calculate

P where

z

a



z

b





x



The variable z is called the standard normal variable.

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Example Assume IQ scores are distributed normally with a mean of 100 and standard deviation of 16.

(A) If the IQ of an individual is x = 124, what z value corresponds to this?

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Example Assume IQ scores are distributed normally with a mean of 100 and standard deviation of 16.

(A) If the IQ of an individual is x = 124, what z value corresponds to this? z=

124 100  1.5 16

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Example (continued) (B) Find the probability that a randomly chosen person has an IQ greater than 120.

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Example (continued) (B) Find the probability that a randomly chosen person has an IQ greater than 120. Step 1. Draw a normal curve and shade appropriate area. State the probability: p(x > 120) where x is IQ.

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Example (continued) Step 2. Convert x score to a standardized z score: z = (120 – 100) / 16 = 20/16 = 5/4 = 1.25 p(x > 120) = p(z > 1.25) Step 3. Use table or TI 83 to find area. Answer: 1 - 0.6021 = 0.1056

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Example 2 A traffic study at one point on an interstate highway shows that vehicle speeds are normally distributed with a mean of 61.3 mph and a standard deviation of 3.3 miles per hour. If a vehicle is randomly checked, find the probability that its speed is between 55 and 60 miles per hour.

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Example 2 (continued) Step 1. Draw a normal curve and shade appropriate area. State the probability: p(55 < x < 60) where x is speed. Step 2. Convert x score to a standardized z score: 60  61.3   55  61.3 p z  = p(-1.91 < z < -0.39) 3.3   3.3

Step 3. Use table or TI 83 to find area. Answer: 0.4719 - 0.1517 = 0.3202 Barnett/Ziegler/Byleen Finite Mathematics 11e

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Mathematical Equation for Bell-Shaped Curves Carl Friedrich Gauss, a mathematician, was probably the first to realize that certain data had bell-shaped distributions. He determined that the following equation could be used to describe these distributions:    ( x )2    2  2   1

f ( x) 

2 

e





where ,  are the mean and standard deviation of the data.

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Using the Normal Curve to Approximate Binomial Probabilities Example: Find the probability of getting 12 or more Heads when you toss a coin 20 times.

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Using the Normal Curve to Approximate Binomial Probabilities Example: Find the probability of getting 12 or more Heads when you toss a coin 20 times. Solution: We have seen that the histogram for a binomial distribution with n = 20 trials and p = 0.50 was shaped like a bell if we join the tops of the rectangles with a smooth curve. To find the probability that x (number of heads) is greater than or equal to 12, we would have to calculate p(x=12) + p(x=13) + p(x=14) + … p(x=20) . The calculations would be very tedious, to say the least. Barnett/Ziegler/Byleen Finite Mathematics 11e

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Using the Normal Curve to Approximate Binomial Probabilities  We could, instead, treat the binomial distribution as a normal curve, since its shape is pretty close to being a bell-shaped curve, and then find the probability that x is greater than 12 using the procedure for finding areas under a normal curve.  = np = 10 = sqrt(np(1-p)) = sqrt(5) = 2.24 Probability that x ≥ 12 = total area in yellow

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Approximating Binomial by Normal (continued) Because the normal curve is continuous and the binomial distribution is discrete, we have to make what is called a correction for continuity.

Since we want p(x ≥ 12), we must include the rectangular area corresponding to x = 12. The base of this rectangle starts at 11.5 and ends at 12.5. Therefore, we must find p(x > 11.5). The rectangle representing the p(x = 12) extends from 11.5 to 12.5 on the horizontal axis.

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Approximating Binomial by Normal (continued) Using the procedure for finding area under a non-standard normal curve we get

11.5  10   p( x  11.5)  p  z    2.24  

= p(z > 0.770) = 0.5 - 0.2794 = 0.2206

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Rule of Thumb Test  Use a normal distribution to approximate a binomial distribution only if the interval

[   3 ,   3 ] lies entirely in the interval from 0 to n. (n is sample size.)

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