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ENGR 224/STAT 224 Probability and Statistics Lecture 10. Definition: Normal or Gaussian. A probabilit distribution is normal if it is symmetric and bell shaped and fits the curve The shape of this curve is completely determined by the parameters m and s. Families of Normal Curves.

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## ENGR 224/STAT 224 Probability and Statistics Lecture 10

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**Definition:Normal or Gaussian**• A probabilit distribution is normal if it is symmetric and bell shaped and fits the curveThe shape of this curve is completely determined by the parameters m and s.**Families of Normal Curves**m = 100,s = 8 m = 25,s = 8 m = 100,s = 15**Examples of Normally Distributed Random Variables**• Grades on a test. • How much coffee is in a coffee jar. • The measurements of distance between two points. • The heights of students in this class.**Definition:Standard Normal Distribution**The Standard Normal Distributionis a normal probability distribution that has a mean of 0 and a standard deviation of 1.m = 0, s = 1 In this way the formula giving the heights of the normal curve is simplified greatly.**Notation: Z-score**Because the standard normal distribution is so important, we will use the letter Z, instead of X, to represent any normally distributed random variable (with mean of 0 and standard deviation of 1), and z to represent the value that this random variable takes on. The value of z is called a z-score. The CDF of the standard normal probability distribution is denoted as F(z)=P(Z ≤ z)**The Standard Normal curve and areas**• P(0 z 1) represents the probability that z takes on values between 0 and 1, which is represented by the area under the curve between 0 and 1. P(0 z 1) = 0.3413**Finding Probabilities when given z-scores.**• For a given z-score, the probability can be found in a table in the back of the text, see inside front cover, which gives values for F(z). • Note: The table only gives the areas under the curve to the left of z. To find other intervals requires some tricks.**Practice Problems**• Use the tables in the back of the book to find the following. • P(0 z 2.43) • P(-2.43 z 0) • P(1.20 z 2.30) • P(-1.50 z 2.4) • P( z 1.8)**Practice Problems**• Assume that the z-scores are normally distributed with a mean of 0 and a standard deviation of 1. Do the following. • If P(0 z a) = 0.4778, find a. • If P(-b z b) = .7814, find b. • If P(z c) =.0062, find c. SOLUTIONS:a = 2.01 b = 1.23 c = -2.50**Non-Standard Normal Distributions**• Most often random variables will not have a mean equal to 0 and a Std. dev. equal to 1. • Examples: • Machine filling 500 g jars of coffee. • Grades on a final exam in large class. • We can convert any non-standard score x to a standard z-score by letting**Example: Ball Bearings**Consider a collection of Ball Bearings known to have a mean diameter of 100 mm and a standard deviation of 15 mm. Find the probability that a ball bearing chosen at random will be between 100 and 130? Solution:**Example: Ball Bearings**Consider a collection of Ball Bearings known to have a mean diameter of 100 mm and a standard deviation of 15 mm. Find the probability that a score chosen at random will be above 125? Solution:**Example: Empirical Rule**Find the following probabilities. That is find the probabilities that a normally distributed random variable x, is situated one standard deviation from the mean and two standard deviations from the mean. Solutions: 0.6826 about 68% 0.9544 about 95%**Example: Manufacturing Bolts**The diameterof a bolt produced by a machine is normally distributed with a mean diameter of 0.82 cm and a standard deviation of 0.004 cm. What percent of bolts will meet the specification that they be between 0.816 cm and 0.826 cm in diameter? Solution:**Example: TV Warranties**A manufacturer of TV sets wants to advertise that if a TV set manufactured by the company lasts for less than a certain number of years, the company will refund the full amount paid for set. The company wants to pick the number of years for the advertisement in such a way that it will not have to give refunds on more than 4 percent of the sets. If, as a statistician for the company, you are asked to provide the number, what would be your answer? You may assume that the life of a TV set is normally distributed with a mean of 8.5 years and a standard deviation of 1.8 years.**Example: TV Warranties**Solution:**Overview**• Standard Normal Distribution • Non-Standard Normal Distributions**Homework**• Practice using the Standard Normal Table • Reread 4.3, problems 36 and 38 • Read 4.4

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