Lecture7_new

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67 Lecture 7: The N ormal and Other Continuous Models (Chapter Reference: 6) Assigning probabilities when your sample space is not a finite set of numbers, but rather a continuous interval is a bit different. We cannot assign individual probabilities to all possible outcomes; so instead, we use the area under a density curve to represent the probability of the random variable taking values in that interval. The probability density function for a continuous random variable X is a curve such that: 1. The total area under the curve is one. 2. The area under the curve for a particular interval is the probability that X is in that interval. Example (Uniform model) Use the density curve below to answer the following question: What percent of the observations lie between 0.5 and 1.2? 0 2 N ormal random variables are the most common type of continuous random variables. The density curve for a normal distribution is a bell-shaped normal curve. Attributes of the N ormal Curve: N otation: • Symmetric: Mean=  • Standard Deviation=  • The entire area under the curve is 100%, or 1

68 Review: Standardizing with Z-Scores: The standard deviation is the most common measure of spread used for normal curves and is a natural ruler for comparing individual values to the mean. To determine how many standard deviations the value is away from the mean, we can standardize this value. So, s X X Z − = or   − = X Z The z-score tells us how many standard deviation an observation is above or below the mean. When we standardize into Z-scores: o Shape of the distribution does not change o The center does change. The mean becomes 0. o The spread does change. The standard deviation becomes 1. The Empirical Rule states that for any normal or approximately normal distribution, approximate percentages or proportions under the curve can be estimated. It is often called the 68-95-99.7% Rule. Empirical Rule 68% of the observations are within one standard deviation of the mean 95% of the observations are within two standard deviations of the mean 99.7% of the observations are within three standard deviations of the mean Example #1: In the last quarter of 2008, a group of 64 mutual funds had a mean return of 2.4% with a standard deviation of 5.6%. Assume the return on the group of funds follows a normal distribution. Use the Empirical Rule (draw and label a curve as part of your answer) a. What percentage of the funds would have returns between -8.8% and 19.2%? b. For the distribution of mutual fund returns, the worst 2.5% will have a return of less than what amount?

69 Practice Problems: 1. A student got her verbal GRE scores back (GRE scores are normally distributed). The student got a 580 and was at the lower 16 th percentile. The report said the standard deviation of the GRE scores was 48. What is the mean of the scores? Use the following information to answer questions 2-4: In 2009, Stat119 exam scores followed a normal distribution with a mean of 75 points. Unfortunately, the value of the standard deviation and all of the other data got lost, except one Z-score. This Z-score was -2, with a corresponding raw exam score of 60. 2. Find the standard deviation of that exam’s results ( show work ) Answer: 3. Assume now that the mean for the exam was 70 with a standard deviation of 7. Using the Empirical Rule, an estimate for the proportion of values between 49 and 77 is: A) 0.68 B) 0.8385 C) 0.975 D) 0.815 E) 0.95 Answer: 4. For the exam with mean 70 and standard deviation 7, what score would a student have to get in order for 97.5% of the scores to be above that value? A) 49 B) 56 C) 63 D) 84 E) 91 Answer:

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70 The Standard N ormal Curve: Standardizing is a technique to help us determine percentages, proportions, probabilities, or area under the curve for any set of data. The new values are unit-less. The requirement is that the data must be normally or approximately normally distributed. We will use TABLE Z from your text. Finding Normal Percentiles using Table Z: Example #1: Use the Z table to find the following: Draw the picture first, shade the region you want and look up the Z in Table 2 to find the proportion to the left of that z-score. The proportion is also known as probability that the value of a particular member of a population will fall in the given interval. a. P(Z<-1.42)= b. P(Z  1.95) = c. P(-1.02<Z  2.57) = Example #2: Women’s heights Assume that college women’s heights follow a normal curve with a mean height of 65 inches and a standard deviation of 2.7 inches. a) Find the probability a college woman, selected at random, is shorter than 62 inches? b) Find the probability a college woman, selected at random, is more than 68 inches tall? c) Find the probability a college woman, selected at random, is between 60 and 68 inches tall?

71 Example #3: According to an article in Newsweek, the rate of water pollution in China is more than twice that measured in the US and more than three times the amount measured in Japan. The mean emission of organic pollutants is 11.7 million pounds per day in China. Assume the water pollution in China is normally distributed throughout the year with a standard deviation of 2.8 million pounds of organic emissions per day. a) What is the probability that on any given day the water pollution in China is at least 15 million pounds per day? b) What is the probability that on any given day the water pollution in China is between 6.2 and 9.3 million pounds per day? Inverse N ormal Probability Calculations: Un-standardizing Sometimes the proportion or percentage is given and you must find the corresponding Z-score and un-standardize the value by finding the X-value STEPS: • Draw the picture • Identify the Z-value from the given value of the proportion - look up the proportion in the MIDDLE of Table A. • Solve for X:   − = X Z  X = Example #1: In a standard normal model, what value(s) of z cut(s) off the region described? Draw a picture. a) The lowest 10% b) The highest 30% c) The middle 70%

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72 Example #2: The distribution of heights of college women is normal, with mean 65 inches and standard deviation 2.7 inches. a) Find the height such that 10% of college women are shorter than that height. b) Determine the two heights that make up the middle 90%. Example #3 : An athletic association wants to sponsor a footrace. The time it takes to run the course is normally distributed with a mean of 58.6 minutes, and a standard deviation of 3.9 minutes a) The association decides to have a tryout run, and eliminate the slowest 30% of the racers. What should the cutoff time be in the tryout run for elimination? b) What is the value of the first quartile for this distribution?

73 How can you assess Normality? N ormal probability plots give a visual way to determine if a distribution is approximately normal. If the distribution is close to normal, the plotted points will lie close to a line. Systematic deviations from a line indicate a non-normal distribution.

74 Practice Problems: 1. For each situation below, find the missing parameter. 15 , 100 % 20 , 22 % 25 , 20 =  =  =  =  below above 2. A World Health Organization study of health in various countries reported that in Canada, systolic blood pressure readings have a mean of 122 and a standard deviation of 16. It is known that the distribution of systolic blood pressure is normal. a) What is the probability a Canadian selected at random has systolic blood pressure between 100 and 135? b) High systolic blood pressures can be very dangerous. What systolic blood pressure represents the boundary for the upper 7% of blood pressures? 3. SAT scores are normally distributed with a mean of 500 and a standard deviation of 100.The SAT testing center claims that the middle 60% of test s cores represent the “typical” or “average” student. Students scoring outside of this range are either mediocre students or advanced students. Find the two test scores representing the middle 60%.

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75 4. Suppose that the distribution for the amount spent by students vacationing for a week in Florida is normally distributed with a mean of $650 and a standard deviation of $120. a) What is the probability that a randomly selected student vacationing for a week in Florida will spend between $500 and $900? b) Only 8% of students will spend more than what amount? 5. Wildlife biologists believe that the weights of adult trout can be described by a normal model with a standard deviation of 1.2 pounds. If only 7% of adult trout weigh more than 5 pounds what is the mean weight (in pounds) of adult trout? A. 3.224 B. 6.776 C. 5.624 D. 4.376 E. 4.928 Work: 6. A radar unit is used to measure the speed of automobiles on an expressway during rush-hour traffic. The speeds of individual automobiles are normally distributed with a mean of 62 mph. Find the standard deviation of all speeds if 3% of the automobiles travel faster than 72 mph.

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