Quiz 2 covered materials in Chapter 5 Chapter 6 Chapter 7. Normal Probability Distribution. Continuous Probability. distribution 11/9/2010.

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1 Quiz covered materials in Chapter 5 Chapter 6 Chapter 7 Normal Probability Distribution Chapter 6 Continuous Probability Distribution Normal Distribution Normal Distribution is symmetrical and bell shaped, implying that most values tend to cluster around the mean, equal to median. Uniform Distribution The uniform distribution is symmetrical and therefore the mean = median. The focus is on a continuous, bell-shaped distribution Discuss the properties Properties 1. The shape is determines by mean, and standard deviation,. The highest point on the normal curve is located at the mean, which is also the median and the mode of the distribution. 3. The normal distribution is symmetrical: the curve s shape to the left of the mean is the mirror image of its shape to the right of the mean. Thus, mean = median 4. The tails of the normal curve extend to infinity in both directions and never touch the horizontal axis. Thus, it has an infinite range. -< X < + 5. However, the tails get close enough to the horizontal axis quickly enough to ensure that the total area under the normal curve equals Since the normal curve is symmetrical, the area under the normal curve to the right of the mean () equals the area under the normal curve to the left of the mean, and each of these areas equals 0.5. When a mathematical expression is available to represent a continuous variable, the probability that various values occur within certain ranges or intervals can be calculated. 1

2 However, the EXACT probability of a particular value from a Normal Distribution is ZERO. e.g. P(X=140) = 0 a b What information do we need to calculate P(a x b)? = The normal probability distribution is defined by the equation f(x) = or all - 1 e x on the 1 x- real line a b What information do we need to calculate P(a x b)? and To calculate P(a x b): a b Random Variable Parameters Normal X= measured e.g. X= a person s weight and Presentation of Solution Follow the normal template page 69 X = For example: X= number of... WRONG µ = Normal = P ( ) = P ( X symbol # ) = Ncd ( L, U,, µ ) = 0.xxxx (words) ( <, >,, ) 4 decimals Must draw a diagram Suppose that the weights of adults are normally distributed with a mean of 170.0lbs and a standard deviation of 5.0 lbs. What is the probability that a person weighs from 150 to 190 lbs? = 5.0 Note: Calculating normal probability = Finding area under a portion of the normal curve 150 X= a person s weight = lbs = 5.0 lbs normal = P( 150 x 190) = Ncd (150,190, 5, 170) Refer to the calculator Lesson 5: Page 169 (Mean,, value, shading) Lower upper

3 Normal Probability Exercises Inverse Normal Page 69 to 70 Q6.1 to Q6.6 Given the value X, find the probability (or the area under the curve) What is the area? Given X µ The Inverse Normal Probability means: Given the probability (area), find the x value? Area= 0.3 X=? µ Inverse Normal Probability The weights of adults are normally distributed with a mean of lbs and a standard deviation of 5.0 lbs. 1) What is the maximum weight of the lightest 30% of the population? Present your solution as follows (use the template) X = person s weight = Normal = 5.0 Area= = 5.0 P( X?)= Area? = Inv = X=? = CASIO Calculator: select STAT mode F5 (DIST) F1 (NORM) F3 (InvN) X = InvN(0.3, 5, 170) = Written statement: The lightest 30% weigh at most lbs Inverse Normal Probability The weights of adults are normally distributed with a mean of lbs and a standard deviation of 5.0 lbs. ) What is the minimum weight to be in the heaviest 30% of the population? Present your solution as follows (use the template) X = person s weight = 5.0 = = 5.0 Normal Area = 0.3 P( X?) = Area = X=?? = Inv = NOTE When using the InvN function, the Area value that is required is the area to the LEFT of the desired X value. CASIO Calculator: select STAT mode F5 (DIST) F1 (NORM) F3 (InvN) X = InvN((1-0.3), 5, 170) = InvN(0.7, 5, 170) Written statement: The heaviest 30% weigh at least lbs. Area= 0.7 Area= 0.3 3

4 Using the same example where the weights of adults are normally distributed with a mean of lbs and a standard deviation of 5.0 lbs. What is the probability that a person weighs at least 160 lbs? P(X160) =? X = person s weight = lbs = 5.0 lbs Normal = 5.0 Area=? P(X160) = Ncd (160, 500, 5, 170) = = Check for +6 = (170)+6(5)=30 Probability Distributions Review Exercise Page 315 Q 7.3 to Q7.4 Q 7.43 to Q 7.47 Lower upper Suppose you are asked to calculate the probability that the average weight of a group of people is at least 160 lbs. How would you go about calculating the probability? Question 6., page 69 A firm s marketing manager believes that total sales for next year can be represented by a normal distribution, with mean of $.5 million and a standard deviation of $300,000. The firm has fixed cost of $1.8 million. a) What is the probability that the firm s sales will be less than $3.0 million? b) What is the probability that the firm will have sufficient sales to cover fixed costs? c) What is the probability that the firm s sales will be within $150,000 of the expected,i.e. mean sales? d) Determine the sales level that has only a 9% chance of being exceeded? Question 6.4, page 69 In the movie Forest Gump, the public school required an IQ of at least 80 to be admitted. a) If IQ test scores are normally distributed with a mean of 100 and a standard deviation of 16, what percent of children would qualify for admittance to the school? b) If the public school wished to have only 5% of all children not qualify for admittance, what is the minimum IQ test score should be required for admittance? Question 6.6, page 69 At a certain university the cumulative grade point average (CGPA) of first year students usually average.73 with a standard deviation It has been found that the marks are usually approximately normally distributed. a) What is the probability that a student will have a CGPA that is between.00 and 3.00? b) What percent of students will be on probation, i.e. their CGPA is less than.00? c) Academic scholarships are awarded to the top 1% of first year students. What is the minimum CGPA is needed to receive a scholarship? Question 7.6, page 315 A large company is currently evaluating 14 cost reducing proposals submitted by employees. Past experience has shown that 30% of such proposals are implemented by the company. a) What is the probability that more than 5 proposals will be implemented? b) What is the probability that at least half of the proposals will be implemented? c) What is the expected number of proposals implemented? 4

5 Question 7.34, page 316 The Long-Life Tire Co. claims that their Super-All-Season tire lasts an average of 110,000 km. It is known that the tire life is normally distributed with a standard deviation of 800 km. Your company has just purchased all new tires for its fleet of 10 cars. Assuming that the manufacturer s claim is true. a) What is the probability that a tire will last longer than 11,000km? d) What is the probability that all the tires of your car will last more than 11,000 km? Question 7.44, page 317 There are 1 agents in one office of a certain real estate firm. Much of the business in this office is conducted by taking a prospective customer out to view a particular property. The time to drive to and view a property is normally distributed and averages 47.3 minutes with a standard deviation of 11.7 minutes. Past studies have shown that 6% of customers that visit properties with an agent will eventually buy a property being shown by that agent. On average each agent visits 3.6 properties a day, with a standard deviation of 1.9 visits. All of the above activities are independent of each other. a) What is the probability that an agent will visit at least 7 properties over the next three days? c) A particular agent currently has 17 prospective customers that are being shown properties for sale. What is the probability that less than 4 of these customers will eventually buy a property being shown by the agent? SAMPLING DISTRIBUTION of the Sample Mean SAMPLING DISTRIBUTION of the Sample Mean Definition: The sampling distribution of the sample mean is the probability distribution of the population of all possible sample means obtained from samples of size n. A simple example concerning a population of four administrative assistants. Each assistants is asked to type the same page of a manuscript. The number of errors made by the assistants are shown below. NumberofAssistant 1 0 Administrative Assistant Number of errors Ann 3 Bob Carla 1 Dave Thenumberoferrorsisuniformlydistributed: =.5errors =1.1errors Numberoferrors Select samples of assistants with replacement from this population. There are 16 possible samples Sample Assistant Number of errors Sample Mean X 1 Ann,Ann Ann, Bob Ann, Carla Ann, Dave Bob, Ann Bob, Bob 7 Bob, Carla Bob, Dave Carla, Ann Carla, Bob Carla, Carla Carla, Dave Dave, Ann Dave, Bob Dave, Carla Dave, Dave

6 Select samples of assistants with replacement from this population. There are 16 possible samples Sample Assistant Number of errors Sample Mean X 1 Ann,Ann Ann, Bob Ann, Carla Ann, Dave Bob, Ann Bob, Bob 7 Bob, Carla Bob, Dave Carla, Ann Carla, Bob Carla, Carla Carla, Dave Dave, Ann Dave, Bob Dave, Carla Dave, Dave x.5 x The mean,, of the population of all possible sample means equals, the mean of the population of individual measurements. The standard deviation, x of the population of all possible sample means is less than, the standard deviation of the population of individual measurements. x n Let us summarize what we have learned about the sampling distribution of x. Assume that the population from which we will randomly select a sample of n measurements has a mean and standard deviation. Then the population of all possible sample means has: Mean x standard deviation x n 3a. If population is normal, x is normal 3b. If population is not normal, x is normal if sample size is large enough (n 30) The Central Limit Theorem Central Limit Theorem States that as the sample size gets large enough, the sampling distribution of sample means is approximately normally distributed regardless of the shape of the distribution of the individual values in the population See Figure 7.5 Sampling Distribution of the mean for different populations for samples of n=, 5, and 30 Conclusion regarding CLT 1. For most population distributions, regardless of shape, the sampling distribution of the sample mean is approximately normally distributed if the samples of at least size 30 are selected.. If the population distribution is fairly symmetric, the sampling distribution of the mean is approximately normal for samples as small as size If the population is normally distributed, the sampling distribution of the mean is normally distributed regardless of the sample size. 6

7 The CLT is very importance in using statistical inference to draw conclusions about a population. It allows you to make inferences about the population mean without having to know the specific shape of the population distribution. Homework (not graded) Central Limit Theorem Homework Page 314 Q 7.19 to 7. Transformation Formula To find normal probabilities is to use the transformation formula. Convert any normal random variable X to standardized normal random variable Z. Z X The Z value is equal to the difference between X and the mean, divided by the standard deviation The random variable X ~ (, ) The standardized normal variable Z ~ (=0, =1) The time to download the Web page is normally distributed with a mean =7 seconds and a standard deviation = seconds What is the Z value for a download time of 9 seconds? 9 7 Z 1 Answer:adownloadtimeof9secondsisequivalentto1standardizedunit The time to download the Web page is normally distributed with a mean =7 seconds and a standard deviation = seconds What is the Z value for a download time of 1 second? 1 7 Z 3 Answer:adownloadtimeof1secondisequivalentto3standardizedunits = Xscale(µ=7,=) Zscale(µ=0,=1) 7

8 Find P(X>9) What is the probability that the download time will be more than 9 seconds? 9 7 Z 1 Given a standardized normal distribution with a mean of 0 and a standard deviation of 1, what is P(Z > 1.08)? P( Z < -0.1)? P(-1.96 < Z < -0.1)? P(X>9)=p(Z>1) =Ncd(10,1,1,0) = Check:P(x>9)=Ncd(100,9,,7)= Zscale(µ=0,=1) Given a normal distribution with mean =50 and standard deviation is 4, the standardized (z score) value corresponding to a data value of 43 would be: Practice Questions Questions 6.1 to 6.17 on page 80 8