Survey
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
* Your assessment is very important for improving the workof artificial intelligence, which forms the content of this project
Statistics and the TI-84 Lesson 8 - Continuous Probability Distributions Normal Distributions 2nd DISTR 1 normalpdf (value x, mean , standard deviation ) gives the functional value (ordinate) of the normal curve at a given value of x. The peak of the normal distribution curve occurs at the mean , that is, when x=. Peak= normalpdf (, , ) 1. Exercise 1. Find the functional value (the ordinate) of the normal function with mean =30 and standard deviation =2.5 at x=24. Find the peak value of the normal function. 2nd DIST 1 normalpdf(24, 30, 2.5) ENTER answer: 0.0089578121 2nd ENTRY (change 24 to 30) ENTER answer: 0.1595769122 Note: when graphing, an appropriate window is needed. Go to WINDOW and use, Xmin: use approximately 4 standard deviations below the mean Xmax: use approximately 4 standard deviations above the mean Ymin=use -.05 Ymax=use a number higher than the peak= normalpdf(mean, mean, standard deviation) Also, go to the gray y= and clear all previous plots. To clear a graph: 2nd DRAW 1 ENTER ClrDraw ENTER Done To store a graph: 2nd DRAW (STO) 1 ENTER (number) nd To recall a graph: 2 DRAW (STO) 2 ENTER Exercise 2 . Draw the graphs of the standard normal distribution function and the normal distribution function with =18 and =1.5. gray Y= Y1=2nd DISTR 1 normalpdf (x, 0, 1) ENTER 2nd DISTR 1 normalpdf(0, 0,1) ENTER answer: 0.3989 is the peak value (use Ymax>0.3989) when selecting an appropriate window for the graph. WINDOW Xmin=-4 Xmax=4 Ymin=-0.05 Ymax=0.5 GRAPH gray Y= deselect Y1 Y2= 2nd DISTR 1 normalpdf(x, 18, 1.5) 2nd DISTR 1 normalpdf(18, 18,1.5) ENTER answer: 0 .26596 (use Ymax=0.4) WINDOW Xmin=12 Xmax=24 Ymin=-.05 Ymax=.4 GRAPH 2. 2nd DISTR 2 normalcdf(lower, upper, mean , Stdv. ) gives the area or probability below the normal curve and between the lower and upper values. Note: if the mean and standard deviations are not given their defaults values are = 0 and stdv = 1 (the standard normal or z-distribution) Exercise 3. Find the area below the standard normal curve, between z=-1.28 and z=2.05 2nd DISTR 2 normalcdf(-1.28, 2.05) ENTER answer: 0 .8795452188 -70- Exercise 4. The monthly rental rates for apartments in a certain city area are normally distributed with an average rent of $450 and a standard deviation of $50. What percentage of those apartments exceed a rent of $375? Hint: use 1E99 to indicate infinite as the upper bound. 2nd DISTR 2 normalcdf (375, 1E99, 450, 50) ENTER answer: 0.933192771 or approximately 93% Exercise 5. In a binomial n=80 and p=.15, use the normal approximation to the binomial to find the probability that the number of successes is at least 17 (use the ±0.5 continuity correction factor) Use =np and = np(1 p) and x≥ 16.5 2nd DISTR 2 normalcdf(16.5,1E99, 80*.15, 80(.15)(.85) ) Answer: 0.079416946 ENTER Exercise 6. In a binomial problem n=100, p=0.8, use the normal approximation to the binomial to find P(x>75) µ = np =100(.8) = 80 , = np(1 p) =4 2nd DISTR ENTER 2 normalcdf(75.5, 1E99, 100*.8, 100(.8)(.2) ) answer: 0.8697054357 3. The sampling distribution of the mean: x distribution n Use: 2nd DISTR 2 normalcdf(lower, upper, , ) ENTER Note: we are assuming that the population size N is large compared to the sample size n. Otherwise use the population size correction factor and obtain n N n N 1 Exercise 7. The average amount of time adults spend watching TV has a mean of 6 hours and standard deviation of 2 hours a day. A random sample of 81 adults is taken. what is the probability that their mean watching time is more than 6.5 hours? Use the Central Limit Theorem. 2nd DISTR 2 normalcdf (6.5, 1E99, 6, 2 )) 64 Answer: 0.0122244334 -71- ENTER Using Shade Norm Exercise 8. In a binomial problem n=100, p=0.8, use the normal approximation to the binomial to find P(x>75) µ = np =100(.8) = 80 , = np(1 p) =4 WINDOW Xmin=60 Xmax=100 Ymin=-.05 Ymax= 2nd DISTR 1 normalpdf(80, 80, 4) 2nd DISTR DRAW 1 ShadeNorm(75.5, 1E99, 80, 4) ENTER Answer: 0.869705, see graph Exercise 9. Use Shade Norm to find the probability that X<35.2 in a normal distribution with mean equal to 28 and standard deviation equal to 3.2. WINDOW Xmin=15 Xmax=41 Ymin=-.05 Ymax= 2nd DIST 1 Normalpdf(28, 28, 3.2) = .12466946 2nd DISTR DRAW 1 ShadeNorm(-1E99, 35.2, 28, 3.2) ENTER Answer: 0.987776 Using the Inverse Normal: InvNorm calculates the x value associated with the area to the left of the given x value. 2nd DISTR 3 invNorm(probability to the left of desired value, mean , stdv) Exercise 10. The grades for a large statistics class are normally distributed with mean of 68 and a standard deviation of 5. If the top 8% of the class will receive A's in the exam, what is the lowest A grade. Note: the area of the left I 1-.08=.92 2nd DISTR 3 invNorm(.92, 68, 5) ENTER Answer: 75.0253578 Exercise 11. Miss Morgan, who is a secretary, drives everyday to work. The time required for the trip from her apartment to the office can be approximated by a normal distribution with mean of 20 minutes and standard deviation of 5 minutes. At what time should Miss Morgan leave her apartment to give herself a 0.98 chance that she will be at her office by 9 a.m.? 2nd DISTR 3 invNorm(.98, 20, 5) ENTER Answer: 30.26874455 min. Answer: she should leave by 8:30 AM to give herself 30 minutes. -72- Exercise 12. A normally distributed random variable x possesses a mean of 14 and a standard deviation of 4. Find the probability that a) x falls between 9 and 23. WINDOW Xmin=-2 Xmax=30 Ymin=-0.05 Ymax=0.1 2nd DISTR 3 DRAW 1 ShadeNorm(9, 23, 14, 4) ENTER answer: 0.882126 b) Find the 90th. percentile of the distribution 2nd DISTR 3 invNorm(0.90, 14, 4) ENTER Answer: 19.12620627 c) Find the value of the distribution below which we find 5% of the values. nd 2 DISTR 3 invNorm (0.05, 14, 4) ENTER Answer: 7.420585496 d) If a sample of size n=4 is taken from this population, what is the probability that the sample mean will be larger than 11? 4 ) ENTER 4 2nd DISTR 2 normalcdf (11, 1E99, 14, Answer: .9331927713 Exercise 13 . In a normal population has mean 65 and standard deviation 4. If sample of size n=4 is taken from the population, a) what is the probability that the mean of the sample is less than 68?. 2nd DISTR 2 normalcdf (-1E99, 68, 65, 4 ) 4 ENTER Answer: 0.933249938 b) Find the third quartile of the distribution of sample means. 2nd DISTR 3 invNorm(.75, 64, 2) ENTER Answer: Q3 65.3489795 Exercise 14. Café "El Delicioso" claims 14 ounces of coffee in each package, with a standard deviation of 0.3 ounces. A government inspector will reject the claim and will have the company fix the settings if a sample of 25 packages shows an average below 13.85 ounces. If the average is really 14 ounces, what is the probability that the inspector will reject the company's claim? 2nd DISTR 2 normalcdf(-E99, 13.85, 14, .06 ) ENTER answer: 0.0062096799 Exercise 15. In problem 14, if the average is really 13.92 ounces of coffee in each package, what is the probability that the inspector will not reject the company's claim? 2nd DISTR 2 normalcdf(13.85,EE99, 13.92, .06 ) ENTER answer: 0.8783274392 -73-