MA677 - problem set 1 1. You model the production of a weaving machine in terms of meters of ribbon produced per minute. Mean production in a one minute interval is 1.2 m with standard deviation of 20 cm. Assuming that the amount of ribbon produced in different 1 minute intervals is independent and identically distributed, estimate the probability that the machine will produce at least 74 meters of ribbon in one hour. 2. In the Gotham metropolitan area, 75 percent of the people live in the city and 25 percent live in the suburbs. A recent concert drew a random sample of 1200 from the metropolitan area. What is the probability that the number of people from the suburbs attending the concert will be fewer than 270? 3. A bolt of canvas is traditionally 39 yards long. You are producing a report the quality of canvas delivered to your customers. If the number of defects on any given bolt of canvas is Poisson distributed with mean 5 and the number of defects on each bolt is counted for a random sample of 125 bolts, what is the probability that the average number of defects per bolt in the sample will be less than 5.5? 4. Suppose that the proportion of defective items in a large manufactured lot is 0.1. What is the smallest random sample that must be taken from the lot in order for the probability to be at least 0.99 that the proportion of defective items in the sample will be less than 0.13? 5. Suppose that three children Alice, Bob, and Cindy throw snowballs at a target. Alice throws 10 times, and the probability that she will hit the target on any given throw is 0.3; Bob throws 15 times, and the probability that he will hit the target on any given throw is 0.2; and Cindy throws 20 times, and the probability that she will hit the target on any given throw is 0.1. What is the probability that the target will be hit at least 12 times. 6. You drawing 16 balls with replacement from a sack of ten balls. Each ball is numbered 0 through 9. What is the probability that the average of the numbers on the balls you have drawn will lie between 4 and 6? 7. You go to a party with a 63-ounce bottle of gin that has a special cap for dispensing drinks. The expected size of each drink dispensed is 2 ounces and the standard deviation of each drink is 1/2 ounce. Dispensed drinks are independent. What is the probability that the bottle will not be empty after 36 drinks have been served. 8. You make 25 independent measurements of a metal rod with an instrument known to produce measurements with standard deviation 1 cm. a. Use Chebyshev's inequality to find a lower bound for the probability that the average of your measurements will differ from the actual length of the metal rod by less than 2.5 mm. b. Now calculate the approximate value of the same probability using the central limit theorem. 9. A random sample of n items is taken from a distribution with mean and standard deviation a. Use the Chebyshev inequality to determine the smallest sample needed so that | | 0.99 4 b. Now use the central limit theorem to determine the smallest sample required for the same criterion. 10. At Whatsamatter U., two parents attend the graduation ceremony for 1/3 of the graduating seniors. One third of the seniors have one parent who attends the ceremony. The remaining third of these seniors have no parents attend. If there are 600 graduating seniors in a particular class, what is the probability that not more than 650 parents will attend the graduation ceremony? 11. Suppose that 1 , , are a random sample from a normal distribution with unknown mean and variance 2 . Assuming that 0, what is the asymptotic distribution of 3 . 12. Suppose that 1 , , are a random sample from a normal distribution with mean 0 and 1 1 unknown variance 2 . Determine the asymptotic distribution of the statistic =1 2 . Answer We have =p=0.1 The variance of the sampling distribution of proportion is 2= p(1p) n = (0.1*0.9)/n By central limit theorem,the z-score is given by z= X n X 0.1=z0.3/ n Now,we have to find value of n such that Pr(Xbar<0.13) 0.99 Pr( (Xbar) 0.130.1 < < 0.99 0.3 n n 0.03 Pr(Z< ( 0.3 ) ) 0.99 N Pr(z<0.1 n ) 0.99 Using normal distribution the z-score corresponding to p=0.99 is 2.33 Pr(Z=12) = Pr( 100.30.7+150.20.8+ 200.10.9=2.51 X 128 128 Z ) =Pr( 2.51 2.51 ) =Pr(Z 1.594) =1- [=NORMSDIST(1.594] =1-0.9445 =0.0555 Answer Digit Probabilit y 0 0.1 1 0.1 2 0.1 3 0.1 4 .1 5 0.1 6 0.1 7 0.1 Finding The expected value = xP(x) = 0*0.1+1*0.1+2*0.1+.....9*0.1 =4.5 8 0.1 9 0.1 E(X2) = 02*0.1+12*0.1+22*0.1+---------92*0.1 = 28.5 Variance =E(X2)-(E(X))2 = 28.5-4.52 = 8.25 Standard deviation = 8.25=2.87 Using the central limit theorem, the mean is 4.5 and standard deviation of drawing 16 ball is 2.87/sqrt(16) = 0.7181 Now, calculating the probability of number lying between 4 and 6 is P(4<=X<=6) =P((4-4.5)/0.7181 0 ,we have 2 Pr( |Xn|t ) n t 2 Since the standard deviation is 1 cm and 1/4th of the value is 2.5 mm,therefore we have to find a lower bound for the probability that the average of measurement is less than of the standard deviation. Thus,we have 2 2 | Xn | Pr( 4 ) n( ) 4 16 2 16 n 2 25 16 1- Pr( |Xn| 4 ) 25 9 Pr( |Xn| 4 ) 25 Thus,a lower bound of the probability is 9/25. is 0.36 b. Now calculate the approximate value of the same probability using the central limit theorem. Answer Pr( |Xn| 4 ) Using central limit theorem Z = (Xbar-mu)/(sigma/sqrt(n)) Z= Xbar Xbar 5(Xbar) = = n 25 ,where Z follows normal distribution Now, probability Pr( |Xn| 4 ) =Pr( =2*0.8944-1 = 0.7888 | | Xnbar 5 4 5 5 ) =Pr( |Z| 4 ) =2Pr (Z<=5/4) -1 We have | /4 0.99 Pr( |Xn | /4 ) Pr( |Xn -------- (1) 2 2 n( ) 4 2 16 2 n 16 n Finding the lower bound of the function,we have | /4 ) 16/n 1- Pr( |Xn | /4 ) 1-16/n Pr( |Xn ----------- (2) From 1 and 2 , 0.99 1- 16/n N = 1600 Answe r Using the central limit theorem,we have | /4 ) Pr( |Xn Xnbar n N n 4 ) = Pr ( |Z| 1 | /4 ) =Pr ( Pr( |Xn ) = 2 4 4 n ( ) 2 ( 4N )1 >= 0.99 ( 4n )(1+ 0.99)/2 ( 4n ) 0.995 Using normal probability, we have n 4 2.567 n = 106 2 A continuous random variable X has a normal distribution with mean and variance and the probability density function is given by x ( 1 2 = e f(x,, 2 2 2 2 ) ,where z = (x-)/sigma and the random variable z follows a standard normal distribution with mean 0 and variance 1 Using delta method where Use the above delta formula to find the asymptotic distribution where the Xn_bar is inverse which is 1/Xn_bar (x) =1/x and and derivative is '(x) = -1/x2 .Thus we can say that the asymptotic 2 n 1 1 ( ) Xnbar is the standard normal distribution with mean 1/ distribution of and variance 2 n 4 Considering the asymptotic distribution of the statistic Z = 1/((Xi^2/n).Rearranging the term Z(y) = 1/Y Derivative of z(y) is z'(y) = -1/y^2 Therefore the asymptotic distribution of Z(y) is the normal distribution with mean Substitute the value of mean = 2 2 2 2 = = 4 4 2 4 n n( ) n 1 2 =>Variance 1 2 2 and variance ( 2 ) n 8 = 2 n 4 A continuous random variable X has a normal distribution with mean u and variance 0'2 and the probability density mction is given by l _ x 'F A e ( 262 ) 2 ,where z = (x-u)/ sigma and the random variable 2 follows a f(x,u, 62) = 0' 27: standard normal distribution with mean 0 and variance 1 Answer We have =p=0.1 The variance of the sampling distribution of proportion is 2= p(1p) n = (0.1*0.9)/n By central limit theorem,the z-score is given by z= X n X 0.1=z0.3/ n Now,we have to find value of n such that Pr(Xbar<0.13) 0.99 Pr( (Xbar) 0.130.1 < < 0.99 0.3 n n 0.03 Pr(Z< ( 0.3 ) ) 0.99 N Pr(z<0.1 n ) 0.99 Using normal distribution the z-score corresponding to p=0.99 is 2.33 Pr(Z=12) = Pr( 100.30.7+150.20.8+ 200.10.9=2.51 X 128 128 Z ) =Pr( 2.51 2.51 ) =Pr(Z 1.594) =1- [=NORMSDIST(1.594] =1-0.9445 =0.0555 Answer Digit Probabilit y 0 0.1 1 0.1 2 0.1 3 0.1 4 .1 5 0.1 6 0.1 7 0.1 Finding The expected value = xP(x) = 0*0.1+1*0.1+2*0.1+.....9*0.1 =4.5 8 0.1 9 0.1 E(X2) = 02*0.1+12*0.1+22*0.1+---------92*0.1 = 28.5 Variance =E(X2)-(E(X))2 = 28.5-4.52 = 8.25 Standard deviation = 8.25=2.87 Using the central limit theorem, the mean is 4.5 and standard deviation of drawing 16 ball is 2.87/sqrt(16) = 0.7181 Now, calculating the probability of number lying between 4 and 6 is P(4<=X<=6) =P((4-4.5)/0.7181 0 ,we have 2 Pr( |Xn|t ) n t 2 Since the standard deviation is 1 cm and 1/4th of the value is 2.5 mm,therefore we have to find a lower bound for the probability that the average of measurement is less than of the standard deviation. Thus,we have 2 2 | Xn | Pr( 4 ) n( ) 4 16 2 16 n 2 25 16 1- Pr( |Xn| 4 ) 25 9 Pr( |Xn| 4 ) 25 Thus,a lower bound of the probability is 9/25. is 0.36 b. Now calculate the approximate value of the same probability using the central limit theorem. Answer Pr( |Xn| 4 ) Using central limit theorem Z = (Xbar-mu)/(sigma/sqrt(n)) Z= Xbar Xbar 5(Xbar) = = n 25 ,where Z follows normal distribution Now, probability Pr( |Xn| 4 ) =Pr( =2*0.8944-1 = 0.7888 | | Xnbar 5 4 5 5 ) =Pr( |Z| 4 ) =2Pr (Z<=5/4) -1 We have | /4 0.99 Pr( |Xn | /4 ) Pr( |Xn -------- (1) 2 2 n( ) 4 2 16 2 n 16 n Finding the lower bound of the function,we have | /4 ) 16/n 1- Pr( |Xn | /4 ) 1-16/n Pr( |Xn ----------- (2) From 1 and 2 , 0.99 1- 16/n N = 1600 Answe r Using the central limit theorem,we have | /4 ) Pr( |Xn Xnbar n N n 4 ) = Pr ( |Z| 1 | /4 ) =Pr ( Pr( |Xn ) = 2 4 4 n ( ) 2 ( 4N )1 >= 0.99 ( 4n )(1+ 0.99)/2 ( 4n ) 0.995 Using normal probability, we have n 4 2.567 n = 106 2 A continuous random variable X has a normal distribution with mean and variance and the probability density function is given by x ( 1 2 = e f(x,, 2 2 2 2 ) ,where z = (x-)/sigma and the random variable z follows a standard normal distribution with mean 0 and variance 1 Using delta method where Use the above delta formula to find the asymptotic distribution where the Xn_bar is inverse which is 1/Xn_bar (x) =1/x and and derivative is '(x) = -1/x2 .Thus we can say that the asymptotic 2 n 1 1 ( ) Xnbar is the standard normal distribution with mean 1/ distribution of and variance 2 n 4 Considering the asymptotic distribution of the statistic Z = 1/((Xi^2/n).Rearranging the term Z(y) = 1/Y Derivative of z(y) is z'(y) = -1/y^2 Therefore the asymptotic distribution of Z(y) is the normal distribution with mean Substitute the value of mean = 2 2 2 2 = = 4 4 2 4 n n( ) n 1 2 =>Variance 1 2 2 and variance ( 2 ) n 8 = 2 n 4 A continuous random variable X has a normal distribution with mean u and variance 0'2 and the probability density mction is given by l _ x 'F A e ( 262 ) 2 ,where z = (x-u)/ sigma and the random variable 2 follows a f(x,u, 62) = 0' 27: standard normal distribution with mean 0 and variance 1