The problems of first TA class of Math144

The problems of first TA class of Math144 T eaching Assistant : Liu Zhi Date : F riday, F eb 2, 27 Q1. According to the journal Chemical Engineering, an important property of a fiber is its water absorbency. A random sample of 2 pieces of cotton fiber is taken and the absorbency on each piece was measured. The following are the absorbency values: 18.71 21.41 2.72 21.81 19.29 22.43 2.17 23.71 19.44 2.5 18.92 2.33 23. 22.85 19.25 21.77 22.11 19.77 18.4 21.12 a). Calculate the sample mean and median for the above sample values. b). Compute the 1% trimmed mean. c). Do a dot plot and constuct a relative frequency histogram of data. Solution: a). By definition, sample mean is: x n i1 x i n 415.35 2 2.7675. Because n is even, so x 1 2 (x n/2 + x n/2+1 ) 1 (2.72 + 2.5) 2.61. 2 b). We can compute the trimmed mean directly: n i1 x tr(1) x i 18.71 23.71 23. 18.4 2.7431. 16 c). Do a dot plot and construct histogram are easy works, complete them by youself please. 1

The problems of second TA class of Math144 T eaching Assistant : Liu Zhi Date : F riday, F eb 9, 27 Q1. Show that P (E F ) 1 P (E ) P (F ). Proof: P (E F ) 1 P ([E F )] ) 1 P (E F ), since P (E F ) P (E ) + P (F ), so we have P (E F ) 1 P (E F ) 1 P (E ) P (F ). Q2. Show that P (A B) P (A) + P (B) 2P (A B). Proof: P (A B) P ([A B] [B A]) Alternative, P (A B) + P (B A) P (A) P (A B) + P (B) P (A B) P (A) + P (B) 2P (A B). P (A B) P (A B) + P (B A) P (A B ) + P (B A ) P (A) P (A B) + P (B) P (A B) P (A) + P (B) 2P (A B). Q3. From a pack of 52 cards, we draw five cards at random. What is the probability that an Ace will appear at the fifth draw? Solution: Denoted by A the event {an ace will appear at fifth draw}. Then P (A) P 4 51P 1 4 P 5 52.769231. Q4. Two dice are thrown, let E be the event that the sum of the dice is odd, let F be the event that the first die lands on 1. And let G be the event that the sum is 5. Describe the events E F, E F, F G, E F, E F G. Answer: Because E {The sum of the dice is odd}, F {First die lands on 1}, G {Sum of the dice is 5}. 2

Then, E F {The sum of the dice is odd and first die lands on 1} {The first die lands on 1 and second die lands on a even number}, E F {First die lands on 1 or the sum of the dice is odd}, E G {The sum of dice is 5}, E F {The first die lands on 1 and the sum of dice is odd}, E F G {The sum is odd and first die lands on 1 and sum of dice is 5}, {First die lands on 1 and second die lands on 4}. Q5. Based on Q4, Find a simple expression for the events: (a). E E, (b). E E, (c). (E F ) (E F ), (d). (E F ) (E F ) (E F ), (e). (E F ) (F G). Answer: (a). E E { The sum of dice is odd or isn t odd} { The sum of dice is a number}. (b). E E {The sum of dice is odd and even also} Φ. (c). (E F ) (E F ) { Sum of the dice is odd or first die lands on 1, sum of the dice is odd or first die don t lands on 1} {The sum of dice is odd} E. (d). (E F ) (E F ) (E F ) F (E F ) {First die lands on 1, sum of dice is odd and first die don t land on 1} {First die lands on 1 and sum of dice is odd} E F. (e). (E F ) (F G) {First die lands on 1 or sum of dice is odd, first die lands on 1 or the sum of dice is 5 } {First die lands on 1 or sum of dice is 5}. 3

The problems of third TA class of Math144 T eaching Assistant : Liu Zhi Date : F riday, F eb 23, 27 Q1. Show that if A and B are independent, then (a). A and B are independent. (b). A and B are independent. Proof: (a). We recall the definition of independence, we say that two events A and B are independent if P (A B) P (A)P (B). Then, P (A B ) P (A) P (A B) so A and B are independent. P (A) P (A)P (B) P (A)(1 P (B)) P (A)P (B ). (b). P (A B ) P (B ) P (AB ) P (B ) P (A)P (B ) P (B )(1 P (A)) P (A )P (B ). This state the independence of A and B. Q2. We draw cards, one at a time, at random and successively from an ordinary deck of 52 cards with replacement. What is the probability that an ace appears before a face card? Solution: Denoted by A the event {an ace appears before a face card} and B n the event {an ace appears at nth draw, while aces and face cards don t appear at first n 1st draw}, then P (A) P (B n ) n1 [( 36 52 )n 1 ( 4 52 )] n1 4 ( 36 52 52 )n 1 n1 1/4. 4

Q3. A die is thrown as long as necessary for an ace or a 6 to turn up. Given that no ace turned up at the first two throwns, what is the probability that at least three throws will be necessary? Solution: Denoted by A the event {no ace turned up at the first two throwns}, by B the event {at least three throws will be necessary} and by C the event {both the result of first throw and second throw are 6}. P (B A) P (B/A) P (A) P (C) P (A) (1/6)2 (5/6) 2 1/25. Q4. In answering a question on a multiple-choice test, a student either knows the answer or she guesses. Let p be the probability that she know the answer and 1 p be the probability that she guesses. Assume that a student who guesses at the answer will be correct with probability 1/m, where m is the number of multiple choice alternatives. What is the conditional probability that a student knew the answer to a question given that she answered it correctly? Answer: Let A {she know the answer}, B {she guess}, C {she answer correctly}. obviously, Ω A B. Then the probability that we want to know is P (A/C). P (A/C) P (A C) P (C) P (C/A)P (A) P (A C) + P (B C) p P (C/A)P (A) + P (C/B)P (B) p P (A) + 1 P (B) m mp (m 1)p + 1. 5

Q5. In an urn are 5 fair coins, two 2-headed coins, and four 2-tailed coins. A coin is to be randomly selected and flipped. Compute the probability that the coin is fair if the result was (a). The flip was head; (b). 2 flips were both heads. Use prior and posterior probabilities. Answer: Let A {The coin is fair}, B {The coin is 2-headed}, Then prior probability is: C {The coin is 2-tailed}. P (A) 5/11, P (B) 2/11, P (C) 4/11. (a). Denoted by x the number of heads appear at first flip. Then obviously, x is or 1. Now we know x 1 by conditions offered by question. So f(1/a) 1/2, f(1/b) 1, f(1/c). The marginal distribution of x 1 can be calculated as: g(1) f(1/a)p (A) + f(1/b)p (B) + f(1/c)p (C) 9/22. Hence the posterior probability of A, given x 1, is π(a/1) f(1/a)p (A) g(1) 5/9. (b). Similar as (a), denoted by x the number of heads appear after 2 flips. x maybe is,1 or 2. by question, we know the result is 2, so f(2/a) 1/4, f(1/b) 1, f(1/c). The marginal distribution of x 2 can be calculated as: g(2) f(2/a)p (A) + f(2/b)p (B) + f(2/c)p (C) 13/44. Hence the posterior probability of A, given x 2, is π(a/2) f(2/a)p (A) g(2) 5/13. 6

Q6. A grocery store sells X Hundred kilograms of rice every day, where the distribution of the random variable X is of the following form: F (x) if x < ; kx 2 if x < 3; k( x 2 + 12x 3) if 3 x < 6; 1 if x 6. Suppose that this grocery store s total sales of rice do not reach 6 kilograms on any given day. (a). Find the value of k. (b). What is the probability that the store sells between 2 and 4 kilograms of rice next Thursday? (c). What is the probability that the store sells over 3 kilograms of rice next Thursday? (d). We are given that the store sold at least 3 kilograms of rice last Friday. What is the probability that is did not sell more than 4 kilograms on that day? Answer: (a). Because suppose that total sales of rice do not reach 6 kilograms on any given day, in other words P (X 6), so P (X 6) P (X < 6) k( 6 2 + 12 6 3) 1, This imply k 1 33. (b). P (2 X 4) P (2 X < 3) + P (X 3) + (3 < X 4) k(3 2 2 2 ) + 5/11 + k(( 4 2 + 12 4 3) ( 3 2 + 12 3 3)) 25/33.757576. (c). P (X > 3) 1 P (X 3) 1 k( 3 2 + 12 3 3) 3/11.272727. (d). P (X 4/X 3) P (X 4,X 3) P (X 3) 5/6. Q7. A random variable X is called symmetric if for all real number x such that P (X x) P (X x). 7

Prove that if X is symmetric, then for all t >, its distribution function F satisfies the following relations: (a). P ( X t) 2F (t) 1, (b). P ( X > t) 2[1 F (t)], (c). P (X t) F (t) + F ( t) 1. Proof: (a). Since t >, so (b). Similar as (a), P ( X t) P ( t X t) P (X t) P (X < t) F (t) (1 P (X t)) F (t) 1 + P (X t) 2F (t) 1. P ( X > t) P (X > t) + P (X < t) 1 F (t) + 1 P (X t) 2 F (t) P (X t) 2[1 F (t)]. (c). P (X t) P (X t) P (X < t) F (t) (1 P (X t)) F (t) 1 + P (X t) F (t) + F ( t) 1. 8

The problems of fourth TA class of Math144 T eaching Assistant : Liu Zhi Date : F riday, M arch 2, 27 Q1. The probability density function of a random variable X is given by: f(x) k (1 x2 ) if 1 < x < ; elsewhere. (a). Calculate the value of k, (b). Find probability distribution function of X. Solution: (a). Because f(x) is probability density function, so and + f(x)dx 1 + f(x)dx 1, k 1 x 2 dx k arcsinx 1 1, Hence k 1/π. (b). Let F (x) is the probability distribution of X, then, F (x) x f(y)dy, if x 1, x 1 1 π 1 x dx 1 (arcsinx + π), if x [ 1, ], 2 π 1, if x. Q2. Let X denote the lifetime of a radio, in years, manufactured by a certain company. The density function of X is given by: f(x) 1 15 e x 15 if x ; elsewhere. What is the probability that, of eight such radios, at least four last more than 15 years? 9

Solution: (a). Let F (x) is the probability distribution of the lifetime of a radio, then 15 1 y F (15) 15 dy 1 1/e, 15 e By similar calculation, we can show that P (X 15) 1 F (15) 1/e, denote p P (X 15), q P (X 15). Hence, P (probability of eight such radios, at least four last more than 15 years) C 4 8p 4 q 4 + C 5 8p 5 q 3 + C 6 8p 6 q 2 + C 7 8p 7 q + C 8 8p 8.33534. Q3. First a point Y is selected at random from the interval (,1). Then another point X is selected at random from the interval (Y,1). Find the probability density function of X. Solution: Let f(x) is the density function of X, G(y) the density function of Y and h(x, y) their joint density function. Clearly, g(y) 1 when y (, 1) and equals otherwise. Now we have following result: h(x, y) f(x y)g(y) f(x y) 1, 1 y if y < x < 1, y (, 1),, otherwise So, f(x) is the marginal distribution of X can be calculated as following: f(x) h(x, y)dy x 1 dy ln(1 x). 1 y Therefore, f(x) ln(1 x), if < x < 1, otherwise. Q4. Let (X, Y ) be a random point from a unit disk centered at the origin. Find P ( X 4/11 Y 4/5). 1 Solution: Obviously, f(x y) 2 1 y, so 2 P ( X 4/11 Y 4/5) 4 11 1 dx 2 1 ( 45 )2 1/33. 1

Q5. Stores A and B, which belong to the same owner, are located in two different towns. If the probability density function of the weekly profit of each store, in thousands of dollars, is given by f(x) x/4 if 1 < x < 3, otherwise. and the profit of one store is independent of the other, what is the probability that next week one store makes at least $ 5 more than the other store? Solution: Let F (x) is the probability distribution of the profit of each store, then F (x) x t dt 1 1 4 8 (x2 1). Hence, P (X Y >.5) 3 3 1.5+y 2.5 3 f(x y)f(y)dxdy x y 1.5+y 4 4 dxdy.175781. Q6. A random variable Y with distribution function F (y) y/5 for < y 5. Determine the conditional expectation of Y, E[Y Y > x] given that Y > x and < x 5. Answer: Since given < x 5, calculate it directly: E[Y Y > x] 5 5 5 yp (Y y Y > x)dy yp (Y y, Y > x) dy P (Y > x) y/5 x 1 x/5 dy (5 + x)/2. Q7. In a given lottery, players pick six different integers between 1 and 49, the order of selection being irrelevant. The lottery commission then select six of these numbers at random as winning numbers. A player wins the grand prize of $1,2, if all six numbers that he has selected match the winning numbers. He wins the second and third prizes of $8 and $35, respectively, if exactly five and four of his selected numbers match the winning numbers. What is the expected value of the amount a player wins in one game? 11

Answer: Let X is the random variable of the amount of a player wins in one game, Now we want to know it s expect value. Denoted by A, B and C the events that the player wins grand prize, second prize and third prize, respectively. Then E[X] 12P (A) + 8P (B) + 35P (C) 1 12 + 8 C5 6C43 1 + 35 C4 6C43 2 C49 6 C49 6 C49 6.134475. 12

The solutions of first quiz of Math144 T eaching Assistant : Liu Zhi Date : F riday, M arch 16, 27 Q1. Following are the informations that we can obtain from problem, Age group Frequency Relative frequency Cumulative frequence 2 24 128.12.12 25 29 28.2.32 3 34 194.18.5 35 39 297.28.78 4 44 24.22 1 So, IQR 75 th percentile 25 th percentile.75.5.25.12 35 + (39 35) [25 + (29 25).78.5.32.12 ] 1.83. Q2. Let a is the number that we want to find, then a C 3 5 2 3 C 1 2 ( 3) C 1 1 4 192. Q3. By the basic property of density function: So k 2 3 e1/3. 1 k + + + + 1 1 + 1 k( 3 2 e 1/3 ). x 3 dx f(x, y)dxdy kx 3 e y/3 dxdy + 1 e y/3 dy 13

Q4. Use the law of probability in continuous case: P (Y < X) 1 3. + + 1 x 1 x P (Y < X X x)f(x)dx P (Y < x)f(x)dx f(y)dyf(x)dx 2ydydx Q5. According to the definition of conditional probability, P (X > 1/2 Y > 1/2) P (X > 1/2, Y > 1/2) P (Y > 1/2) 1 2 1/2 2 1/2 43/52. 1/2 (x2 + xy )dydx 3 1 (x2 + xy )dxdy 3 14

The problems of fifth TA class of Math144 T eaching Assistant : Liu Zhi Date : F riday, M arch 23, 27 Q1. Let the probability density function of a random variable X be given by find E(X 2 + X). Solution: EX 2 f(x) x 1, if x 2. otherwise. EX 1 1 x 2 (1 x)dx + (x 2 x 3 )dx + 2 1 2 1 x 2 (x 1)dx (x 3 x 2 )dx 1/3 1/4 + 4 1/4 8/3 + 1/3 3/2. 1 1 x(1 x)dx + (x x 2 )dx + 2 1 2 1/2 1/3 + 7/3 3/2 1, So, E(X 2 + X) 1 + 3/2 5/2. 1 x(x 1)dx (x 2 x)dx Q2. Let the joint probability density function of random variables X and Y be f(x, y) 2e (x+2y) if x, y. otherwise. Find E(X) and E(Y ), and E(X 2 + Y 2 ). 15

Solution: The marginal density functions of X and Y are: f(x) f(y) hence, E(X) E(Y ) 1, E(X 2 + Y 2 ) 1/2, 2 + 1/2 5/2. 2e (x+2y) dy e x 2e (x+2y) dx 2e 2y, xf(x)dx xe x dx yf(y)dy 2ye 2y dy x 2 f(x)dx + x 2 e x dx + y 2 f(y)dy 2y 2 e 2y dx Q3. Consider repeated sampling with replacement from the box containing five red balls, three white balls, and two blue balls. Let the sampling continue until blue balls have shown five times. Determine the probability that, by then, red balls will have shown exactly eight times and white balls will have shown six times. Soluction:The samling is repeated with replacement, because it will continue until blue balls appear five time, this imply last sample must be a blue ball, let p 1, p 2, p 3 are the probabilities of a blue ball, a red ball and white ball appears at a sampling, respectively. Then p 1 1/5, p 2 1/2, p 3 3/1 and the probability which we want to know is: P P (X 1 4, X 2 8, X 3 6)p 1 18! 4!8!6! p 1 4 p 2 8 p 36 p 1.8. Q4. In a drawer are 4 black socks, 6 gray socks, and 1 white socks. One 16

reaches in and randomly grabs 3 socks. Compute the probability of a matching pair. Soluction: Use the Hypergeometric distribution, P (the matching pair) P (a pair of black socks appear) + P (a pair of gray socks appear) +P (a pair of white socks appear) [ C2 4C 1 16 C 3 2.7895. + 4 3 2 2 19 18 ] + 6C 1 [C2 14 + 6 5 4 C2 3 2 19 18 ] + 1C 1 [C2 1 + 1 9 8 C2 3 2 19 18 ] Q5. A father asks his sons to cut their backyard lawn. Since he does not specify which of the three sons is to do the job, each boy tosses a coin to determine the odd person, who must then cut the lawn. In the case that all three get heads or tails, they continue tossing until they reach a decision. Let p be the probability of heads and q 1 p, the probability of tails. (a). Find the probability that they reach a decision in less than n tosses. (b). If p 1/2, what is the minimum number of tosses required to teach a decision with probability.95? Solution: (a). Let p k is the probability of they reach a decision at kth tossing, then this is a geometric distribution, p k P k 1 (all three get heads or tails) P (one s result is different of other s) (p 3 + q 3 ) k 1 (C 1 3pq 2 + C 1 3qp 2 ), Hence, P (reach a decision in less than n tosses) n 1 p k k1 (3pq 2 + 3qp 2 ) 1 (p3 + q 3 ) n 1 1 (p 3 + q 3 ) 1 (p 3 + q 3 ) n 1. (b). Since p 1/2, we just need to know what is minimum integer which satifying 1 ((1/2) 3 +(1/2) 3 ) n 1.95, this implies that n 4, therefore, 4 is minimum number of tosses required to teach a decision with probability.95. 17

Q6. If the average number of claims handled daily by an insurance company is 5, what proportion of days have less than 3 claims? What is the probability that there will be 4 claims in exactly 3 of the next 5 days? Assume that the number of claims on different days is independent, and the company insures a large number of clients, each having a small probability of making a claim on any given day. Solution:According the assumptions of problem, we can see the claims process as a poisson process with with rate λ 5, So P (N < 3) P (N ) + P (N 1) + P (N 2) e 5 + 5e 5 + 52 e 5 2!.125. Since the number of claims of different days is independent, then P (there will be 4 claims in exactly 3 of the next 5 days) C 3 5[P (N 4)] 3 [1 P (N 4)] 2 1 ( 54 e 5 ) 3 (1 54 e 5 4! 4!.37. Q7. Suppose that a Scottish soldiers chest size is normally distributed with mean 39.8 and standard deviation 2.5 inches, respectively. What is the probability that of 2 randomly selected Scottish soldiers, exactly five have a chest of at least 4 inches? ) 2 Solution: Let X is the random variable of soldiers chest size, and Z is the standard normal random variable, then P (five have a chest of at least 4 inches of 2 soldiers) C2[P 5 (X 4)] 5 [P (X < 4)] 15 C2[P 5 ( X 39.8 4 39.8 )] 5 [P ( X 39.8 < 2.5 2.5 2.5 C2[P 5 (Z.98)] 5 [P (Z <.98)] 15 C 5 2.46 5.54 15.3. 4 39.8 ] 15 ) 2.5 Q8: According to electrical circuit theory, the voltage drop across a resistor is related to the current flowing through the resistor by the equation VIR, where R is the resistance level measured in ohms, I is the current in amperes, and V is the voltage in volts. Suppose that the measured voltage in a certain electrical circuit has a normal distribution 18

with mean 12 and standard deviation 2 and five measurements of the voltage are taken. Determine the probability that exactly two of the measurements lie outside the range 118-122. Solution: Let Z is the standard normal random variable, then P (two of the five measurements lie outside the range 118-122) C5[P 2 (X < 118 or X > 122)] 2 [P (118 X 122)] 3 C5[P 2 ( X 12 < 1 or X 12 > 1] 2 [P ( 1 X 12 1)] 3 2 2 2 1[P (Z < 1 or Z > 1)] 2 [P ( 1 Z 1)] 3 1 (.3174) 2 (.6826) 3.32. Q9. Suppose that of all the clouds that are seeded with silver iodide, 58% show splendid growth. If 6 clouds are seeded with silver iodide, (a). What is the probability that exactly 35 show splendid growth? (b). What is the probability that at least 35 show splendid growth? (c). What is the probability that the number of splendid growth in [35,4]? (make the continuity correction of.5) Solution: p P (splendid growth).58, hence, (a). P (exactly 35 show splendid growth of 6 clouds) C6p 35 35 (1 p) 25 P ( x.5 nq np(1 p) Z x +.5 np np(1 p) ) 35.5 6.58 P ( Z 6.58.42.1. (b). P (X 35) 6 k35 P (Z C k 6p k (1 p) 6 k 35.5 nq np(1 p) ) 35 +.5 6.58 6.58.42 ) P (Z 35.5 6.58 6.58.42 ).5313. 19

(c). P (35 X 4) 4 k35 P (.4633. C k 6p k (1 p) 6 k 35.5 6.58 6.58.42 Z 4 +.5 6.58 6.58.42 ) 2

The problems of sixth TA class of Math144 T eaching Assistant : Liu Zhi Date : F riday, M arch 23, 27 Q1. A statistician is trying to determine a model for claim size from historical data on a particular group of policies for an insurance company. The statistician observes that by considering the cube of the claim sizes, one obtains an exponential distribution with mean 2. Determine the probability that a future claim on a policy similar to the ones being studied will exceed 1.5. Solution: Because X 3 have a exponential distribution with parameter λ 1 2, then (X3 /α) β /λ Weibull(α, β), let α 2 3, β 1/3, we know X Weibull(2 3, 1/3), therefore, the probability of it exceed 1.5 is: 1.5 ( P (X > 1.5) 1 F (1.5) e 2 3 )1/3.5642. Q2. A particular high-technology stock whose current price is $1 either increase 5% or decrease 3% in any given day. There is a 5% chance that the stock price will increase on any given day independent of price movements on previous days. Determine the probability that the stocks price is more than five times its current price 1 trading days from now. Solution: Because 1(1 +.5) k (1.3) 1 k 5 k 59, this show that if the stock price is more than five times its current price after 1 trading days, then it is increasing at least 59 days of the 1 days. Hence P (The stocks price is more than five times its current price after 1 trading days) C1(1/2) k 1 59 k 1 P (Z P (Z 1.7).445655. 59.5 5 1 (1/2) 2 ) Q3. A professor takes early retirement on July 1, exactly 3 months after his 6 th birthday, and purchases a term life insurance policy for $25. The maturity date of the policy coincides with his 65 th birthday. According to a mortality table for similarly situated men, the probability that a male of exact age 6 dies before reaching his 61 th birthday 21

is.13764. Determine the probability that the professor dies before reaching his 61 th birthday if deaths are assumed to be exponentially distributed between ages. Solution: Because assumed the deaths to be exponentially distributed, i.e, let X is the lifetime of a male of exact age 6, then X exp(λ), now, we need to calculate the value of λ, since P (X < 61 X > 6) P (X < 1) 1 e λ 1.13764 λ ln.14839, 1.13764 therefore: P (X <.75) 1 e.14839.75.1587. 22

The problems of seventh TA class of Math144 T eaching Assistant : Liu Zhi Date : F riday, April 13, 27 Q1. Let X and Y be the coordinates of a random point selected uniformly from the unit disk {(x, y) : x 2 + y 2 1}. Are X and Y independent? Are they uncorrelated? Why or why not? Solution: The marginal density function of X is: f(x) 1 π 1 π + + 1 f(x, y)dy 1 π 1(x2 + y 2 1)dy 1 1 x 2 1( 1 x 2 y 1 x 2 )dy 1 x 2 dy 2 1 x 2, π Similarly, the density function of Y is g(y) 2 P (X x, Y y) 1 π y x 1 1 y x f(t, u)dtdu 1 1 y x 1 u 2 1 dtdu π 1 1 u 2 / P (X x)p (Y y). 1 y 2 π. So, 1( 1 u 2 t 1 u 2 )dtdu This show that X and Y are not independent. And because Cov(X, Y ) E[XY ] E[X]E[Y ] 1 π 1 π 1 1 1 1 1 1 xy1( 1 y 2 x + 1 y 2 )dxdy E[X]E[Y ] 1 y 2 xydxdy E[X]E[Y ] 1 y 2 23

1 π 1 1 [ 2 π. y( 1 2 x2 1 1 1 y 2 1 y Hence, X and Y are uncorrelated. x 1 x 2 dx] 2 2)dy E[X]E[Y ] Q2. A right triangle has hypotenuse of length 9. If the probability density function of one side s length is given by: f(x) x, 6 if 2 < x < 4;, elsewhere. What is the expected value of the length of the other side? Solution: BecauseX 2 +Y 2 9 2, and X has the density function f(x). Then, E[Y ] E[ 81 X 2 ] 4 81 x2 x 2 6 dx 8.42. Q3. An urn consists of 4 red balls and 6 green balls. What is the probability of getting exactly k red balls in a sample size of 1 if the sampling is done (a) with replacement. (b) without replacement. Solution: Let X is the exactly numbers of red calls in a sample size of 1. (a). Because sampling is done with replacement, So P (X k) C k 1( 4 1 )k ( 6 1 )1 k. (b). If the sampling is done without replacement, then P (X k) Ck 4C6 1 k. C1 Q4. The cdf of χ 2 v is given by, for degrees of freedom v, F v (x) P (χ 2 v x) x Show that, for v > 2, F v (x) F v 2 (x) 2f v (x). 24 f v (u)du, for x >

Solution: Use the property of Γ function: Γ(u + 1) uγ(u), then when v > 2, x (1/2) v/2 F v (x) Γ( v) u v 2 1 e u/2 du 2 2 x (1/2) v/2 Γ( v 2 ) u v 2 1 de u/2 2 (1/2)v/2 Γ( v) x v 2 1 e x/2 + 2 2 x x (1/2) (v 2)/2 1/2 2f v (x) + 2 Γ( v 2) v 2 ( v 2 2 2 F v 2 (x) 2f v (x). (1/2) v/2 Γ( v 2 ) e u/2 du v 2 1 1)u v 2 2 1 e u/2 du Now, you can download the notes from http : //ihome.ust.hk/ liuzhi/math144.html 25

The solutions of midterm examination T eaching Assistant : Liu Zhi Date : F riday, April 2, 27 Q1. What are the mean, median, and mode, respectively, for the density function f(x) 1 for < x <? π(1+x 2 ) (A).,, (B). none exist (C). not exist, not exist, (D). not exist,, not exist (E).not exist,,. Solution: Let X is a random variable with the density function, it is easy to verify that it s mean doesn t exist, and because it density function is symmetric about, so both the median and mode are. Q2. What is the standard deviation of the distribution: f(x) Cx 4 C6 x 6, x C1 6, 1, 2, 3, 4? (A). 8/3 (B). 1/3 (C). 4/5 (D). 16/81 (E).81/25 Solution: Calculate it directly, or by the mean of hypergeometric distrubtion. Q3. Let X and Y be continuous random variables with joint density function: f(x, y) 1, for < y < 1 x and 1 < x < 1;, elsewhere. What is the variance of X? (A). 1/18 (B). 1/6 (C). 2/9 (D). 11/18 (E). 2/3 Solution: The density funtion of X is: So, EX f(x) 1 1 1 1 x dx 1 x, 1 < x < 1. xf(x)dx x(1 + x)dx + 1 x(1 x)dx 1 2 x2 1 + 1 3 x3 1 + 1 2 x2 1 + 1 3 x3 1, 26

EX 2 1 x 2 (1 + x)dx + 1 1/3 1/4 + 1/3 1/4 1/6. var(x) EX 2 (EX) 2 1/6. x 2 (1 x)dx Q4. Let X and Y be continuous random variables with joint density function: f(x) 3 (2 x y), 4 for < x < 2, < y < 2, and x + y < 2,, elsewhere. What is the conditional probability P (X < 1 Y < 1)? (A). 1/2 (B). 3/4 (C). 49/64 (D). 6/7 (E).7/8 Solution: The density function of Y is f(y) 3 4 2 3 (2 x y)1(x + y < 2)dx 4 2 y (2 x y)dx 3 4 [4 2y 1 2 x2 2 y y(2 y)] 3 3 2 y 3 8 (2 y)2 3 2 y + 3 4 y2 3 8 y2 3 2 y + 3 2. So: P (Y < 1) P (X < 1, Y < 1) 1 7/8. 1 1 6/8, ( 3 8 y2 3 2 y + 3 2 )dy 3 (2 x y)1(x + y < 2)dxdy 4 Hence, P (X < 1 Y < 1) P (X < 1, Y < 1) P (Y < 1) 6/7. 27

Q5. Let X and Y be independent random variables with binomial distributions Bin(k; n, a) and Bin(k; n, b) respectively. Let Z X + 2Y. Then for all ɛ >, Tchebycheff s inequality guarantees that P ( Z na 2nb ɛ) is always less than or equal to what? (A). (na(1 a) + nb(1 b))/ɛ 2. (B). (na(1 a) + 2nb(1 b))/ɛ 2. (C). (na(1-a)+4nb(1-b))/ɛ 2. (D). (na(1 a) + 2nb(1 b))/ɛ 3. (E). (na(1 a) + 4nb(1 b))/ɛ 3. Solution: Recalled that the Tchebycheff s inequality is: P ( X EX ɛ) V ar(x) ɛ 2. Because V ar(z) V ar(x) + V ar(2y ) na(1 a) + 4nb(1 b), So P ( Z na 2nb ɛ) (na(1 a) + 4nb(1 b))/ɛ 2. Q6. Simplify the expression of N kn Cn k. Solution: Because: So, N kn C n k Ck n k! n!(k n)! k!(k + 1 k + n) (n + 1)!(k n)! (k + 1)! (n + 1)!(k n)! k! (n + 1)!(k n 1)! C n+1 k+1 Cn+1 k, k n + 1, n + 1,. Cn n + Cn+1 n + + CN n 1 + Cn+1+1 Cn+1 n+1 + Cn+2+1 n+1 Cn+2 n+1 + + C n+1 N 1+1 Cn+1 N 1 + Cn+1 N+1 Cn+1 N C n+1 N+1. Q7. A man invites his fiancée to a fine hotel for a Sunday brunch. They decide to meet in the lobby of the hotel between 11 : 3 am. and 12 : moon. If they arrive at random times during this period, what is the probability that they will meet within 1 minutes? 28

Solution: Let X and Y is the time of the man and his fiancée arrive at the hotel, start from 11 : 3, then X and Y are independent and have identical density function: f(x) 1, 3 if < x < 3,, elsewhere. Then, P ( X Y 1) P ( 1 X Y 1) + 3 1 3 1 3 1 9 [ 1 9 [ P ( 1 X Y 1 Y y)f(y)dy P ( 1 X Y 1 Y y)f(y)dy 3 P (y 1 X 1 + y Y y)dy 3 3 (y+1) (y 1) 1 y+1 1 dy + 1 3 dy (y + 1)dy + 2 y+1 1 y 1 2 1 1 [1 + 5 + 6 25] 9 5 9. 2dy + dy + 3 3 2 3 2 y 1 dy] (4 y)dy] Q8. Two subcontractors have been hired to build different parts of an experimental aircraft. The expected completion times for the two subcontractors are 1 months and 12 months, respectively. Both parts are necessary before final assembly of the aircraft can take place. How long should the primary contractor expect to wait to complete final assembly if at least one of the subcontractors is expected to be finished in 6 months? Solution: Because E[Min(X,Y)]6, so E[Max(X,Y)]E[X]+E[Y]-E[Min(X,Y)]16. Q9. Calculate the expected number of aces in a randomly selected five cards from a pool of 52 card that is found to have exactly twol jacks. 29

Solution: Let X is the number of aces in five cards and the set A {It have exactly two jacks in these 5 cards}. Then E[X A] P (X A) + 1 P (X 1 A) + 2 P (X 2 A) + 3 P (X 3 A) P (A, X 1) 2P (A, X 2) 3P (A, X 3) + +, P (A) P (A) P (A) We calculate the needed probability as following: P (A) C3 48C 2 4 C 5 52 P (A, X 1) C2 44C 2 4C 1 4 C 5 52 P (A, X 2) C1 44C 2 4C 2 4 C 5 52 P (A, X 3) C3 4C 2 4 C 5 52,,,, Therefore, E[X A] 1 [C44C 2 4C 2 4 1 + 2C44C 1 4C 2 4 2 + 3C4C 3 4] 2 1 4. C 3 48C 2 4 Q1. (a.) Given X is a normal variable with mean equals 2 and variance equals 4, compute P (X 2 X < 6). Solution: X N(2, 4) implies X 2 2 N(, 1). Now we can compute the probability: P (X 2 X < 6) P ((X + 2)(X 3) < ) P ( 2 < X < 3) P ( 2 2 < X 2 2 2.6687. < 3 2 2 ) (b.) Given X is a normal random variable with mean equals 1 and variance equals 2; Y is a normal random variable with mean equals 3 and variance equals 4. Both X and Y are independent of each other. Compute P (2X + 3Y > 9). 3

Solution: Because X N(1, 2), Y N(3, 4), then 2X N(2, 8), Y N(9, 36), also since X and Y are independent, hence 2X + 3Y N(11, 44). and let Z is the standard normal variable, Then P (2X + 3Y > 9) P (Z > 1 11 ) P (Z <.3).6179. 31

The problems of eighth TA class of Math144 T eaching Assistant : Liu Zhi Date : F riday, April 27, 27 Q1. Civil engineers believe that W, the amount of weight(in units of 1, pounds) that a certain span of a bridge can withstand without structural damage resulting is normally distributed with mean 4 and standard deviation 4. Suppose that the weight of a car is a random variable with mean 3 and standard deviation.3. How many cars would have to be on the bridge span for the probability of structural damage to exceed.1? Solution: Let X is random variable of weight that a certain span of a bridge can withstand without structural damage resulting(in units of 1, pounds), X N(4, 4 2 ) and Y i is the weigth of a car, then by CLT, n i1 Y i N(3n,.9n), this imply that n i1 Y i X N(3n 4, 16 +.9n), therefore the probability of structural damage when n cars on the bridge span: n p P ( Y i > X) P ( i1 n Y i X > ) i1 P ( ( n i1 Y i X) (3n 4) 16 +.9n > P (Z > 4 3n 16 +.9n ), (3n 4) 16 +.9n ) p >.1 imply n 16 by check the standard normal distribution table. Q2. If T has a t distibution with 8 degrees of freedom, find (a). P (T 1), (b). P (T 2), (c). P ( 1 < T < 1). Solution: (a). Check the t distibution table, we find P (t 8.889).2, P (t 8 1.18).15,then P (T 1) 1.889.2 + (.15.2) 1.18.889.174658. 32

(b). Similarly, because P (t 8 1.86).5, P (t 8 2.36).25, so P (T 2) 1 P (T > 2) 2 1.86 1 [.5 + (.25.5) 2.36 1.86 ].957848. (c). P ( 1 < T < 1) P (T < 1) P (T < 1) 1 P (T 1) P (T 1) 1 2.174658.65684. Q3. (a). Find a 95% confident interval for the mean of a normal distribution with unknown variance based fon the 2 samples data: (7, 9, 3, 2, 3, 8, 4, 6, 2, 6, 4, 3, 8, 3, 2, 7, 9, 5, 8, 8). (b). Find the confidence coefficient (1-α) of the t interval (3.738, 6.962) about population mean (µ). Solution: (a). Because the variance is unknown, we should use the t estimation: I. Two tails: x 1 2 2 i1 x 1 i 5.35, s ( 2 19 i1 x2 i 2 x2 ) 2.5188. So, in this case, the confident interval of mean(µ) with confident coefficient.95 is [5.35 t.5/2,19 s s 2, 5.35 + t.5/2,19 2 ] [5.35 2.93 2.5188 2, 5.35 + 2.93 2.5188 2 ] [3.71497, 6.9853]. II. One tail: the upper one-sided bound is: s x + t.5,19 5.35 + 2 and the lower one-sided bound is: s x t.5,19 5.35 2 1.729 2.5188 2 6.32381, 1.729 2.5188 2 4.37619. s (b). 5.35 + t α/2,19 2 6.962 t α/2,19 2.86211 α.1. Q4. Find a 95% confident interval for µ 1 µ 2 with x 1 1, x 2 9, s 2 1 9, s 2 2 4, and n 1 n 2 1 when sampling from two independent normal distribution. 33

Solution: Case I: Suppose that two normal distributions have same 2.55, then the 95% confident in- variances, s p terval for µ 1 µ 2 is: (n 1 1)s 2 1 +(n 2 1)s 2 2 n 1 +n 2 2 [ x 1 x 2 t α/2,n1 +n 2 s p 1 n 1 + 1 n 2, x 1 x 2 + t α/2,n1 +n 2 s p 1 n 1 + 1 n 2 ] [.293178, 1.7682]. Case II. When two normal distributions have different variances, v ( s2 1 n 1 + s2 2 n 2 ) 2 ( s2 1 n ) 2 1 n 1 1 + ( s 2 2 n ) 2 2 n 2 1 µ 1 µ 2 is: 172. then in this case, the 95% confident interval for [ x 1 x 2 t α/2,v [.293312, 1.7669]. s 1 1 + s2 2 s 1 1, x 1 x 2 + t α/2,v + s2 2 ] n 1 n 2 n 1 n 2 Q5. Find a 95% confident interval for µ 1 µ 2 with x 1 1, x 2 9, s 2 1 9, s 2 2 4, and n 1 n 2 1 in sampling from two independent normal distributions with same but unknown variances. Solution: We can obtain the answer by change the n 1, n 2 of Q4 from 1,1 to 1,1, respectively. Then s p 2.55, same as Q4. So the 95% confident interval for µ 1 µ 2 is [.671124, 1.93289]. 34

The problems of ninth TA class of Math144 T eaching Assistant : Liu Zhi Date : F riday, M ay 4, 27 Q1. Let us define S 2 1 n n i1 (X i X) 2. Show that E(S 2 ) [(n 1)/n]σ 2, and hence S 2 is a biased estimator for σ 2. Solution: Because S 2 1 n n i1 (X i X) 2 1 n n i1 X2 i X 2, so E(S 2 ) E[ 1 n Xi 2 ] E[ n X 2 ] 1 n 1 n i1 n E[Xi 2 ] E( 1 n X i ) 2 n i1 n E[Xi 2 ] 1 n E[X 2 n 2 i ] 2 E[X n 2 i X j ] i1 i1 i1 n 1 n E[X2 ] n 1 n [EX]2 n 1 n (E[X2 ] [EX] 2 ) n 1 n σ2. This is required result and hence, S 2 is a biased estimator for σ 2. Q2. If X is a binomial random variable, show that (a). ˆp X/n is an unbiased estimator of p; (b). p X+ n/2 n+ is a biased estimator of p; (c). Show that the estimator p becomes unbiased as n n. Solution: (a). E[ˆp] E[X]/n np/n p. This show that ˆp is an unbiased estimator of p. (b). E[p ] E[X]+ n/2 n+ np+ n/2 n n+ /p, so p is a biased estimator of p. n np + n/2 (c). lim n n + n p lim n p. 1 + n n + lim n 1 2(1 + n) 35 i<j

so, the required result follows. Q3. An electrical firm manufactures light bulbs that have a length of life that is approximately normally distributed with a standard deviation of 4 hours. If a sample of 3 bulbs has an average life of 78 hours, (a). Find a 96% confidence interval for population mean of all bulbs produced by this firm; (b). How large a sample is needed if we wish to be 96% confident that our sample mean will be within 1 hours of the true mean? Solution: The sample mean x is 78, and the standard deviation of population is 4, using the Normal table, we can get z.2 2.5. Hence the 96% confidence interval for µ is: [ x z.2 ( σ 3 ), x + z.2 ( σ 3 )] [765.29, 794.971]. (b). We can be 96% confident that this error will less than t α/2 ( s n ), i.e, n ( z.2 4 ) 2 67.24 e Therefore, a sample with size 68 is needed if we wish be 96% confident that our sample mean will be within 1 hours of the true mean. Q4. Many cardiac patients wear implanted pacemakers to control their hearbeat. A plastic connector module mounts on the top of the pacemaker. Assuming a standard deviation of.15 and an approximate normal distribution, (a). Find a 95% confidence interval for the mean of all connector modules made by a certain manufacturing company. A random sample of 75 modules has an average of.31 inch. (b). How large a sample is needed if we wish to be 95% confident that our sample mean will be within.5 inch of true mean? Solution: (a).from the problem, we know: n 75, x.31, σ.15, hence the 95% confidence interval for the mean of all modules is: [.31 z.25 (.15 75 ),.31 + z.25 (.15 75 )] [.39661,.31339]. (b). n ( z.25.15 ) 2 35.8225..5 36

Therefore, a sample with size 38 is needed if we wish to be 95% confident that our sample mean will be within.5 inch of true mean. Q5. The heights of a random sample of 5 college students showed a mean of 174.5 centimeters and a standard deviation of 6.9 centimeters. (a). Construct a 98% confidence interval for the mean height of all college students. (b). What can we assert with 98% confidence about the possible size of our error it we estimate the mean height of all college students to be 174.5 centimeters? Solution: Because standard deviation of the population is unknown, so we should use the t distribution: (a). s 6.9, x 174.5, n 5, so 98% confidence interval for the mean height of all college students is: [174.5 t.1,49 ( 6.9 s ), 174.5 + t.1,49 ( )] [173.229, 175.771]. 5 5 (b). We can be 98% confident that this error will less than t α/2 ( s n ), i.e, e t.1,49 ( 6.9 ) 1.27148. 5 Q6. A random sample of 1 automobile owners shows that, in the state of Virginia, an automobile is driven on the average 23,5 kilometers per year with a standard deviation of 39 kilometers. Assume the distribution of measurements to be approximately normal. (a). Construct a 99% confidence interval for the average number of kilometers an automobile is driven annually in Virginia. (b). What can we assert with 99% confidence about the possible size of our error if we estimate the average number of kilometers dirven by car owners in Virginia to be 23,5 kilometers per year? Solution: We use the t distribution, x 23, 5, s 39, n 1, then the 99% confidence interval for population mean µ is: [235 t.5,99 ( 39 1 ), 235 + t.5,99 ( 39 1 )] [22479.4, 2452.6]. (b). e t.5,99 ( 39 1 ) 12.63. 37

Q7. An efficiency expert wishes to determine the average time that it takes to drill three holes in a certain metal clamp. How large a sample will he need to be 95% confident that his sample mean will be within 15 seconds of the true mean? Assume that it is known from previous studies that σ 4 seconds. Solution: The population standard deviation is σ 4, Then by Theorem 9.2 of textbook, n ( (1.96)(4) ) 2 27.318, 15 Therefore, a sample with size 28 be needed to be 95% confident that his sample mean will be within 15 seconds of the true mean. 38

The problems of Tenth TA class of Math144 T eaching Assistant : Liu Zhi Date : F riday, M ay 11, 27 Q1. A manufacturer of car batteries claims that his batteries will last, on average, 3 years with a variance of 1 year. If 5 of this batteries have lifetimes of 1.9, 2.4, 3., 3.5 and 4.2 years, construct a 95% confidence interval for σ 2 and decide if the manufacturer s claim that σ 2 1 is valid. Assume the population of battery lives to be approximately normally distributed. Solution: Because P ( (n 1)S2 χ 2 α/2 < σ 2 < (n 1)S2 ) 1 α, n 5, α.5, χ 2 1 α/2 S 2 1 n 1 n (x i x) 2 n n i1 x2 i ( n i1 x i) 2 n(n 1) i1 (5)(4826) 225 2.815, v 4, χ 2.25 11.143, χ 2 1.25.484, then the 95% confidence interval for σ 2 is: [ (n 1)S2, χ 2 α/2 (n 1)S2 ] [.29256, 6.73554], χ 2 1 α/2 since 1 belong in the interval, so the claim that σ 2 1 is valid. Q2. A random sample of 2 students obtained a mean of x 72 and a variance of S 2 16 on a college placement test in mathematics. Assuming the scores to be normally distributed, contruct a 98% confidence interval for σ 2. Solution: Similar calculation steps as last question, n 2, S 2 16, v 19, χ 2.1 36.191, χ 2 1.1 7.633, so the 98% confidence interval for σ 2 is: [ (n 1)S2, χ 2 α/2 (n 1)S2 ] [8.39988, 39.8271]. χ 2 1 α/2 Q3. A random sample of 64 bags of white cheddar popcorn weighed, on average, 5.23 ounces with a standard deviation of.24 ounces. Test the hypothesis that µ 5.5 ounces against the alternative hypothesis, µ < 5.5 ounces at the.5 level of significance. Solution: 1. H : µ 5.5 ounces. 39

2. H 1 : µ < 5.5 ounces. 3. α.5. 4. Critical region: z < 1.645. 5. Computations: x 5.23 ounces, σ.24 ounces, and z 5.23 5.5.24/ 64 9. 6. Decision: Reject H, accept H 1. Q4. Past experience indicated that the time for high school seniors to complete a standardized test is a normal random variable with a mean of 35 minutes. If a random sample of 2 high school seniors took an average of 33.1 minutes to complete this test with a standard deviation of 4.3 minutes, test the hypothesis at the.5 level of significance that µ 35 minutes against the alternative that µ < 35 minutes. Solution: 1. H : µ 35 minutes. 2. H 1 : µ < 35 minutes. 3. α.5. 4. Critical region: z < 1.645. 5. Computations: x 33.1 minutes, σ 4.3 minutes, and z 33.1 35 4.3/ 2 1.9766. 6. Decision: Reject H, accept H 1. 4