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**Unformatted text preview: **Statistics 265 Elements of Probability Theory Spring Session 2000 Assignment 1
Due: Thursday May 11, 2000 Solutions 1.5 The following results on summations will help us in calculating the sample variance s : For any constant c; n P a. c = nc:
2 b. c. i n P n P =1 i P xi + yi : i i i Use (a), (b), and (c) to show that
=1 =1 cyi = c n P (xi + yi ) = i =1 yi : n P n =1 =1 n X s = n 1 1 4 yi i
2 = 2 2 n 1 Xy ni i
=1 !2 3 5 : Solution: From the de nition of s we have n X s = n 1 1 (yi y) i "n n n 1 X y 2y X y + X y =n 1 i i i# i i "n X = n1 1 yi ny i
2 2 2 =1 2 =1 =1 =1 2 2 #
2 = n1 1 yi i
= 2 n X 4 =1 2 n 1 Xy ni i
=1 !2 3 5 : 1.14 Prove that the sum of the deviations of a set of measurements about their mean is equal to zero; that is
n X i
Solution:
=1 (yi y ) = 0:
n X n X n X i
=1 From the de nition of the mean we have
n X i
=1 (yi y ) = = = i yi n X i
=1 y =1 i yi ny yi
n X i
=1 =1 yi =0 1.24 Let k 1: Show that, for any set of n measurements, the fraction included in the interval y ks to y + ks is at least (1 1=k ): Hint: ! n 1 X(y y ) : s =n 1 i
2 2 2 =1 In this expression, replace all deviations for which jyi y j ks with ks. Simplify.] This result is known as Tchebyshe 's theorem. Solution: Let A = fi : jyi y j ks g and B = fi : jyi y j < ksg; then
X X (yi y ) + (yi y ) s =n1 1 i2A i2B 1 X(y y ) n 1 i2A i 1 Xk s n 1 i2A = n 1 1 jAj k s ; where jAj is the cardinality of A; that is, number of indices in A; that is, the number of measurements outside the interval jyi y j < ks: Since s 6= 0 (s = 0 if and only if all the yi 's are equal), then we have
2 2 2 2 2 2 2 2 " # 1 1 n 1 1 k jAj > n k jAj:
2 2 Therefore, which is what we wanted to show. n jAj = 1 jAj 1 1 ; n n k
2 2.4 Suppose two dice are tossed and the numbers on the upper faces are observed. Let S denote the set of all possible pairs that can be observed. These pairs can be listed, for example, by letting (2; 3) denote that a 2 was observed on the rst die and a 3 on the second.] De ne the following subsets of S : A: The number on the second die is even. B: The sum of the two numbers is even. C: At least one number in the pair is odd. List the points in A; C; A \ B; A \ B; A B ; and A \ C: Solution: We have A = f(i; 2j ) : 1 i 6; 1 j C = f(2i; 2j ) : 1 i 3; 1 j A \ B = f(2i; 2j ) : 1 i 3; 1 j A \ B = f(2i 1; 2j ) : 1 i 3; 1 A B = A \ B = f(2i 1; 2j ) : 1 A \ C = f(i; 2j 1) : 1 i 6; 1 3g 3g 3g j 3g i 3; 1 j 3g f(2i; j ) : 1 i 3; 1 j 6g j 3g 2 2.8 A vehicle arriving at an intersection can turn right, turn left, or continue straight ahead. The experiment consists of observing the movement of a single vehicle through the intersection. a. List the sample space for this experiment. b. Assuming that all sample points are equally likely, nd the probability that the vehicle turns.
Solution: a. The sample space is given by S = f RT; LT; ST g: b. If all sample points are equally likely, then we can assign the probability to each of them. The event A; that the vehicle turns, is A = f RT; LT g; so P (A) = P (fRT g) + P (fLT g) = :
1 3 2 3 2.12 Suppose two balanced coins are tossed and the upper faces observed. a. List the sample points for this experiment. b. Assign a reasonable probability to each point. (Are the points equally likely?) c. Let A denote the event that exactly one head is observed abd B the event that at least one head is observed. List the sample points in A and B: d. From your answer to (c), nd P (A); P (B ); P (A \ B ); P (A B ); and P (A B ):
Solution: a. The sample space is S = fHH; HT; TH; TT g: b. Since the coins are balanced, the outcomes should be equally likely, and we can assign a probability of to each of them. c. If A is the event that exactly one head is observed, then A = fHT; TH g; while if B is the event that at least one head is observed, then B = fHH; HT; TH g: d. From part (c), we have
1 4 P (B ) = P (fHH g) + P (fHT g) + P (fTH g) = 3 4 1 P (A \ B ) = P (fHT g) + P (fTH g) = 2 3 P (A B ) = P (fHH g) + P (fHT g) + P (fTH g) = 4 P (A B ) = P (HH; HT; TH; TT ) = P (S ) = 1 1 P (A) = P (fHT g) + P (fTH g) = 2 = 2 4 3 2.18 A retailer sells only two styles of stereo consoles, and experience shows that these are in equal demand. Four customers in succession come into the store to order stereos. The retailer is interested in their preferences. a. List the possibilities for preference arrangements among the four customers (that is, list the sample space). b. Assign probabilities to the sample points. c. Let A denote the event that all four customers prefer the same style. Find P (A):
Solution: a. The sample space is S = f (x ; x ; x ; x ) : xi = 1 or 2; i = 1; 2; 3; 4 g: b. Since each of the two styles is in equal demand, then each of these outcomes is equally likely, and we can assign the probability 4 = to each of these outcomes. c. Let A be the event that all four customers prefer the same style, then A = f (1; 1; 1; 1); (2; 2; 2; 2) g; and P (A) = = :
1 2 3 4 1 1 2 16 2 1 8 16 2.20 Patients arriving at a hospital outpatient clinic can select one of three stations for service. Suppose that physicians are assigned randomly to the stations and that the patients therefore have no station preference. Three patients arrive at the clinic and their selection of stations is observed. a. List the sample points for the experiment. b. Let A be the event that each station receives a patient. List the sample points in A: c. Make a reasonable assignment of probabilities to the sample points and nd P (A):
Solution: a. Three patients enter the hospital and randomly choose stations 1, 2,or 3 for service. Let the triple (a; b; c) represent the simple event that patients 1, 2, and 3 choose stations a; b; and c; respectively. The sample space is then given by S = f (a; b; c) : a; b; c = 1; 2; 3 g: b. Let A be the event that each station receives a patient, then an outcome (a; b; c) is in A if and only if a; b and c are all distinct, therefore A = f (1; 2; 3); (1; 3; 2); (2; 1; 3); (2; 3; 1); (3; 1; 2); (3; 2; 1) g: c. Since the patients select stations at random, each sample point is equally likely, and is assigned a probability : Therefore, P (A) = = :
1 6 2 9 27 27 2.38 A manufacturer has nine distinct motors in stock, two of which came from a particular supplier. The motors must be divided among three production lines, with three motors going to each line. If the assignment of motors to lines is random, nd the probability that both motors from the particular supplier are assigned to the rst line. Solution: The total number of ways to divide the 9 motors into 3 groups of size 3 is the multinomial coe cient : If the both of the motors from a particular supplier are assigned to the rst line, there are only 7 motors to be assigned , one to line 1 and 3 each to lines 2 and 3. This can be done in ways. Therefore, the probability that both motors from the particular supplier are assigned to the rst line is 7! 1 p = 1!3!3! = 7!3! = 12 : 9! 9! 3!3!3!
9! 3!3!3! 7! 1!3!3! 4 2.44 Show that, for any integer n 1; a. n = 1: Interpret this result. n n = 1: Interpret this result. b. c. n = nn r : Interpret this result. r
0 d. n P i Hint: Consider the binomial expansion of (x + y)n with x = y = 1:]
=0 n = 2n : i a. For an n 1; we have n = n!(nn! n)! = 1: The number of ways to choose n objects from n distinct n objects is 1, the only way to do this is to choose all of the objects. b. For any n 1; we have n = 0!(nn! 0)! = 1: The number of ways to choose 0 objects from n distinct objects is 1, the only way to do this is to choose none of the objects. c. For any n 1; we have
0 Solution: n = n! = n! n r r!(n r)! (n r)!(n (n r))! = n r :
There are as many ways to choose r out of n objects as there are to choose n r out of n objects. Instead of choosing the r objects each time, we simply choose the other n r objects. d. For n 1; the binomial theorem gives (x + y)n = and setting x = y = 1; we have
n X i
=0 n X i
=0 n xiyn i i n = (1 + 1)n = 2n: i
1 2 2.46 Consider the situation where n items are to be partitioned into k < n distinct subsets. The multinomial coe cients n1 n2n nk provide the number of distinct partitions where n items are in group 1, n are in group 2, : : : , nk are in group k: Prove that the total number of distinct partitions equals kn: Hint: Recall Exercise 2.44(d).] Solution: From the multinomial theorem, we have (x + x +
1 2 + xk )n = X n1 + + nk n n n
= 1 n 2 n1 n2 nk x x
1 2 xnk ; k and setting x = x =
1 2 = xk = 1; then we have
X n1 + + nk n n n
= 1 n
2 nk = (1 + 1 + + 1)n = kn : 5 ...

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