MATH 511 Homework Solutions: Expected Values and Probability Distributions | Assignments Mathematics

MATH 511, Meade HW Solutions

3.2 –3 (omit c), 5, 14, 18, **3c, **8 2/13/04

a. f(x) = x/10, x = 1,2,3,4

E(x) = (1) (1/10) + (2) (2/10) + (3) (3/10) + (4) (4/10)

= (1 + 4 + 9 + 16 )/10 = 30/10 = 3

b. f(x) = x/55 , x = 1, 2, 3, 4 ,5 , 6, 7, 8, 9, 10

E(x) = x=1∑10 x2/55 = [(10)(11)(21)/(6)]/55 = 385/55 = 7

**c. f(x) = 3(1/4)x , x = 1, 2, 3, …

E(x) = x=1∑∞ 3x (1/4)x = x=1∑∞ 3x/4x ≤ x=1∑∞ 3x/4x = 3

This assures that the series does indeed converge. The series, x=1∑∞ x/4x ,

can now be approximated. After about ten approximations it is clear that

the series converges to 0.44444… which is 4/9.

So the E(X) = 3(4/9) = (4/3)

d. f(x) = (1/30)(x+1)2 , x = 0, 1, 2, 3

E(x) = (1/30) [ 0 + 4 + 18 + 48] = 70/30 = 7/3

e. f(x) = 2/n(n+1)x , x = 1, 2, 3, …, n

E(x) =2/n(n+1) i=1∑n x2 = [2/n(n+1)] [n(n+1)(2n+1)/6)] = (2n+1)/3

5. E(x) = [2,987,994(0) + 12,000(25) + 4(10,000) + 1(50,000) + 1(200,000)] - 0.50

3,000,000

= 19 ⅔ cents – 50 cents = - 30⅓ cents

**8. f(x) = 6/(x2), x = 1, 2, 3, …

E(x) = (6/) x=1∑∞ x/x2 = (6/) x=1∑∞ 1/x ; this is the harmonic series, which

diverges. Therefore E(x) does not exist. One way to show this is the

integral test for infinite series.

14.

a. [16(25) + 3(100) + 1(300)]/20 = 1,000/20 = 50. Note that this makes sense:

1000 students divided into 20 classes; the average class size must be

1000/20=50.

b. f(x) = 0.4, x = 25

= 0.3, x = 100

= 0.3, x = 300. Note that here you are selecting a random student, not

a random class.

c. E(x) = 25(0.4) + 100(0.3) + 300(0.3) = 130

18.

a. E[(X - E(X - E(X) – E( – 0

b. E{[(X – ) / ]2} = (1/2) (E[(X – )2]) = (1/2) (E(X2) – 2E(X) + 2)

= (1/2) (E(X2) – 22 + 2) = (1/2) (E(X2) – 2)

= (1/2) 2 = 1

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MATH 511, Meade HW Solutions 3.2 –3 (omit c), 5, 14, 18, **3c, **8 2/13/

a. f(x) = x/10, x = 1,2,3, E(x) = (1) (1/10) + (2) (2/10) + (3) (3/10) + (4) (4/10) = (1 + 4 + 9 + 16 )/10 = 30/10 = 3 b. f(x) = x/55 , x = 1, 2, 3, 4 ,5 , 6, 7, 8, 9, 10 E(x) = (^) x=1∑^10 x^2 /55 = [(10)(11)(21)/(6)]/55 = 385/55 = 7 **c. f(x) = 3(1/4)x^ , x = 1, 2, 3, … E(x) = (^) x=1∑ ∞ 3x (1/4)x^ = (^) x=1∑ ∞ 3x/4x^ ≤ (^) x=1∑ ∞ 3 x/4x^ = 3 This assures that the series does indeed converge. The series, (^) x=1∑ ∞ (^) x /4x^ , can now be approximated. After about ten approximations it is clear that the series converges to 0.44444… which is 4/9. So the E(X) = 3(4/9) = (4/3) d. f(x) = (1/30)(x+1)^2 , x = 0, 1, 2, 3 E(x) = (1/30) [ 0 + 4 + 18 + 48] = 70/30 = 7/ e. f(x) = 2/n(n+1)x , x = 1, 2, 3, …, n E(x) =2/n(n+1) (^) i=1∑n^ x^2 = [2/n(n+1)] [n(n+1)(2n+1)/6)] = (2n+1)/

E(x) = [2,987,994(0) + 12,000(25) + 4(10,000) + 1(50,000) + 1(200,000)] - 0. 3,000, = 19 ⅔ cents – 50 cents = - 30⅓ cents **8. f(x) = 6/(x^2 ), x = 1, 2, 3, … E(x) = (6/) (^) x=1∑ ∞ x/x^2 = (6/) (^) x=1∑ ∞ 1/x ; this is the harmonic series, which diverges. Therefore E(x) does not exist. One way to show this is the integral test for infinite series.
a. [16(25) + 3(100) + 1(300)]/20 = 1,000/20 = 50. Note that this makes sense: 1000 students divided into 20 classes; the average class size must be 1000/20=50. b. f(x) = 0.4, x = 25 = 0.3, x = 100 = 0.3, x = 300. Note that here you are selecting a random student, not a random class. c. E(x) = 25(0.4) + 100(0.3) + 300(0.3) = 130
a. E[(X - E(X - E(X) – E( –  0 b. E{[(X – ) / ]^2 } = (1/^2 ) (E[(X – )^2 ]) = (1/^2 ) (E(X^2 ) – 2E(X) + ^2 ) = (1/^2 ) (E(X^2 ) – 2^2 + ^2 ) = (1/^2 ) (E(X^2 ) – ^2 ) = (1/^2 ) ^2 = 1

3.3 – 3, 5, 7(abc), 14, **7(de)

There is six-question multiple choice test with five possible answers, only one of which is correct. If a student guesses randomly and independently, then what is the probability that: a. The student gets questions 1 and 4 correct. (0.2)^2 (0.8)^4 = 0. b. The student gets exactly two questions right. 6 C 2 (0.2)^2 (0.8)^4 = 0.
The number of people who believe that the IRS abuses their power has a binomial distribution. b(25,0.7). In order to solve this problem using Appendix C, Table II, one needs to convert the problems so that p ≤ 0.5. So Let Y = the number who do not believe the IRS abuses their power. Y has a distribution of b(25, 0.3). a. P(X ≥ 13) = P(Y ≤ 12) = 0. b. P(X ≤ 11) = P(Y ≥ 14) = 1 – P(Y < 14) = 1 – P(Y ≤ 13) = 1 – 0.9940 = 0. c. P(X = 12) = 25 C 12 (0.7)^12 (0.3)^13 = 0. d. (x) = np = (25)(0.7) = 17. ^2 (x) = np(1 – p) = (25)(0.7)(0.3) = 5. (x) = √^2 = 2.
a. W has a binomial distribution b(2000, The probabilty comes from that the fact that exactly one fourth of the unit circle sits inside the unit square. The area of the unit circle is Acircle(r = 1) = (1)^2 = . b. (w) = np = (2000)() = 1,570. ^2 (w) = np(1 – p) = (2000)()(1 - = 337. (w)= √^2 = 18. c. E(W/500) = (w)/500 =  **d. Click on the Excel link on the solutions column for a spreadsheet concerning parts d and e of this problem. **e. Notice that exactly one eighth of the unit sphere is contained in the unit cube The volume of the unit sphere is (4/3). So the probability that a point on The unit cube is also on the unit sphere is (1/8) (4/3) = 6
a. X = b(8, 0.90) , Let Y = number of mints weighing less than 20.7g = b(8, 0.10) b. i. P(X = 8) = 8 C 8 (0.90)^8 (0.10)^0 = 0. ii. P(X ≤ 6) = P(Y ≥ 2) = 1 – P(Y < 2) = 1 – P(Y ≤ 1) = 1 – 0.8131 = 0. iii. P(X ≥ 6) = P(Y ≤ 2 = 0.

MATH 511 Homework Solutions: Expected Values and Probability Distributions, Assignments of Mathematics

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