Virtual Laboratories > Finite Sampling Models > 1 2 3 4 5 6 7 8 9 [10]

10. Notes

Simulating a Random Samples

It's very easy to simulate a random sample of size n, with replacement from D = {1, 2, ..., N}. Recall that the ceiling function ceil(x) gives the smallest integer that is at least as large as x.

1. Let U_i be a random number for i = 1, 2, ..., n. Show that X_i = ceil(NU_i), i = 1, 2, ..., n simulates a random sample, with replacement, from D.

It's a bit harder to simulate a random sample of size n, without replacement, since we need to remove each sample value before the next draw.

2. Show that the following algorithm generates a random sample of size n, without replacement, from D.

For i = 1 to N, let b_i = i.
For i = 1 to n,
let j = N – i + 1;
let U_i = random number;
let J = ceil(j U_i);
let X_i = b_J;
let k = b_j;
let b_j = b_J;
let b_J = k.
Return (X₁, X₂, ..., X_n).

Related Topics

• Sampling with replacement (or sampling from an infinite population) leads to independent, identically distributed random variables. The chapter on Random Samples is a general study of such variables.
• Card games are based on sampling without replacement; dice games are based on sampling with replacement. The chapter on Games includes a number of such games.
• Multinomial Trials are based on sampling with replacement from a multi-type population.
• The problem of estimating parameters based on a random sample is studied in the chapter on Point Estimation.

Books

• Perhaps the best general reference for combinatorial probability remains the classic text An Introduction to Probability Theory and Its Applications, by William Feller.
• An amazing number of probability problems can be formulated in terms of ball and urn models. The authoritative reference for this theory is the book Urn Models and Their Application by Johnson and Kotz.

Selected Answers for Section 1

1.17.

1. 1 / 4
2. 1 / 221
3. 4 / 17
4. 1 / 52

1.19. 0.000547

Selected Answers for Section 2

2.15. Y = number of defective chips in the sample

1. P(Y = k) = C(10, k) C(90, 5 - k) / C(100, 5) for k = 0, 1, 2, 3, 4, 5.
2. E(Y) = 0.5, var(Y) = 0.432
3. P(Y > 0) = 0.416

2.16. Y = number of women, Z = 10 - Y = number of men

1. E(Y) = 6, var(Y) = 1.959
2. E(Z) = 4, var(Z) = 1.959
3. P(Y = 0) + P(Y = 10) = 0.00294

2.22. Y = number of tagged fish in the sample

1. P(Y 2) = 0.6108
2. P(Y 2) = 0.6083
3. Relative error: 0.0042.

2.23. 0.6331

Selected Answers for Section 3

3.5. 20

3.6. 2000

3.7.

3.9.

3.11.

1. Reject the lot when Y 2.
2. Reject the lot when Y 1.

3.14. 2000

Selected Answers for Section 4

4.15.

1. 0.2386
2. 0.0741
3. 0.0180
4. 0.2385

4.16.

1. 3.25
2. 1.864
3. -0.6213
4. -1 / 3

4.17.

1. 0.2370
2. 0.2168

4.18.

1. 0.0753
2. 0.3109

Selected Answers for Section 5

5.6.

1. P(X₍₃₎ = k) = C(k - 1, 2) C(25 - k, 2) / C(25, 5) for k = 3, 4, ..., 22
2. E(X₍₃₎) = 13
3. var(X₍₃₎) = 125 / 7.

5.17. 1437

5.19. 2322

Selected Answers for Section 6

6.5. 1,334,961

6.9.

6.12.

6.22.

1. 1 / 100
2. 1 / 16

Selected Answers for Section 7

7.5. 0.6029

7.7. 0.2778

7.9. 0.6181

7.11. 0.3024

7.14. 9

Selected Answers for Section 8

8.9. 0.3041

8.11. 0.2218

8.14. 0.3415

8.16. 0.3174

8.20. 87.576, 2.942

8.21. 22.952, 1.826

8.21. 9.894, 1.056

8.25. Let V denote the number of distinct answers.

1. j	1	2	3
   P(V = j)	1/16	9/16	6/16

2. P(V = 1) = 1/16
3. E(V) = 37/16
4. sd(V) = 0.6830

8.25. Let V denote the number of ducks killed.

1. j	1	2	3	4	5
   P(V = j)	1/10000	927/2000	9/50	63/127	189/625

2. E(V) = 4.095
3. sd(V) = 0.727

Selected Answers for Section 9

9.6. 0.9104

9.7. 0.8110

9.8. 0.0456

9.12. 10.988, 1.130

9.13. 14.700, 6.244

9.14. 29.290, 11.211

9.15. 2364.646, 456.207

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