0
$\begingroup$

Is there a way to find a statistical distribution for the below data? This is derived from frequency count of different topics in a large corpus of news stories. I wonder if this follows a Poisson Distribution or some other distribution or "none at all".

As an example n=2 and Count=22939 means that there are 22939 different topics that appear exactly twice in a corpus of news stories. The last non-zero value n=687 and Count=1 means that there is 1 topic that appears in 687 news stories.

n   Count
1   115323
2   22939
3   6538
4   4841
5   2193
6   1682
7   928
8   909
9   608
10  654
11  422
12  417
13  286
14  313
15  179
16  189
17  108
18  133
19  72
20  113
21  65
22  99
23  43
24  66
25  40
26  58
27  27
28  51
29  32
30  50
31  27
32  30
33  20
34  23
35  13
36  28
37  16
38  19
39  6
40  20
41  13
42  21
43  10
44  16
45  12
46  10
47  14
48  18
49  7
50  8
51  5
52  10
53  6
54  11
55  5
56  8
57  9
58  11
59  3
60  12
61  4
62  3
63  6
64  8
65  5
66  3
67  3
68  11
69  3
70  3
71  6
72  8
73  1
74  2
75  5
76  5
77  3
78  2
79  1
80  0
81  1
82  6
83  2
84  5
85  4
86  2
87  3
88  2
89  2
90  4
91  0
92  0
93  4
94  3
95  2
96  2
97  3
98  2
99  0
100 3
101 2
102 0
103 1
104 0
105 1
106 0
107 2
108 2
109 2
110 1
111 0
112 0
113 0
114 0
115 1
116 1
117 0
118 2
119 2
120 0
121 0
122 0
123 2
124 2
125 1
126 1
127 0
128 1
129 0
130 1
131 1
132 1
133 0
134 1
135 0
136 0
137 0
138 0
139 0
140 1
141 0
142 1
143 2
144 1
145 1
146 1
147 0
148 1
149 1
150 0
151 0
152 0
153 1
154 0
155 0
156 0
157 1
158 0
159 0
160 0
161 0
162 1
163 0
164 2
165 0
166 0
167 1
168 2
169 1
170 2
171 0
172 0
173 1
174 0
175 0
176 1
177 0
178 0
179 1
180 0
181 2
182 1
183 0
184 0
185 1
186 1
187 0
188 0
189 0
190 1
191 0
192 0
193 0
194 0
195 1
196 0
197 0
198 0
199 0
200 0
201 0
202 0
203 0
204 1
205 0
206 0
207 0
208 0
209 1
210 0
211 1
212 0
213 0
214 0
215 0
216 2
217 0
218 0
219 0
220 0
221 0
222 1
223 0
224 0
225 0
226 1
227 0
228 0
229 2
230 0
231 0
232 0
233 0
234 0
235 0
236 0
237 0
238 0
239 0
240 0
241 0
242 0
243 0
244 1
245 0
246 0
247 0
248 0
249 0
250 0
251 0
252 0
253 0
254 0
255 0
256 0
257 0
258 0
259 1
260 0
261 0
262 1
263 0
264 1
265 0
266 1
267 0
268 0
269 0
270 0
271 0
272 0
273 0
274 0
275 0
276 0
277 0
278 0
279 0
280 0
281 0
282 0
283 0
284 0
285 0
286 1
287 0
288 0
289 0
290 0
291 0
292 0
293 0
294 0
295 0
296 0
297 0
298 1
299 0
300 0
301 0
302 0
303 1
304 0
305 0
306 0
307 0
308 0
309 0
310 0
311 0
312 0
313 1
314 0
315 0
316 0
317 0
318 0
319 0
320 0
321 0
322 1
323 0
324 1
325 0
326 0
327 0
328 0
329 0
330 0
331 0
332 0
333 0
334 0
335 0
336 0
337 0
338 0
339 0
340 0
341 0
342 0
343 0
344 0
345 0
346 0
347 0
348 0
349 0
350 1
351 0
352 0
353 0
354 0
355 0
356 0
357 0
358 0
359 0
360 1
361 0
362 0
363 0
364 0
365 0
366 0
367 0
368 0
369 0
370 0
371 0
372 0
373 0
374 0
375 0
376 0
377 0
378 0
379 0
380 0
381 0
382 0
383 0
384 0
385 0
386 0
387 0
388 0
389 0
390 1
391 0
392 0
393 0
394 0
395 0
396 0
397 0
398 1
399 0
400 0
401 0
402 0
403 0
404 0
405 0
406 0
407 0
408 0
409 0
410 0
411 0
412 0
413 0
414 0
415 0
416 0
417 0
418 0
419 0
420 0
421 0
422 0
423 0
424 0
425 1
426 0
427 0
428 0
429 0
430 0
431 0
432 0
433 0
434 0
435 0
436 0
437 0
438 0
439 0
440 0
441 0
442 0
443 0
444 0
445 1
446 0
447 0
448 0
449 0
450 0
451 0
452 1
453 0
454 0
455 0
456 0
457 0
458 0
459 0
460 0
461 0
462 0
463 0
464 0
465 0
466 0
467 0
468 0
469 0
470 0
471 0
472 0
473 0
474 0
475 0
476 0
477 0
478 0
479 0
480 0
481 0
482 0
483 0
484 0
485 0
486 0
487 0
488 0
489 0
490 0
491 0
492 0
493 0
494 0
495 0
496 0
497 0
498 0
499 0
500 0
501 0
502 0
503 0
504 0
505 0
506 0
507 0
508 0
509 0
510 0
511 0
512 0
513 0
514 0
515 0
516 0
517 0
518 0
519 0
520 0
521 0
522 0
523 0
524 0
525 0
526 0
527 0
528 0
529 0
530 0
531 0
532 0
533 0
534 0
535 0
536 0
537 0
538 0
539 1
540 0
541 1
542 0
543 0
544 0
545 0
546 0
547 0
548 0
549 1
550 0
551 0
552 0
553 0
554 0
555 0
556 0
557 0
558 0
559 0
560 0
561 0
562 0
563 0
564 0
565 0
566 0
567 0
568 0
569 0
570 0
571 0
572 0
573 0
574 0
575 0
576 0
577 0
578 0
579 0
580 0
581 0
582 0
583 0
584 0
585 0
586 1
587 0
588 0
589 0
590 0
591 0
592 0
593 0
594 0
595 0
596 0
597 0
598 0
599 0
600 0
601 0
602 0
603 0
604 0
605 0
606 0
607 0
608 0
609 0
610 0
611 0
612 0
613 0
614 0
615 0
616 0
617 0
618 0
619 0
620 0
621 0
622 0
623 1
624 0
625 0
626 0
627 0
628 0
629 0
630 0
631 0
632 0
633 0
634 0
635 0
636 0
637 0
638 0
639 0
640 0
641 0
642 0
643 0
644 1
645 0
646 0
647 0
648 0
649 0
650 0
651 0
652 0
653 0
654 0
655 1
656 0
657 0
658 0
659 0
660 0
661 0
662 0
663 0
664 0
665 0
666 0
667 0
668 0
669 0
670 0
671 0
672 0
673 0
674 0
675 0
676 0
677 0
678 0
679 0
680 0
681 0
682 0
683 0
684 0
685 0
686 0
687 1
688 0
689 0
690 0
691 0
692 0
693 0
694 0
695 0
696 0
697 0
698 0
699 0
700 0
$\endgroup$
3
  • $\begingroup$ Well, it's certainly not Poisson, since there are no zeros. A shifted Poisson or Negative Binomial might work. $\endgroup$ – Stephan Kolassa Feb 27 '20 at 16:17
  • 2
    $\begingroup$ Could you clarify what you mean by "find a statistical distribution for"? The data, after all, have a distribution, as given in the question. More to the point, could you articulate a reason for this fitting exercise, whatever it might be? What do you hope to learn from it or do with it? $\endgroup$ – whuber Feb 27 '20 at 16:41
  • $\begingroup$ Pareto distribution (80 20 rule)? $\endgroup$ – Ed Rigdon Feb 28 '20 at 1:22

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