Frequency Distribution Explanation with an Example

Problem

Data were collected on the number of passengers at each train station in Melbourne. The numbers

for the weekday peak time, 7 am to 9:29 am, are given below. Construct a frequency distribution for

these data. What does the frequency distribution reveal about train usage in Melbourne?

401 1181 1178 637 2830 1000 2958 962 697 401 1442 115 454 2837 789 1396 603 2400 970 596 1841 1005 603 1386 867 692 245 1187 1548 697 656 1494 103 750 2199 647 987 528 337 874 1508 1630 148 883 1086 256 1443 309 487 278 1101 2286 1559 477 981 422 694 174 1245 647 497 1042 875 1028 208 1435 1420 1136 641 401 104 472 934 574 388 242 1222 95 742 1393 930 707 1117 339 1430 1616 1284 856 1079 362 1461 1120 909 1199 677 163 349 513 2447 593 669 244 1115 1112 367 721 640 287 746 364 254 239 348 514 519 1864 695 187 698 221 346 1180 643 4604 803 434 563 1365 661 649 528 1131 547 663 631 3074 767 1058 216 1406 814 1209 303 380 515 2521 343 2746 1018 686 750 537 1554 438 380 1292

Solution

To draw a frequency distribution table, we need maximum, minimum and Range Values

Maximum Value = 7729

Minimum Value = 95

Range = Maximum Value — Minimum Value = 7729–95 = 7634

I want to draw a table with 10 bins, So 7634 / 10 = 763

So 763 will be the difference between each bin’s maximum and minimum value

Explanation:

116 train stations have 95 passengers between 7 am to 9:29 am

68 train stations have 858 passengers between 7 am to 9:29 am

only 1 station has 7729 passengers between 7 am to 9:29 am

I hope it will be useful for 1 person in the universe. Thank you so much for reading and claps.