Question

enrollments for selected independent religiously controlled 4-year colleges listed below are the enrollments for selected independent religiously controlled 4-year colleges that offer bachelor’s degrees only for a recent year. construct a grouped frequency distribution with six classes and find the mean and modal class. 1530 2819 2595 2131 2750 1445 2970 2785 2528 2922 1999 1170 2391 3315 2046 2643 3275 1037 3049 3465 3182 1780 1610 2813 2016 2176 2225 3385 2688 3067 3152 3490 2379 part: 0 / 6 part 1 of 6 (a) what is the class width for a frequency distribution with 6 classes? the class width is .

Explanation:

Step1: Find the range of the data

First, we need to find the maximum and minimum values in the data set.
The data points are: 1530, 2819, 2595, 2131, 2750, 1445, 2970, 2785, 2528, 2922, 1999, 1170, 2391, 3315, 2046, 2643, 3275, 1037, 3049, 3465, 3182, 1780, 1610, 2813, 2016, 2176, 2225, 3385, 2688, 3067, 3152, 3490, 2379.

The minimum value ($\min$) is 1037 and the maximum value ($\max$) is 3490.

The range ($R$) is calculated as $R=\max - \min$. So, $R = 3490 - 1037=2453$.

Step2: Calculate the class width

The formula for class width ($w$) when we have $k$ classes is $w=\frac{\text{Range}}{k}$, and we usually round up to the nearest whole number (or a convenient number). Here, $k = 6$.

So, $w=\frac{2453}{6}\approx408.83$. Since we need to cover the entire range, we round up to 409. Wait, but let's check again. Wait, maybe I made a mistake in the range. Wait, let's re - calculate the range:

Wait, maximum value: 3490, minimum value: 1037. $3490 - 1037 = 2453$. Then class width $=\frac{2453}{6}\approx409$. But sometimes, we can also use a more convenient number. Wait, maybe I miscalculated the data points. Let's count the number of data points: Let's list them again:

1. 1037
2. 1170
3. 1445
4. 1530
5. 1610
6. 1780
7. 1999
8. 2016
9. 2046
10. 2131
11. 2176
12. 2225
13. 2379
14. 2391
15. 2528
16. 2595
17. 2643
18. 2688
19. 2750
20. 2785
21. 2813
22. 2819
23. 2922
24. 2970
25. 3049
26. 3067
27. 3152
28. 3182
29. 3275
30. 3315
31. 3385
32. 3465
33. 3490

Yes, min is 1037, max is 3490. Range is 3490 - 1037 = 2453.

Class width $w=\lceil\frac{2453}{6}
ceil=\lceil408.83
ceil = 409$. Wait, but maybe the problem expects us to use a different approach? Wait, sometimes, we can also adjust the range slightly. Wait, maybe I made a mistake in the range. Wait, 3490 - 1037 = 2453. Then 2453 divided by 6 is approximately 409. But let's check with a different method. Let's see, if we take class width as 400, then 6 classes would cover from 1000 - 1400, 1400 - 1800, 1800 - 2200, 2200 - 2600, 2600 - 3000, 3000 - 3400, 3400 - 3800. But our max is 3490, so 3400 - 3800 would cover it. Wait, maybe my initial calculation of the range is wrong. Wait, 3490 - 1037 = 2453. Let's try class width of 409:

First class: 1037 - 1037 + 409 - 1=1037 - 1445

Second: 1446 - 1854

Third: 1855 - 2263

Fourth: 2264 - 2672

Fifth: 2673 - 3081

Sixth: 3082 - 3490

Yes, this covers the range. But let's check the formula again. The class width is usually calculated as $\lceil\frac{\text{Range}}{\text{Number of classes}}
ceil$. So, $\frac{2453}{6}\approx408.83$, so we round up to 409.

Wait, but maybe I made a mistake in the range. Let's re - check the maximum and minimum:

Looking at the data: 1037 is the smallest, 3490 is the largest. So range is 3490 - 1037 = 2453. Then class width $=\frac{2453}{6}\approx409$.

Answer:

409