I can't seem to figure out the class interval for this grouped frequency distribution table:
Highest Score=684
Lowest Score=56
& the professor has instructed us to use an interval size of 61
The starting point needs to be divisible by 61 due to convention.
But the lowest score 56 is not divisible by 61. Help!
How about starting at zero score and having 13 intervals?:
0-61
62-123
124-185
etc.
620-681
682-743
To determine the class interval for a grouped frequency distribution table, you need to consider the range of the data (the difference between the highest and lowest scores) and the desired interval size.
In this case, the highest score is 684, and the lowest score is 56. The interval size you need to use is 61.
Since the lowest score (56) is not divisible by 61, you will need to adjust the starting point of the class interval to ensure it is divisible by 61. To do this, you can round down the lowest score to the nearest multiple of 61 without going below the minimum value.
Let's calculate this:
The floor function can be used to round down the lowest score to the nearest multiple of 61. Using the floor function: floor(56/61) = 0.
Therefore, since 56 is less than 61, you can adjust the starting point of the class interval to be zero.
Now, substitute the starting point and interval size into the equation for the class interval:
Class Interval = Starting Point + (k * Interval Size),
where k denotes the sequential number of each class interval.
In this case, the starting point is 0, and the interval size is 61. So, the class intervals will be:
Class Interval 1: 0 - 60
Class Interval 2: 61 - 121
Class Interval 3: 122 - 182
...
and so on, until you reach the highest score.
Remember to adjust the upper limit of the last class interval to include the highest score (684 in this case).