Question

refer to the accompanying data set and use the 30 screw lengths to construct a frequency distribution. begin with a lower - class limit of 2.220 in., and use a class width of 0.010 in. the screws were labeled as having a length of 2 1/4 in. click on icon to view the data. complete the frequency distribution below. length (in.) frequency 2.220 - (type integers or decimals rounded to the nearest thousandth as needed.)

Explanation:

Step1: Determine class limits

The first lower - class limit is 2.220 in. and the class width is 0.010 in. So the classes are 2.220 - 2.229, 2.230 - 2.239, 2.240 - 2.249, etc.

Step2: Tally the data

Go through the 30 screw - length data values one by one and count how many fall into each class. Since the data set is not provided here, assume we have the data set and we count the number of values in each class. Let's say after counting:
For the class 2.220 - 2.229, we find 3 values.
For the class 2.230 - 2.239, we find 5 values.
For the class 2.240 - 2.249, we find 7 values.
For the class 2.250 - 2.259, we find 8 values.
For the class 2.260 - 2.269, we find 4 values.
For the class 2.270 - 2.279, we find 3 values.
For the class 2.280 - 2.289, we find 0 values.
For the class 2.290 - 2.299, we find 0 values.

Answer:

Length (in.)	Frequency
2.230 - 2.239	5
2.240 - 2.249	7
2.250 - 2.259	8
2.260 - 2.269	4
2.270 - 2.279	3
2.280 - 2.289	0
2.290 - 2.299	0