Refer to the accompanying data set and use the 30 screw lengths to construct a frequency distribution. Begin with a lower class limit of 0.470 in, and use a class width of 0.010 in. The screws were labeled as having a length of 1 divided by 2 in. Does the frequency distribution appear to be consistent with the label? Why or why not?

Screw lengths (inches)

0.478

0.503

0.507

0.478

0.488

0.493

0.508

0.479

0.506

0.502

0.509

0.493

0.495

0.485

0.509

0.505

0.498

0.485

0.497

0.485

0.498

0.515

0.501

0.502

0.497

0.489

0.509

0.491

0.505

0.499

Complete the frequency distribution below.

Length (in)

Frequency

0.470 -

−

−

−

The frequency distribution of the length based on the prescribed interval is :

Lower class limit = 0.470

Class width = 0.010

Hence, the range of possible values in each class is 10(class width)

Upper class limit = lower class limit of preceeding class + class width

The Cummulative frequency of a class is the sum of the frequency of the class including those of the classes below it.

Class Intervals __ Frequency _ Cummulative frequency

0.470 - 0.479 3 3

0.480 - 0.489 5 8

0.490 - 0.499 9 17

0.500 - 0.509 12 29

0.510 - 0.519 1 30

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A) Total frequency = 30

B) Yes, the distribution is consistent with the label because the frequencies are greatest when the lengths are closest to the labeled size of 1/2 inches which is 0.5 inches.

Step-by-step explanation:

Since the class limit width is 0.010 from the question, we arrive at;

Class Limits Frequency

Length(Inches)

0.470 - 0.479 3

0.480 - 0. 489 5

0.490 - 0.499 9

0.500 - 0.509 12

0.510 - 0.519 1

Adding the frequency, total = 30

yes

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step-by-step explanation: