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 -

-

-

-

Question

Mathematics

Toby Dickson

28 July, 00:14

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 -

-

-

-

+4

Answers (1)

Know the Answer?

Zachary Franco · Answer 1

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