Question

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 -
−
−
−

Answer 1

The data on the screw length of 30 screws is provided in the question. The lower-class limit is 0.470 inches and the class width is 0.010 inches. The maximum and minimum length is 0.515 inches and 0.478 inches, respectively. Therefore, the total number of classes is calculated as follows: Maximum value - Minimum value Total number of class Class width 0.519 - 0.478 0.010 = 4.1 * 5 (Approximated to nearest integer) The lower limit of the first class is 0.470. 0.010 is added with 0.470 to get the lower limit of the next class and the process is continued till the lower limit of fifth class is obtained. The lower limits of all five classes are 0.470, 0.480, 0.490, 0.500, and 0.510. The upper limit of first class can be determined by subtracting one unit from the lower limit of the second class. Then, the class width is added to get the upper limits of other classes. The upper limits are 0.479, 0.489, 0.499, 0.509, and 0.519. Frequency of each class can be determined by counting the number of screws whose length lies within a particular class limit. The frequency distribution is obtained as follows: Length Frequency 0.470-0.479 3 0.480-0.489 5 0.490-0.499 0.500-0.509 12 0.510-0.519 The frequency dis same as the original label length. oution can be considered as consistent when the mean label length is The necessary calculation for obtaining the mean is done using Excel software. The screenshot is shown below. в 1 Lower Limit Upper Limit 0.479 0.489 0.499 0.509 0.519 Midpoint=(Lower limit+Upper limit)/2 =(A2+B2)2 НА3-В3)2 (A4+B4)/2 =(A5+B5)2 НА6+В6)2 Frequency Frequency*Midpoint =D2*C2 =D3*C3 =D4*C4 =D5*C5 =D6*C6 2 0.47 3 0.48 4 0.49 12 5 0.5 6 0.51

Therefore, the mean length of the screw is calculated as follows: E(Frequency x Midpoint) Mean Total frequency 1.424 +2.423 + 4.451+6.054 + 0.515 30 14.865 30 = 0.496 2 0.50 The screws are labelled as 1 in 2 inches, that is 1/2 = 0.50. Since, the mean label length is same as the original label length, it can be concluded that the frequency distribution is consistent.