ari313
03.03.2020 •
Mathematics
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 -
−
−
−
Solved
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Ответ:
The frequency distribution of the length based on the prescribed interval is :
Class Interval _ Frequency _ Cumm/ frequency 0.470 - 0.479 3 30.480 - 0.489 5 80.490 - 0.499 9 170.500 - 0.509 12 290.510 - 0.519 1 30Lower 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
Learn more : link
Ответ:
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
Ответ:
Let me explain, so when you have income, imagine the “in” part which means the money is coming in, and expenses are like bills, grocery’s. Basically anything you spend the money on.
Hope I helped!