A manufacturing company produces steel housings for electrical equipment. The main component part of the housing...

A manufacturing company produces steel housings for electrical equipment. The main component part of the housing is a steel trough that is made out of a 14-gauge steel coil. It is produced using a 250-ton progressive punch press with a wipe-down operation that puts two 90-degree forms in the flat steel to make the trough. The distance from one side of the form to the other is critical because of weatherproofing in outdoor applications. The company requires that the width of the trough be between 8.31 inches and 8.61 inches. The widths of the troughs, in inches, are collected from a sample of 49 troughs and stored in Trough and shown here:

8.312 8.343 8.317 8.383 8.348 8.410 8.351 8.373 8.481 8.422 8.476 8.382 8.484 8.403 8.414 8.419 8.385 8.465 8.498 8.447 8.436 8.413 8.489 8.414 8.481 8.415 8.479 8.429 8.458 8.462 8.460 8.444 8.429 8.460 8.412 8.420 8.410 8.405 8.323 8.420 8.396 8.447 8.405 8.439 8.411 8.427 8.420 8.498

8.409

a. Construct a frequency distribution and a percentage distribution

b. Construct a cumulative percentage distribution. (2 points)

c. What can you conclude about the number of troughs that will meet the company’s requirements of troughs being between 8.31 and 8.61 inches wide?

answer please

Solutions

Expert Solution

Solution:

Given:Sample size=49

a)Frequency is the number of observations in a particular interval.

Percentage frequency distribution=(Frequency*100)/ Total frequency

Let us find the width of interval=Range/Number of desired class groupings

Let us assume number of desired class groupings be 6 and range=Maximum value-minimum value=8.498-8.312=0.186

So width of interval=0.186/6=0.031

In first class interval,the lower class limit is 8.312 and upper class limit is 8.312+0.031=8.343

In second class interval,the lower class limit is 8.344 and upper class limit is 8.344+0.031=8.375 and so on.

The frequency distribution and percentage distribution is calculated as follows.

Class interval	Frequency	Percentage distribution
8.312-8.343	4	(4/49)*100=8.16
8.344-8.375	3	(3/49)*100=6.12
8.376-8.407	7	(7/49)*100=14.29
8.408-8.439	19	(19/49)*100=38.78
8.440-8.471	8	(8/49)*100=16.33
8.472-8.503	8	(8/49)*100=16.33
Total	49

b)Cumulative percentage distribution=(corresponding percentage frequency)+(Preceding percentage frequency)

The table below shows the calculation of cumulative percentage distribution.

Percentage distribution	cumulative percentage distribution
8.16	8.16
6.12	8.16+6.12=14.28
14.29	14.28+14.29=28.57
38.78	28.57+38.78=67.35
16.33	67.35+16.33=83.68
16.33	83.68+16.33=100

c)The company requirement that the width of the trough be between 8.31 inches and 8.61 inches is satisfied.

It can be concluded that all the given troughs will meet the requirement of being between 8.31 and 8.61 inches.

orchestra answered 1 year ago

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