Question
In: Statistics and Probability
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.
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