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Since the number of observations is even, the median will be the average of the two numbers closest to the middle.
Count the number of observations in each interval.
How many peaks are there? How about outliers?
Compare the general location of the observations and the spread of the distributions.
Could it be that the cans have been overfilled?
See solution.
See solution.
See solution.
See solution.
The cans contain too much fluid.
To create a boxplot, we need to find the following information for the two data sets.
&Minimum value &1^(st) Quartile &Median &3^(rd) Quartile &Maximum value
Since the observations have been ordered from least to greatest, we can immediately identify the minimum and maximum value as 361 and 380. Also, the number of values in the data set is 30, an even number, which means the median must be the average of the observations that are closest to the middle. These are the 15^(th) and 16^(th) observation.
|ccccc|
361, & 362, & 363, & 365, & 366,
366, & 367, & 367, & 367, & 368,
368, & 368, & 369, & 369, & 369,
369, & 370, & 370, & 370, & 370,
371, & 371, & 371, & 371, & 373,
375, & 375, & 376, & 376, & 380,
Again the observations have been ordered from least to greatest, which means we can immediately identify the minimum and maximum value as 349 and 366. The number of values in the data set is 30, which means the median must be the average of the observations that are closest to the middle, the 15^(th) and 16^(th) observation. |ccccc| 349, & 349, & 350, & 351, & 353, 353, & 353, & 354, & 354, & 354, 354, & 355, & 355, & 355, & 356, 357, & 358, & 361, & 361, & 361, 361, & 361, & 361, & 362, & 362, 363, & 364, & 365, & 366, & 366, The average of the values closest to the middle is 356+3572=356.5. To find the 1^(st) and 3^(rd) Quartile, we have to identify the middle value of the lower and upper half. This will be the 8^(th) and 23^(rd) value for the lower and upper half, respectively. |ccccc| 349, & 349, & 350, & 351, & 353, 353, & 353, & 354, & 354, & 354, 354, & 355, & 355, & 355, & 356, 357, & 358, & 361, & 361, & 361, 361, & 361, & 361, & 362, & 362, 363, & 364, & 365, & 366, & 366, As we can see, the upper and lower quartile are 354 and 361, respectively. Let's summarize what we have found for the diet cans. &Minimum value=349 &1^(st) Quartile = 354 &Median = 356.5 &3^(rd) Quartile = 361 &Maximum value=366
To make the combination histogram and boxplot, we need to count the number of observations in each interval.
Let's count the number of observations in each interval for the regular cans.
r|l
Interval & Observations
348-352 &
352-356 &
356-360 &
360-364 & 361, 362, 363,
364-368 & 365, 366, 366, 367, 367, 367
& 368, 368, 368
368-372 & 369, 369, 369, 369, 370, 370
& 370, 370, 371, 371, 371, 371
372-376 & 373, 375, 375, 376, 376
376-380 & 380
380-384 &
Let's count the number of observations in each interval for the diet cans. r|l Interval & Observations 348-352 & 349, 349, 350, 351 352-356 & 353, 353, 353, 354, 354, 354 & 354, 355, 355, 355, 356 356-360 & 357, 358, 360-364 & 361, 361, 361, 361, 361, 361 & 362, 362, 362, 363, 364 364-368 & 365, 366, 366 368-372 & 372-376 & 376-380 & 380-384 & Using our information from Part A, we can create our combination histogram and boxplot.
Comparing the diagrams from Part A, we notice that neither dataset contains any outlier. As for the regular cans, the observations have a peak in the middle, while the diet cans have two peaks.
Examining the diagrams, we notice that the observations of the regular cans are generally higher than that of the diet cans. However, the spread of the diet cans is greater than that of the regular cans.
The cans must have been filled with too much fluid.