A dealer in recycled paper places empty trailers at various sites. The trailers

Question

A dealer in recycled paper places empty trailers at various sites. The trailers are gradually filled by individuals who bring in old newspapers and magazines, and are picked up on several schedules. One such schedule involves pickup every second week. This schedule is desirable if the average amount of recycled paper is more than 1600 cubic feet per 2-week period. The dealer’s records for eighteen 2-week periods show the following volumes (in cubic feet) at a particular site: recycle=c(1660,1820,1590,1440,1730,1680,1750,1720,1900,1570,1700,1900,1800,1770,2010,1580,1620,1690)

a. Calculate the following:

i. mean:

ii. median

iii. Mode:

iv. variance:

v. standard deviation:

vi. Q1

vii. Q3

viii. IQR

b. Make a histogram of the data using hist()

c. Make a boxplot of the data using boxplot()

d. Describe the graphs you made (skewed, unimodal, etc.). Are there any outliers? What are their values? Remember, any value that is less than Q1 (1.5)I QR or greater than Q3 + (1.5)I QR where I QR = Q3 Q1, is considered to be an outlier.

Explanation / Answer

Solution

Although all that required can be computed from the given data directly, subtracting 1600 from each value would ease computations. If x represents the given data, the new variable is y = x – 1600. Then, we just add 1600 to the mean, median and mode of y-values to get these for the x-values. Other parameters do not change.

Mean(y) = sum of y values/18 = 2230/18 = 123.8889. So, mean of x = 1723.8889

Median is middle-most value in the ordered set. So, we first arrange y-values in ascending order as: - 160, - 30, - 20, - 10, 60, 80, 90, 100, 120, 120, 130, 150, 170, 200, 220, 300, 300, 410.

There are two middle values since18 is even. So, median of y = average of (100, 120) = 110

Hence median of x = 1710

Mode is the value that occurs maximum number of times. For y-values, the mode is 120 as well as 300 and hence modes of x are 1720 and 1900[just for additional information, this situation is refrred to as 'bi-modal']

Variance of x = variance of y = {sum(y – mean y)2}/18 = 312427.8/18 = 17357.1

Standard Deviation = sq.rt of variance = 131.7463

Q1 = first quartile = the value below which ¼ th of the values lie. In our ordered set 4th value is – 10 and 5th value is 60. So Q1 = 4.5th (i.e.,18/4) value = (- 10 + 60)/2 = 25

Similarly, Q3 = third quartile = the value below which ¾ th of the values lie. In our ordered set 13th value is 170 and 14th value is 200. So Q3 = 13.5th {i.e.,18(3/4) value} = (170 + 200)/2 = 185

Finally, IQR = Inter-Quartile Range = Q3 – Q1 = 185 – 25 = 160

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