Question

A sample of six weights for packages delivered by a mailing company (in lbs.) is shown below:

7.0, 7.3, 7.5, 7.7, 8.0, 12.4

a) Find the mean for the data above.

b) Which one of the following interpretations of the mean calculated in a) is true (state True or False):

1) It represents the "average" weight. If each package would weigh the same, they each would have a weight equal to the mean _________

2) It is the middle point in the set of weights. That is, half the weights are higher than the mean, and half the weights are lower than the mean ___________

c) Find the median for the data above

d) Which one of the 2 measures of central tendency (mean or median) best describes the data? Explain.

e) Find the standard deviation for the data above

f) Find the 67^th percentile for the data above

g) Find the IQR for the data above

Answer 1

Sr. no	X	(X - X̄)²
1	7	1.734
2	7.3	1.034
3	7.5	0.667
4	7.7	0.380
5	8	0.100
6	12.4	16.674

	X	(X - X̄)²
total sum	49.9	20.58833333
n	6	6

a)

mean =    ΣX/n =    8.317

b)

1) It represents the "average" weight. If each package would weigh the same, they each would have a weight equal to the mean true

2) It is the middle point in the set of weights. That is, half the weights are higher than the mean, and half the weights are lower than the mean (it is called median) so, answer is False

c)

Median=0.5(n+1)th value =    3.5   th value of sorted data
=   7.6

d)

mean and median are not equal . data seems to be skewed (mean>median), so median is best measure of central tendency of data

e)

sample std dev =   √ [ Σ(X - X̄)²/(n-1)] = √(20.588/5) = 2.0292

f)

67th percentile = 0.67(n+1)th value = 4.69th value = 7.7+0.67*(8-7.7)=7.901

g)

quartile , Q1 = 0.25(n+1)th value=   1.75   th value of sorted data
=   7.225  
      
Quartile , Q3 = 0.75(n+1)th value=   5.25   th value of sorted data
=   9.1  
      
IQR = Q3-Q1 =    1.875