Question

5. For a sample of 11 employers, the most recent hourly wage increases were 18, 30, 25, 5, 7, 2, 20, 12, 15, 55, and 40 cents per hour. For these sample data, determine the following.

a) The mean, median, and range.

b) The mean absolute deviation.

c) The variance and standard deviation.

Answer 1

1. The mean

= Sum of all values / 11 number of observations

= 229/11

= 20.82 ( rounded to 2 decimal places )

Median is the “middle” value of a sorted list of numbers.

We therefore sort the list of numbers from smallest to largest. The middle value is the sixth number . The middle value thus is 18

MEDIA = 18

Range is difference between highest value and lowest value

= 55 – 2

= 53

1. We define Absolute deviation as Absolute difference between the individual value and the mean value .

Mean absolute deviation is sum of Absolute deviationdivided by 11 which is number of observations .

Please refer below table for relevant calculations

Value	Mean	Absolute deviation ( AD)
18	20.82	2.82
30	20.82	9.18
25	20.82	4.18
5	20.82	15.82
7	20.82	13.82
2	20.82	18.82
20	20.82	0.82
12	20.82	8.82
15	20.82	5.82
55	20.82	34.18
40	20.82	19.18
	sum =	133.46

Accordingly , sum of absolute deviations = 133.46

Therefore , Mean absolute deviation = 133.46/11 = 12.13

1. The variance can be used by applying the formula VAR,s( )

We thus place all the data in excel and apply the formula VAR.P ( ) and get value = 255.36

Thus Variance of the sample = 255.36

Standard deviation of the sample = Square root ( Variance ) = Square root ( 255.36) =15.98