Question

2. Consider the following observations on shear strength (MPa) of a joint bonded in a particular manner:

22.2	40.4	16.4	73.7	36.6	109.9
30.0	4.4	33.1	66.7	81.5

a. Calculate the value of the sample mean and median. Why is the median so different from the mean? Explain. [3 points]

b. Calculate the sample standard deviation. [2 points]

c. Are the smallest and largest observations outliers? Justify your answer with appropriate calculations. [5 points]

1. Are the smallest and largest observations outliers? Justify your answer with appropriate calculations. [5 points]

Answer 1

a)

Mean = (22.2 + 40.4 + 16.4 + 73.7 + 36.6 + 109.9 + 30.0 + 4.4 + 33.1 + 66.7 + 81.5)/11
= 514.9/11
Mean = 46.8091

The median is the middle number in a sorted list of numbers. So, to find the median, we need to place the numbers in value order and find the middle number.

Ordering the data from least to greatest, we get:

4.4   16.4   22.2   30.0   33.1   36.6   40.4   66.7   73.7   81.5   109.9   

So, the median is 36.6 .

A mean is computed by adding up all the values and dividing by total values. The Median is the number found at the exact middle of the set of values.

b)

Standard Deviation σ = √(1/11 - 1) x ((22.2 - 46.8091)² + (40.4 - 46.8091)² + (16.4 - 46.8091)² + (73.7 - 46.8091)² + (36.6 - 46.8091)² + (109.9 - 46.8091)² + (30.0 - 46.8091)² + (4.4 - 46.8091)² + (33.1 - 46.8091)² + (66.7 - 46.8091)² + (81.5 - 46.8091)²)
= √(1/10) x ((-24.6091)² + (-6.4091)² + (-30.4091)² + (26.8909)² + (-10.2091)² + (63.0909)² + (-16.8091)² + (-42.4091)² + (-13.7091)² + (19.8909)² + (34.6909)²)
= √(0.1) x ((605.60780281) + (41.07656281) + (924.71336281) + (723.12050281) + (104.22572281) + (3980.46166281) + (282.54584281) + (1798.53176281) + (187.93942281) + (395.64790281) + (1203.45854281))
= √(0.1) x (10247.32909091)
= √(1024.732909091)
= 32.0114

c)