# QNT 561 ch 2-4

Sample mean = (3.2+2.5+2.1+3.7+2.8+2.0)/6 = 16.3/6 = 2.717

The median is the middle value when the data set is ordered.

Here since we have even number of observations we need to find the average of the two middle values.

Ordered data: 2.0, 2.1, 2.5, 2.8, 3.2, 3.7

Median = (2.5+2.8)/2 = 2.65

2.38

a. Mean = Sum/n = 85/10 = 8.5

b. Mean = Sum/n = 400/16 = 25

c. Mean = Sum/n = 35/45 = 0.778

d. Mean = Sum/n = 242/18 = 13.444

2.44

a. Sample mean = (462.20+458.84+445.32+240.58+192.68)/5 = 1799.62/5 = 359.92

The average amount charged is approximately equal to 359.92

b. Ordered data: 192.68, 240.58, 445.32, 458.84, 462.20

Median = Middle value = 445.32

Half of the banks charged lower than 445.32 and the other half charged greater than 445.32

2.46

a. The sample mean = Sum of all the observations/ 20  = 37.62/20  = 1.881

Sample average for surface roughness is 1.881

b. Median = Average of the two middle observations (since even number of data)

= (10th observation + 11th observation)/2 = (2.03+2.05)/2 = 2.04

Half of the sample roughness measurements fall below 2.04 and the other half fall above 2.04.