Answer:
Mean = 2891.4
Varaince = 8009.84
Step-by-step explanation:
Given that the accompanying observations are on stabilized viscosity (cP) for specimens of a certain grade of asphalt with 18% rubber added:
To find mean we add all those and divide by number of entries and we get
mean = 2891.4
To find variance= sum of squares of each entry from 2891.4
Since 2891.4 is in decimal we fix assumed mean as 2891 and find variance
We find as 8010
To this we subtract 8010 - 0.4^2 = 8009.84
(where 0.4 is the difference between assumed mean and real mean)
d d^2
2783 -108 11664
2925 34 1156
3046 155 24025
2845 -46 2116
2858 -33 1089
total 14457 40050
Mean 2891.4 Var 8010
Assumed mean 2891
Median is the middle entry after arranging in ascending order
Median = 2858