27 February, 03:32

# The accompanying observations are on stabilized viscosity (cP) for specimens of a certain grade of asphalt with 18% rubber added:2783 2925 3046 2845 2858(a) What are the values of the sample mean x and sample median x tilde?(b) Calculate the sample variance using the computational formula.[Hint: First subtract a convenient number from each observation.] (Round your answer to the nearest whole number.)

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1. 27 February, 03:45
0
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