Three sets of data on comparable situationsare available as under: Set Number Number of data Mean Value Standard Deviation 1 9 8 1.6 2 12 7 1.2 3 15 9 1.4 Considering all the data sets together, the overall mean value would be:

Statistics Practice Test 1

Question : 6 of 37

Marks: +1, -0

Three sets of data on comparable situationsare available as under:

Set Number	Number of data	Mean Value	Standard Deviation
1	9	8	1.6
2	12	7	1.2
3	15	9	1.4

Considering all the data sets together, the overall mean value would be:

Solution:

Sum of number of data = 9 + 12 + 55 = 36
Sum of mean value = 9 × 8 + 12 × 7 + 15 × 9 = 72 + 84 + 135 = 291
Mean value

=291/36=8.08

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