Consider the following information: Observations 1 2 3 4 5 6 Num. of defects 10 18 13 15 9 12 The number of runs above and below the sample median is:

There are 3 runs above and below the sample mean

Step-by-step explanation:

From  the given information; we have a table that can be computed as:

Observations              1      2       3      4     5     6

Number of defects     10     18    13    15     9     12

If we sort the order in ascending order; we have:

Observations              5      1       6      3     4     2

Number of defects     9     10    12    13     15     18

The median is the  middle number; here the integers are even :

Thus, the median is  (12 + 13)/2 = 12.5

We need to make counts from the table given above to determine if they are below or above the median

Observations              1      2       3      4     5     6

Number of defects     10     18    13    15     9     12

Above(A) /Below (B)    B     A       A     A    B      B

To count for the number of runs from the  beginning and when there is a change of runs from A to B; we have:  

Observations              1      2       3      4     5     6

Number of defects     10     18    13    15     9     12

Above(A) /Below (B)    B     A       A     A    B      B

Runs                             1        2      -      -      3      -

Therefore, from the analysis, there are 3 runs above and below the sample mean

i think it's either c or d

step-by-step explanation: the entire question is extremely odd. 04 obviously refers to the interest rate. honestly i will say the answer is c. because the (2 x 12) part would mean it's compounded semiannually. i'll say choose c.

19

step-by-step explanation: