Question

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:

Answer 1

Answer:

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

Answer 2

The median element of a dataset is the middle element of the dataset.

There are 3 runs above and below the sample median

The information is given as:

Observations  1      2       3      4     5     6

Defects           10     18    13    15     9     12

In ascending order, the information is represented as:

Defects           9   10   12   13   15   18

Observations  5    1     6    3     4    2

The total number of observations is:

$\mathbf{n = 5 + 1 + 6 + 3+ 4 + 2}$

$\mathbf{n = 21}$

The median element is calculated as:

$\mathbf{Median = \frac{n + 1}{2}th}$

Substitute 21 for n

$\mathbf{Median = \frac{21 + 1}{2}th}$

$\mathbf{Median = \frac{22}{2}th}$

$\mathbf{Median = 11th}$

The 11th element is 12.

So, the median is:

$\mathbf{Median = 12}$

Considering the given information;

Observations  1      2       3      4     5     6

Defects           10     18    13    15     9     12

The defects greater than the median (i.e. 12) are above (A) the median, while the defects less than the median are below (B)

So, we have:

Observations  1      2       3      4     5     6

Defects           10     18    13    15     9     12

Status               B     A       A     A    B      -

Next, we take a count of the number of times the status changes, i.e. from B to A, from A to B, from A or B to -

The count is 3.

Hence, there are 3 runs above and below the sample median

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