Here for detecting outlier we will first write the table in ascending order
0 |
0 |
0 |
65 |
71 |
151 |
184 |
226 |
257 |
280 |
304 |
320 |
332 |
355 |
389 |
459 |
567 |
587 |
595 |
3027 |
Here as n = 20
Qquartile 1 = Q1 = (5th number + 6th number)/2 = (71 + 151)/2 = 111
Quartile 2 = Q3 = (15th number + 16th number) = (389+459)/2 = 424
Inter quartile range = 424 -111 = 313
so here as we know that any number will be an outlier if it is outside the limits of Q1 - 1.5 IQR or Q3 + 1.5 IQR
so here
Q1 - 1.5 IQR = 111 - 1.5 * 313 = -358.5
Q3 + 1.5IQR = 424 + 1.5 * 313 = 893.5
so here as we see that number 3027 is an outlier as it is outside the limit.
(b) Here option b graph is correct.
(c) Here part (c) is not given.
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