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#### karush

##### Well-known member

- Jan 31, 2012

- 3,119

(a) I presume the frequency row has to equal 100

so

\(\displaystyle k=100 - (26 + 10 + 20 + 29 + 11)= 4\)

(b)(i) again presume the median is based on frequency and on ordered list

so

median of $4\ 10\ 11 \ 20\ 26\ 29 = \frac{31}{2}$ or $15.5$

(ii) interquartile range? isn't this data list 100 numbers long?

or is $Q_1=10$ and $Q_3=26$ so interquartile range$=26-10=13$