A teacher recorded the individual time taken by $200$ students to complete a cross-country race. T

Question

A teacher recorded the individual time taken by $200$ students to complete a cross-country race. The results are presented in the form of a cumulative frequency curve. Use the curve to estimate the median time taken to run the race,[1] the interquartile range of the distribution,[2] the percentage of students who took longer than $66$ min to finish the race,[2] the probability of choosing a student who took $45$ min or less time to finish the race.[2] The frequency table relates to the same cross-country race results. FrequencyTime ($t$ min) $13$$30 < t \le 40$ $r$$40 < t \le 50$ $90$$50 < t \le 60$ $s$$60 < t \le 70$ $9$$70 < t \le 80$ $4$$80 < t \le 90$ Find the value of $r$ and the value of $s$.[2] Showing your method clearly, estimate the mean time required to run the race.[3] Estimate the standard deviation of the distribution.[2]

Accepted Answer

(a)(i) $52$ min (a)(ii) $10$ min (a)(iii) $10\%$ (a)(iv) $\frac{19}{100}$ (b)(i) $r=62,\ s=22$ (b)(ii) $53.2$ min (b)(iii) $10.0$ min

Frequency	Time (\(t\) min)
\(13\)	\(30 < t \le 40\)
\(r\)	\(40 < t \le 50\)
\(90\)	\(50 < t \le 60\)
\(s\)	\(60 < t \le 70\)
\(9\)	\(70 < t \le 80\)
\(4\)	\(80 < t \le 90\)

Statistics - Mean and Standard Deviation for Grouped Data

Statistics - Mean and Standard Deviation for Grouped Data

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