Question

Dan has a choice of two buses to get to school: Number 50 or Number 63. Over a month he kept a record of the number of minutes each bus was late when it set off from his home bus stop: no. 50-4,2,0,6,4,8,8,6,3,9 no.63-3,4,0,10,3,5,13,1,0,1 a. For each bus calculate the mean number of minutes late. b. Find the range for each bus. c. Which bus would you advise dan to catch? Give a reason for your answer.

Answer 1

Bus No. 50
The mean is calculated as:

10 = 4 +2 +0+ ... + 9 10 50 10 = 5 10

$\small Range=Max(x_i)-Min(x_i)=9-0=9$

Bus No. 63
$\small \overline{x}=\frac{\sum_{i=1}^{10}x_i}{10}=\frac{3+4+0+...+1}{10}=\frac{40}{10}=4$
$\small Range=Max(x_i)-Min(x_i)=13-0=13$

It is observed that, though, on average Bus No. 63 is late by only 4 minutes on an average, whereas, Bus No. 50 is late by 5 minutes on an average.
However, the Bus No. 63 has a higher range of ariving late (13 minutes) whereas, Bus No. 50 has a lesser range of  arriving late (9 minutes).
Thus, it is observed that Bus No. 50 and Bus No. 63 has a closer mean late time, whereas their range has a large difference. So, it is wiser to choose Bus No. 50 and be late by one minute more, than to choose a bus which have a higher dispersion.

Hence, the feasible choice is Bus No. 50.

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