# Case Study Question 05

## Chapter 14: Statistics

### Class 10

Distance Analysis of Public Transport Buses

Transport department of a city wants to buy some Electric buses for the city. For which they want to analyse the distance travelled by existing public transport buses in a day.
The following data shows the distance travelled by 60 existing public transport buses in a day.

Question.1.
The upper limit of a class and lower limit of its succeeding class is differ by

(a) 9
(b) 1
(c) 10
(d) none of these

(b) : The upper limit of a class and the lower class of its succeeding class differ by 1.

Question.2.
The median class is

(a) 229.5-239.5
(b) 230-239
(c) 220-229
(d) 219.5-229.5

(d) : Here, class intervals are in inclusive form. So, we first convert them in exclusive form. The frequency distribution table in exclusive form is as follows :

Question.3.
The cumulative frequency of the class preceding the median class is

(a) 14
(b) 18
(c) 26
(d) 10

(b) : Clearly, the cumulative frequency of the class preceding the median class is 18.

Question.4.
The median of the distance travelled is

(a) 222 km
(b) 225 km
(c) 223 km
(d) none of these

(d) : Median = l+\left[ \frac{\frac{N}{2}-c.f.}{f}\right] \times h
= 219.5+\left[ \frac{30-18}{26}\right] \times 10
= 219.5+\left[ \frac{12\times10}{26}\right]
= 219.5 + 4.62
= 224.12
\therefore Median of the distance travelled is 224.12 km

Question.5.
If the mode of the distance travelled is 223.78 km, then mean of the distance travelled by the bus is

(a) 225 km
(b) 220 km
(c) 230.29 km
(d) 224.29 km

(d) : We know, Mode = 3 Median – 2 Mean
\therefore Mean = \frac{1}{2}(3 Median – Mode)
= \frac{1}{2}(672.36-223.78)
= 224.29 km

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