Construct the following frequency distribution :

Class	Frequency
0 - 5	13
6 - 11	10
12 - 17	15
18 - 23	8
24 - 29	11

The upper limit of the median class is

1. 17

2. 17.5

3. 18

4. 18.5

Converting the discontinuous interval into continuous interval.

Adjustment factor = (Lower limt of one class - Upper limit of previous class) / 2

$= \dfrac{11 - 10}{2} \\[1em] = \dfrac{1}{2} \\[1em] = 0.5$

We construct the cumulative frequency distribution table as under :

Here n (total no. of observations) = 57.

As n is odd,

$\therefore \text{Median } = \dfrac{n + 1}{2} \text{th observation} \\[1em] = \dfrac{57 + 1}{2} \\[1em] = \dfrac{58}{2} \\[1em] = 29 \text{th observation}$

As observation from 24^th to 38^th lies in the class 11.5 - 17.5

∴ Median class = 11.5 - 17.5, with upper limit = 17.5

Hence, Option 2 is the correct option.