Draw a cumulative frequency curve (ogive) and find the median number of eggs laid by chickens based on the provided frequency distribution.

To draw the cumulative frequency curve, plot the cumulative frequency ($cf$) against the upper class boundaries.
The points to plot are: $(44.5, 0)$, $(49.5, 10)$, $(54.5, 46)$, $(59.5, 110)$, $(64.5, 162)$, $(69.5, 190)$, and $(74.5, 200)$.
Connect these points with a smooth curve. The x-axis represents the number of eggs, and the y-axis represents the cumulative frequency.

The median is the value at the $\frac{N}{2}$-th position.
Median position $= \frac{200}{2} = 100$.
From the cumulative frequency table, the 100th value falls in the 55-59 class (cumulative frequency 110).
Median class: 55-59 (boundaries 54.5 - 59.5).
Lower boundary of median class ($L$) = 54.5
Frequency of median class ($f_m$) = 64
Cumulative frequency of class before median class ($cf_b$) = 46
Class width ($c$) = 59.5 - 54.5 = 5

The first quartile is the value at the $\frac{N}{4}$-th position