(1) PROBABILITY
(2) FREQUENCY
(3) HISTOGRAM
(4) DATA
(5) EVENT

(i) The proportion of the sector for hours:
in sleeping = \(Number\ of\ sleeping\ hours\over Whole\ day\)
\(={8\ hours\over 24\ hours}={1\over3}\)
in school =\({6\ hours\over 24\ hours}={1\over4}\)
in home work = \({4\ hours\over 24\ hours}={1\over 6}\)
in play = \({3\ hours\over 24\ hours}={1\over 8}\)
in others = \({3\ hours\over 24\ hours}={1\over 8}\)
(ii) ∵ Total angle at the centre of a circle
= 360°
i.e., Central angle for 1 (whole) = 360°
∴ Central angle for play
(\(1\over8\)th part of Whole) = \(1\over8\) x 360° = 45°
Similarly, central angle for home work
(\(1\over6\) th part of Whole) = \(1\over6\) x 360° = 60°

(a) 16%
(b) 63000

(a) 70000
(b) 80000

(a) Percentage of unutilised capacity = \(\frac{Production\quad capacity-production}{production\quad capacity}\times 100\)
Onida = \(\frac{20}{200}\times\) 100 = 10%
Zenith = \(\frac{70}{250}\times\) 100 = 28%
Excel = \(\frac{20}{300}\times\) 100 = 6.67%
Videocon = \(\frac{60}{320}\times\)100 = 18.75%
BPL  = \(\frac{50}{150}\times\) 100 = 33.3%
Hence, BPL had maximum unutilised capacity.
(b) Total production = 180+ 180 + 280 + 260 + 100 = 1000
:. Required percentage = \(\frac{280}{1000}\times\) 100 = 18%

(a) Angle made by the expenditure on bricks
= 360^o - (45^o + 100^o + 90^o + 75^o)
= 360^o - 310^o = 50^o
Thus, expenditure on bricks
= \(\frac{50}{360}\times \) 540000 = Rs 75000
(b) Expenditure on bricks = Rs 75000
Expenditure on timber = \(\frac{90}{360}\times 540000\)
= \(\frac{540000}{4}\) = Rs 135000
\(\therefore\) Difference between expenditure on bricks and expenditure on timber = 135000 - 75000 = Rs 60000
Hence, expenditure on bricks is Rs 60000 less than expenditure on timber.

(a) P (Suhail winning) = \(\frac{1}{7}\)
(b) P (Suhail winning now) = \(\frac{4}{15}\)
                         [\(\therefore\) there are 4 face cards]
    P (Reshma winning) = \(\frac{4}{15}\)
                       [\(\therefore\) there are four 4 numbers card]

Central angle for oceans = \((\frac{Value\quad of\quad the\quad component}{Sum\quad of\quad the\quad components}\times 360)^o\)
For pacific ocean = 40% = \(\frac{40}{100}\times 360^o\) = 144^o
For atlantic ocean = 30% = \(\frac{30}{100}\times 360^o\) = 108^o
For Indian ocean = 20% = \(\frac{20}{100}\times 360^o\) = 72^o
For others ocean = 10% = \(\frac{10}{100}\times 360^o\) = 36^o
On the basis of above data we can draw the following pie chart:

Here, the types of shoppers are Wom.an (W), Man (M), Boy (B) or Girl (G). Then, frequency distribution table for the above data is given below:

Types of shoppers	Tally marks	Frequency   (Number of shoppers)
W		28
M		15
B		5
G		12
Total		60

Now, draw two perpendicular axes OX and OY and take types of shoppers along OX and frequency i.e. number of shoppers along OY. Choose a suitable scale for OX and OY and draw bars of equal gaps in between. Thus, we get the following bar graph:

Here, to represent the given information, we draw the bar graph. So, draw two perpendicular axes OX and OY. Take months on OX and number of watches sold on OY. Choose a suitable scale for OX and OY and drawbars of equal width with equal gaps in between.
Thus, we get the following bar graph