The following table shows the distribution of marks obtained by some students in an examination. (a) Construct a cumulative frequency table for the distribution (b) Draw an ogive for the distribution (c) Use your graph in (b) to determine : (i) semi- interquartile range ; (ii) number of students who failed, if the pass mark for the examination is 37 ; (iii) probability that a student selected at random scored between 20% and 60%.

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The following table shows the distribution of marks obtained by some students in an examination.

Marks	0-9	10-19	20-29	30-39	40-49	50-59	60-69	70-79	80-89	90-99
Frequency	50	50	40	60	100	100	50	25	15	10

(a) Construct a cumulative frequency table for the distribution
(b) Draw an ogive for the distribution
(c) Use your graph in (b) to determine : (i) semi- interquartile range ; (ii) number of students who failed, if the pass mark for the examination is 37 ; (iii) probability that a student selected at random scored between 20% and 60%.

Explanation

(a)

Marks(x)	No of students(f)	Class boundaries	Cum Freq $(\sum f)$
0-9	50	-0.5 - 9.5	50
10-19	50	9.5 - 19.5	100
20-29	40	19.5 - 29.5	140
30-39	60	29.5 - 39.5	200
40-49	100	39.5 - 49.5	300
50-59	100	49.5 - 59.5	400
60-69	50	59.5 - 69.5	450
70-79	25	69.5 - 79.5	475
80-89	15	79.5 - 89.5	490
90-99	10	89.5 - 99.5	500

(c)(i) Position of lower quartile,

Q_{1} = \frac{\sum f + 1}{4} = \frac{500 + 1}{4}

\frac{501}{4} = 125.25 t h

position.

∴ Q_{1} = 26.0

Position of upper quartile,

Q_{3} = 3 (\frac{\sum f + 1}{4}) = 3 (125.25) = 375.75 t h

position.

∴ Q_{3} = 56.3

Semi-interquartile range =

\frac{Q_{3} - Q_{1}}{2} = \frac{56.3 - 26}{2}

\frac{30.3}{2} = 15.15

(ii) If the pass mark is 37, then 190 students failed.
(iii) Those who scored between 0% and 20% = 100
Those who scored between 0% and 60% = 400
Those who scored between 20% and 60% = 400 - 100 = 300.
P(20% < x < 60%) =

\frac{300}{500} = \frac{3}{5}

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