The following table gives the frequency distribution of 800 candidates in an examination:
0-10
10-20
20-30
30-40
40-50
50-60
Marks:
60-70
70-80 80-90 90-100
10
40
80
140
170
130
100
70
40
20
No. of candidates
Draw less than ogive and answer the following questions from the graph:
(1) If the minimum works required for passing are 35, what percentage of candidates pass
the examination?
(ii) If it is decided to allow 80% of the candidates to pass, what should be the minimum
marks for passing?
Answers
Answer:
Here we have, the cumulative frequency distribution. So, first we convert it into an ordinary frequency distribution. we observe that are 80 students getting marks greater than or equal to 0 and 77 students have secured 10 and more marks. Therefore, the number of students getting marks between 0 and 10 is 80-77= 3.
Similarly, the number of students getting marks between 10 and 20 is 77-72= 5 and so on. Thus, we obtain the following frequency distribution.
Marks Number of students
0-10 3
10-20 5
20-30 7
30-40 10
40-50 12
50-60 15
60-70 12
70-80 6
80-90 2
90-100 8
Now, we compute mean arithmetic mean by taking 55 as the assumed mean.
Computative of Mean
Marks
(x
i
) Mid-value (f
i
) Frequency u
i
10
x
i
−55
f
i
u
i
0-10 5 3 -5 -15
10-20 15 5 -4 -20
20-30 25 7 -3 -21
30-40 35 10 -2 -20
40-50 45 12 -1 -20
50-60 55 15 0 0
60-70 65 12 1 12
70-80 75 6 2 12
80-90 85 2 3 6
90-100 95 8 4 32
Total
∑f
i
=80 ∑f
i
u
i
= -26
We have,
N= sumf
i
=80,∑f
i
u
i
=−26, A= 55 and h= 10
∴
X
=A+h[
N
1
∑f
i
u
i
]
⇒
X
=A+h[
N
1
∑f
i
u
i
]
⇒
X
=55+10×
80
−26
=55−3.25=51.75Marks.
Answered By anshika Awasthi