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Q1. Construct a cumulative frequency distribution table of the following distribution: C.I. 12.5-17.5 17.5-22.5 22.5-27.5 27.5-32.5 32.5-37.5 F 2 22 19 14 13   C.I. 12.5-17.5 17.5-22.5 22.5-27.5 27.5-32.5 32.5-37.5 F 2 22 19 14 13
Solution
C.I. F C.F. 12.5-17.5 2 2 17.5-22.5 22 24 22.5-27.5 19 43 27.5-32.5 14 57 32.5-37.5 13 70
Q2. Find the mode of the following data. X 10 12 14 16 18 20 f 5 3 10 3 2 1
Solution
We observe that the value 14 has the maximum frequency i.e. 10. Hence, the modal value is 14.
Q3. Write the frequency distribution table for the following data: Marks Above 0 above 10 above 20               Above 30 Above 40 Above 50 No. of students 30 28 21 15 10 0
Solution
The frequency distribution table is as follows: Marks No. of students 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 2 7 6 5 10 Total 30
Q4. Convert the given frequency distribution to a more than type cumulative frequency distribution. Marks 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90 No. of students 2 2 3 4 6 6 5 2 4
Solution
Marks No. of students More than or equal to 80 4 More than or equal to 70 6 More than or equal to 60 11 More than or equal to 50 17 More than or equal to 40 23 More than or equal to 30 27 More than or equal to 20 30 More than or equal to 10 32 More than or equal to 0 34
Q5. Write any two merits and demerits of arithmetic mean.
Solution
Merits: (1) It is based on all observations (2) It is simple to understand and calculate Demerits: (1) it is affected by extreme values (2) it cannot be determined graphically.
Q6. Construct a more than cumulative frequency distribution table for the given data: Class Interval 50 - 60 60 - 70 70 - 80 80 - 90 90 - 100 100 - 110 Frequency 12 15 17 21 23 19
Solution
Class Interval Cumulative Frequency More than 50 More than 60 More than 70 More than 80 More than 90 More than 102 107 95 80 63 42 19
Q7. The following distribution gives the daily income of 50 workers of a factory. Daily income (in Rs) 10-12 12-14 14-16 16-18 18-20 No. of workers 11 4 2 3 5 Convert the above distribution to a less than type cumulative frequency distribution.
Solution
Daily income (in Rs) C.F. Less than 12 11 Less than 14 15 Less than 16 17 Less than 18 20 Less than 20 25

Q8. In the following distribution: Monthly income range (In Rs.) No. of families Income more than Rs 10000 100 Income more than Rs 13000 85 Income more than Rs 16000 69 Income more than Rs 19000 50 Income more than Rs 22000 33 Income more than Rs 25000 15   Find the number of families having income range (in Rs.) 16000-19000?
Solution
Monthly income range (In Rs.) No. of families 10000-13000 15 13000-16000 16 16000-19000 19 19000-22000 17 22000-25000 18 25000-28000 15 No. of families having income range (in Rs.) 16000-19000 is 19.
Q9. The following table shows the heights of 50 boys: Height (cm) 120 121 122 123 124 Frequency 5 8 18 10 9 Find the mode of heights.
Solution
We observe that the height 122 cm has the highest frequency of 18. Hence, the modal value of height is 122 cm.
Sample Paper (extra circulation)
1.   Write the empirical relation between mean, mode and median. (AI CBSE 2009 C)
2.   Write the median class of the following distribution :( CBSE 2009)
Class
Frequency
0-10
4
10-20
4
20-30
8
30-40
10
40-50
12
50-60
8
60-70
4
3.   What is the modal class of the following distribution? (CBSE 2009)
Age (in years)
Number of patients
0-10
16
10-20
13
20-30
6
30-40
11
40-50
27
50-60
18
4.   Find the median (AI CBSE 2008)
Marks
Frequency
0-10
8
10-20
10
20-30
12
30-40
22
40-50
30
50-60
18
5.   Find the class marks of classes 10-25 and 35-55 (AI 2008F)
6.   What is the median class of the grouped data (AI CBSE 2008 C?)
Class
Frequency
128-135
8
135-142
5
142-149
9
149-156
12
156-163
5
163-170
1
7.   Find the missing frequencies when the mean of dats is 53. (CBSE 2008)
Age (in
years)
0-20
20-40
40-60
60-80
80-100
Total
Number of
people
15
F1
21
F2
17
100
8.   The following table gives production yield per hectare of wheat of 100 farms of a village. (AI CBSE 2009 C)
Production yield
Number of farms
40-45
4
45-50
6
50-55
16
55-60
20
60-65
30
65-70
24
        Change the distribution to a ‘more than type’ distribution and draw its ogive.
9.   The distribution below gives the weights of 30 students of a class. Find the mean and median weight of students (CBSE 2009 C)
Weight (in kgs)
Number of students
40-45
2
45-50
3
50-55
8
55-60
6
60-65
6
65-70
3
70-75
2
10.   The lengths of 40 leaves of a plant are measured correct up to the nearest millimeter and the data is as under:
Length (in mm)
Number of leaves
118-126
4
126-134
5
134-142
10
142-150
12
150-158
4
158-166
5
               Find the mean and median length of the leaves (CBSE 2009 C)

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Q 1. Verify whether 2 is a zero of the polynomial x 3  + 4x 2  - 3x - 18 or not? Solution p(x) = x 3  + 4x 2  - 3x - 18 Now, 2 will be a zero of the polynomial if p(2) = 0. We have: p(2) = (2) 3  + 4(2) 2  - 3(2) - 18        = 8 + 16 - 6 - 18        = 24 - 24        = 0 Thus, 2 is a zero of the polynomial x 3  + 4x 2  - 3x - 18. Q 2. Complete the following. (i) A polynomial of degree three is called a …….polynomial. (ii) A polynomial of degree zero is called a …….polynomial. (iii) A polynomial of degree two is called a …….polynomial. (iv) A polynomial of degree one is called a …….polynomial. Solution (i) A polynomial of degree three is called a cubic polynomial.   (ii) A polynomial of degree zero is called a constant polynomial.   (iii) A polynomial of degree two is called a quadratic polynomial.   (iv) A polynomial of degree one is called a linear polynomial. ...