Your cart is empty.
Practice each question, then open the answer or worked solution.
The blood groups of 30 students are recorded as follows:
A, B, O, A, AB, O, A, O, B, A, O, B, A, AB, B, A, AB, B,
A, A, O, A, AB, B, A, O, B, A, B, A
Prepare a frequency distribution table for the data.
The frequency distribution table for the data is as follows:
| Blood group | Frequency |
| A | 12 |
| AB | 4 |
| B | 8 |
| O | 6 |
| Total | 30 |
The value of π upto 35 decimal places is given below:
3. 14159265358979323846264338327950288
Make a frequency distribution of the digits 0 to 9 after the decimal point.
The frequency distribution of the digits 0 to 9 after the decimal point is as follows:
| Digits | Frequency |
| 0 | 1 |
| 1 | 2 |
| 2 | 5 |
| 3 | 6 |
| 4 | 3 |
| 5 | 4 |
| 6 | 3 |
| 7 | 2 |
| 8 | 5 |
| 9 | 4 |
The scores (out of 100) obtained by 33 students in a mathematics test are as follows:
69, 48, 84, 58, 48, 73, 83, 48, 66, 58, 84 000
66, 64, 71, 64, 66, 69, 66, 83, 66, 69, 71
81, 71, 73, 69, 66, 66, 64, 58, 64, 69, 69
Represent this data in the form of a frequency distribution.
The given data in the question in a frequency distribution table can be represented as:
| Scores | N/O students
Frequency |
| 48 | 3 |
| 58 | 3 |
| 64 | 4 |
| 66 | 7 |
| 69 | 6 |
| 71 | 3 |
| 73 | 2 |
| 81 | 1 |
| 83 | 2 |
| 84 | 2 |
Prepare a continuous grouped frequency distribution from the following data:
Also find the size of class intervals.
If m is mid-point of a class and h is the class size,
We know that,
Lower limits of the class intervals = m – h/2
Upper limits of the class intervals = m + h/2.
Class size (h) = 15 – 5 = 10 …(i)
Then,
The class interval formed for the mid-point 5 = (5 – 10/2) – (5 + 10/2)
= 0 – 10
Continuing in the same manner, the continuous class table is:
| Mid-point | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 |
| Frequency | 4 | 8 | 13 | 12 | 6 |
The size of class intervals = 10 (from equation 1)
Convert the given frequency distribution into a continuous grouped frequency distribution:
In which intervals would 153.5 and 157.5 be included?
Consider the classes 150 – 153 and 154 – 157.
The lower limit of 154 – 157 = 154
The upper limit of 150 – 153 = 153
The difference = 154 – 153 = 1
Half the difference = 1/2 = 0.5
Therefore,
The new class interval formed from 150 – 153 = (150 – 0.5) – (153 + 0.5)
= 149.5 – 153.5.
Continuous classes formed can be represented in a table as:
| Class Interval | Frequency |
| 149.5 – 153.5 | 7 |
| 153.5 – 157.5 | 7 |
| 157.5 – 161.5 | 15 |
| 161.5 – 165.5 | 10 |
| 165.5 – 169.5 | 5 |
| 169.5 – 173.5 | 6 |
Hence, from the table,
It is clear that 153.5 belongs to the class interval 153.5 – 157.5 and 157.5 belongs to the class interval 157.5 – 161.5.
The expenditure of a family on different heads in a month is given below:
Draw a bar graph to represent the data above.
Expenditure on Education of a country during a five year period (2002-2006), in crores of rupees, is given below:
Represent the information above by a bar graph.
The given data can be represented by a bar graph as follows:
The following table gives the frequencies of most commonly used letters a, e, i, o, r, t, u from a page of a book:
Represent the information above by a bar graph.
The given data can be represented by a bar graph as follows:
If the mean of the following data is 20.2, find the value of p:
According to the question,
Mean = 20.2
So,
∑fixi/∑fi=20.2
So we get,
∑fixi = (10×6)+(15×8)+(20×p)+(25×10+)(30×6)
=610+20p
We also get,
∑fi=6+8+p+10+6
=30+p
Therefore,
∑fixi/∑fi=20.2
(610+20p)/30+p=20.2
20.2(30+p) =610+20p
606+20.2p=610+20p
20.2p-20p=610-606
0.2p=4
p=4/0.2
p=20
Obtain the mean of the following distribution:
| xi | fi | xifi |
| 4 | 4 | 16 |
| 6 | 8 | 48 |
| 8 | 14 | 112 |
| 10 | 11 | 110 |
| 12 | 3 | 36 |
| Total | 40 | 322 |
Mean = ∑xifi/∑fi = 322/40 = 8.05