Your cart is empty.
Practice each question, then open the answer or worked solution.
Find the mean of the distribution :
Class 1-3 3-5 5-7 7-10
Frequency 9 22 27 17
We first, find the class mark xi of each class and then proceed as follows.
| Class | Class Marks (xi) | Frequency (fi) | fixi |
| 1-3 | 2 | 9 | 18 |
| 3-5 | 4 | 22 | 88 |
| 5-7 | 6 | 27 | 162 |
| 7-10 | 8.5 | 17 | 144.5 |
| Σfi = 75 | Σfixi = 412.5 |
Mean,
Therefore, mean of the given distribution = 5.5.
Calculate the mean of the scores of 20 students in a mathematics test :
Marks 10-20 20-30 30-40 40-50 50-60
Number of students 2 4 7 6 1
We first, find the class mark xi of each class and then proceed as follows
| Class | Class Marks (xi) | Frequency (fi) | fixi |
| 10-20 | 15 | 2 | 30 |
| 20-30 | 25 | 4 | 100 |
| 30-40 | 35 | 7 | 245 |
| 40-50 | 45 | 6 | 270 |
| 50-60 | 55 | 1 | 55 |
| Σfi = 20 | Σfixi = 700 |
Mean,
Therefore, mean of scores of 20 students in mathematics test = 35.
Calculate the mean of the following data :
Class 4 – 7 8 –11 12– 15 16 –19
Frequency 5 4 9 10
The given data is not continuous.
So, we subtract 0.5 from the lower limit and add 0.5 in the upper limit of each class.
| Class | Class Marks (xi) | Frequency (fi) | fixi |
| 3.5 – 7.5 | 5.5 | 5 | 27.5 |
| 7.5 – 11.5 | 9.5 | 4 | 38 |
| 11.5 – 15.5 | 13.5 | 9 | 121.5 |
| 15.5 – 19.5 | 17.5 | 10 | 175 |
| Σfi = 28 | Σfixi = 362 |
Mean,
Therefore, mean of the given data = 12.93.
The following table gives the number of pages written by Sarika for completing her own book for 30 days :
Number of pages written per day 16-18 19-21 22-24 25-27 28-30
Number of days 1 3 4 9 13
Find the mean number of pages written per day.
| Class Marks | Mid – Value (xi) | Number of days (fi) | fixi |
| 15.5 – 18.5 | 17 | 1 | 17 |
| 18.5 – 21.5 | 20 | 3 | 60 |
| 21.5 – 24.5 | 23 | 4 | 92 |
| 24.5 – 27.5 | 26 | 9 | 234 |
| 27.5 – 30.5 | 29 | 13 | 377 |
| Σfi = 30 | Σfixi = 780 |
The given data is not continuous.
Hence, we subtract 0.5 from the lower limit and add 0.5 in the upper limit of each class.
Therefore, the mean of pages written per day = 26.
The daily income of a sample of 50 employees are tabulated as follows :
Income (in Rs) 1-200 201-400 401-600 601-800
Number of employees 14 15 14 7
Find the mean daily income of employees.
| C.I | xi | di = (xi – a) | Fi | fidi |
| 1 – 200 | 100.5 | – 200 | 14 | – 2800 |
| 201 – 400 | 300.5 | 0 | 15 | 0 |
| 401 – 600 | 500.5 | 200 | 15 | 2800 |
| 601 – 800 | 700.5 | 400 | 7 | 2800 |
| Σfi = 50 | Σfixi = 2800 |
∴ Assumed mean, a = 300.5 and di = (xi – a)
= 300.5 + 2800/50
= 356.5
Hence, the average daily income of employees = Rs.356.5
An aircraft has 120 passenger seats. The number of seats occupied during 100 flights is given in the following table :
Number of seats 100-104 104-108 108-112 112-116 116-120
Frequency 15 20 32 18 15
Determine the mean number of seats occupied over the flights.
| Class Interval | Class Marks (xi) | Frequency (fi) | Deviation (di = xi – a) | fidi |
| 100 – 104 | 102 | 15 | – 8 | – 120 |
| 104 – 108 | 106 | 20 | – 4 | – 80 |
| 108 – 112 | 110 | 32 | 0 | 0 |
| 112 – 116 | 114 | 18 | 4 | 72 |
| 116 – 120 | 118 | 15 | 8 | 120 |
| N = Σfi = 100 | Σfidi = – 8 |
∴ Assumed mean, a = 110
Class width, h = 4
And total observations, N = 100
Hence, finding mean,
= 110 + (-8/100)
= 110 – 0.08
= 109.92
But we know that the seats cannot be in decimal.
Therefore, the number of seats = 109.
The weights (in kg) of 50 wrestlers are recorded in the following table :
| Weight (in kg) | 100-110 | 110-120 | 120-130 | 130-140 | 140-150 |
| Number of wrestlers | 4 | 14 | 21 | 8 | 3 |
Find the mean weight of the wrestlers.
| Weight (in kg) | Number of Wrestlers (fi) | Class Marks (xi) | Deviation
(di = xi – a) |
fidi |
| 100 – 110 | 4 | 105 | – 20 | – 80 |
| 110 – 120 | 14 | 115 | – 10 | – 140 |
| 120 – 130 | 21 | 125 | 0 | 0 |
| 130 – 140 | 8 | 135 | 10 | 80 |
| 140 – 150 | 3 | 145 | 20 | 60 |
| N = Σfi = 50 | Σfidi = – 80 |
∴ Assumed mean, (a) = 125
Class width, (h) = 10
and total observations, (N) = 50
By step deviation method,
= 125 – 16
= 123.4kg
Hence, mean weight of wrestlers = 123.4kg
The mileage (km per litre) of 50 cars of the same model was tested by a manufacturer and details are tabulated as given below :
Mileage (km/l) 10-12 12-14 14-16 16-18
Number of cars 7 12 18 13
Find the mean mileage.
The manufacturer claimed that the mileage of the model was 16 km/litre. Do you agree with this claim?
| Mileage (km L-1) | Class –Marks (xi) | Number of cars (fi) | fixi |
| 10 – 12 | 11 | 7 | 77 |
| 12 – 14 | 13 | 12 | 156 |
| 14 – 16 | 15 | 18 | 270 |
| 16 – 18 | 17 | 13 | 221 |
| Total | Σfi = 50 | Σfixi = 724 |
Here, Σfi = 50
Σfixi = 724
= 724/50 = 14.48
Hence, mean mileage = 14.48 km/h
No, I don’t agree with the claim because the manufacturer is claiming mileage 1.52 km/h more than average mileage.
The following is the distribution of weights (in kg) of 40 persons :
Weight (in kg) 40-45 45-50 50-55 55-60 60-65 65-70 70-75 75-80
Number of persons 4 4 13 5 6 5 2 1
Construct a cumulative frequency distribution (of the less than type) table for the data above.
| Weight (in kg) | Cumulative frequency |
| Less than 45 | 4 |
| Less than 50 | 4 + 4 = 8 |
| Less than 55 | 8 + 13 = 21 |
| Less than 60 | 21 + 5 = 26 |
| Less than 65 | 26 + 6 = 32 |
| Less than 70 | 32 + 5 = 37 |
| Less than 75 | 37 + 2 = 39 |
| Less than 80 | 39 + 1 = 40 |
The following table shows the cumulative frequency distribution of marks of 800 students in an examination:
Marks Number of students
Below 10 10
Below 20 50
Below 30 130
Below 40 270
Below 50 440
Below 60 570
Below 70 670
Below 80 740
Below 90 780
Below 100 800
Construct a frequency distribution table for the data above.
The frequency distribution table for the given data is:
| Class Interval | Number of students |
| 0-10 | 10 |
| 10-20 | 50 – 10 = 40 |
| 20-30 | 130 – 50 = 80 |
| 30-40 | 270 – 130 = 140 |
| 40-50 | 440 – 270 = 170 |
| 50-60 | 570 – 440 = 130 |
| 60-70 | 670 – 570 = 100 |
| 70-80 | 740 – 670 = 70 |
| 80-90 | 780 – 740 = 40 |
| 90-100 |
|