EXERCISE 15.2
Page No 275:
Question 1:
The following table shows the ages of the patients admitted in a hospital during a year:
age (in years) | 5 − 15 | 15 − 25 | 25 − 35 | 35 − 45 | 45 − 55 | 55 − 65 |
Number of patients | 6 | 11 | 21 | 23 | 14 | 5 |
Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.
Answer:
To find the class marks (x_{i}), the following relation is used.
Taking 30 as assumed mean (a), d_{i} and f_{i}d_{i}are calculated as follows.
Age (in years) | Number of patients f_{i} | Class mark x_{i} | d_{i} = x_{i} − 30 | f_{i}d_{i} |
5 − 15 | 6 | 10 | − 20 | − 120 |
15 − 25 | 11 | 20 | − 10 | − 110 |
25 − 35 | 21 | 30 | 0 | 0 |
35 − 45 | 23 | 40 | 10 | 230 |
45 − 55 | 14 | 50 | 20 | 280 |
55 − 65 | 5 | 60 | 30 | 150 |
Total | 80 | 430 |
From the table, we obtain
Mean of this data is 35.38. It represents that on an average, the age of a patient admitted to hospital was 35.38 years.
It can be observed that the maximum class frequency is 23 belonging to class interval 35 − 45.
Modal class = 35 − 45
Lower limit (l) of modal class = 35
Frequency (f_{1}) of modal class = 23
Class size (h) = 10
Frequency (f_{0}) of class preceding the modal class = 21
Frequency (f_{2}) of class succeeding the modal class = 14
Mode =
Mode is 36.8. It represents that the age of maximum number of patients admitted in hospital was 36.8 years.
Video Solution
Question 2:
The following data gives the information on the observed lifetimes (in hours) of 225 electrical components:
Lifetimes (in hours) | 0 − 20 | 20 − 40 | 40 − 60 | 60 − 80 | 80 − 100 | 100 − 120 |
Frequency | 10 | 35 | 52 | 61 | 38 | 29 |
Determine the modal lifetimes of the components.
Answer:
From the data given above, it can be observed that the maximum class frequency is 61, belonging to class interval 60 − 80.
Therefore, modal class = 60 − 80
Lower class limit (l) of modal class = 60
Frequency (f_{1}) of modal class = 61
Frequency (f_{0}) of class preceding the modal class = 52
Frequency (f_{2}) of class succeeding the modal class = 38
Class size (h) = 20
Therefore, modal lifetime of electrical components is 65.625 hours.
Video Solution
Question 3:
The following data gives the distribution of total monthly household expenditure of 200 families of a village. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure.
Expenditure (in Rs) | Number of families |
1000 − 1500 | 24 |
1500 − 2000 | 40 |
2000 − 2500 | 33 |
2500 − 3000 | 28 |
3000 − 3500 | 30 |
3500 − 4000 | 22 |
4000 − 4500 | 16 |
4500 − 5000 | 7 |
Answer:
It can be observed from the given data that the maximum class frequency is 40, belonging to 1500 − 2000 intervals.
Therefore, modal class = 1500 − 2000
Lower limit (l) of modal class = 1500
Frequency (f_{1}) of modal class = 40
Frequency (f_{0}) of class preceding modal class = 24
Frequency (f_{2}) of class succeeding modal class = 33
Class size (h) = 500
Therefore, modal monthly expenditure was Rs 1847.83.
To find the class mark, the following relation is used.
Class size (h) of the given data = 500
Taking 2750 as assumed mean (a), d_{i}, u_{i}, and f_{i}u_{i}are calculated as follows.
Expenditure (in Rs) | Number of families f_{i} | x_{i} | d_{i} = x_{i} − 2750 | f_{i}u_{i} | |
1000 − 1500 | 24 | 1250 | − 1500 | − 3 | − 72 |
1500 − 2000 | 40 | 1750 | − 1000 | − 2 | − 80 |
2000 − 2500 | 33 | 2250 | − 500 | − 1 | − 33 |
2500 − 3000 | 28 | 2750 | 0 | 0 | 0 |
3000 − 3500 | 30 | 3250 | 500 | 1 | 30 |
3500 − 4000 | 22 | 3750 | 1000 | 2 | 44 |
4000 − 4500 | 16 | 4250 | 1500 | 3 | 48 |
4500 − 5000 | 7 | 4750 | 2000 | 4 | 28 |
Total | 200 | − 35 |
From the table, we obtain
Therefore, mean monthly expenditure was Rs 2662.50.
Video Solution
Page No 276:
Question 4:
The following distribution gives the state-wise teacher-student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret the two measures.
Number of students per teacher | Number of states/U.T |
15 − 20 | 3 |
20 − 25 | 8 |
25 − 30 | 9 |
30 − 35 | 10 |
35 − 40 | 3 |
40 − 45 | 0 |
45 − 50 | 0 |
50 − 55 | 2 |
Answer:
It can be observed from the given data that the maximum class frequency is 10 belonging to class interval 30 − 35.
Therefore, modal class = 30 − 35
Class size (h) = 5
Lower limit (l) of modal class = 30
Frequency (f_{1}) of modal class = 10
Frequency (f_{0}) of class preceding modal class = 9
Frequency (f_{2}) of class succeeding modal class = 3
It represents that most of the states/U.T have a teacher-student ratio as 30.6.
To find the class marks, the following relation is used.
Taking 32.5 as assumed mean (a), d_{i}, u_{i}, and f_{i}u_{i} are calculated as follows.
Number of students per teacher | Number of states/U.T (f_{i}) | x_{i} | d_{i} = x_{i} − 32.5 | f_{i}u_{i} | |
15 − 20 | 3 | 17.5 | − 15 | − 3 | − 9 |
20 − 25 | 8 | 22.5 | − 10 | − 2 | − 16 |
25 − 30 | 9 | 27.5 | − 5 | − 1 | − 9 |
30 − 35 | 10 | 32.5 | 0 | 0 | 0 |
35 − 40 | 3 | 37.5 | 5 | 1 | 3 |
40 − 45 | 0 | 42.5 | 10 | 2 | 0 |
45 − 50 | 0 | 47.5 | 15 | 3 | 0 |
50 − 55 | 2 | 52.5 | 20 | 4 | 8 |
Total | 35 | − 23 |
Therefore, mean of the data is 29.2.
It represents that on an average, teacher−student ratio was 29.2.
Video Solution
Question 5:
The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches.
Runs scored | Number of batsmen |
3000 − 4000 | 4 |
4000 − 5000 | 18 |
5000 − 6000 | 9 |
6000 − 7000 | 7 |
7000 − 8000 | 6 |
8000 − 9000 | 3 |
9000 − 10000 | 1 |
10000 − 11000 | 1 |
Find the mode of the data.
Answer:
From the given data, it can be observed that the maximum class frequency is 18, belonging to class interval 4000 − 5000.
Therefore, modal class = 4000 − 5000
Lower limit (l) of modal class = 4000
Frequency (f_{1}) of modal class = 18
Frequency (f_{0}) of class preceding modal class = 4
Frequency (f_{2}) of class succeeding modal class = 9
Class size (h) = 1000
Therefore, mode of the given data is 4608.7 runs.
Video Solution
Question 6:
A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised it in the table given below. Find the mode of the data:
Number of cars | 0 − 10 | 10 − 20 | 20 − 30 | 30 − 40 | 40 − 50 | 50 − 60 | 60 − 70 | 70 − 80 |
Frequency | 7 | 14 | 13 | 12 | 20 | 11 | 15 | 8 |
Answer:
From the given data, it can be observed that the maximum class frequency is 20, belonging to 40 − 50 class intervals.
Therefore, modal class = 40 − 50
Lower limit (l) of modal class = 40
Frequency (f_{1}) of modal class = 20
Frequency (f_{0}) of class preceding modal class = 12
Frequency (f_{2}) of class succeeding modal class = 11
Class size = 10
Therefore, mode of this data is 44.7 cars.