EXERCISE 14.1
1. A survey was conducted by a group of students as a part of their environment awareness programme, in which they collected the following data regarding the number of plants in 20 houses in a locality. Find the mean number of plants per house.
Number of Plants 
02 
24 
46 
68 
810 
1012 
1214 
Number of Houses 
1 
2 
1 
5 
6 
2 
3 
Which method did you use for finding the mean, and why?
Answer:
No. of plants
(Class interval) 
No. of houses (f_{i}) 
Midpoint (x_{i}) 
f_{i}x_{i } 
02 
1 
1 
1 
24 
2 
3 
6 
46 
1 
5 
5 
68 
5 
7 
35 
810 
6 
9 
54 
1012 
2 
11 
22 
1214 
3 
13 
39 

Sum f_{i }= 20 

Sum f_{i}x_{i} = 162 
Mean = xÌ„ = âˆ‘f_{i}x_{i} /âˆ‘f_{i }= 162/20 = 8.1
We would use direct method because the numerical value of f_{i} and x_{i} are small.
2. Consider the following distribution of daily wages of 50 workers of a factory.
Daily wages (in Rs.) 
100120 
120140 
140160 
160180 
180200 
Number of workers 
12 
14 
8 
6 
10 
Find the mean daily wages of the workers of the factory by using an appropriate method.
Answer:
Here, the value of midpoint (x_{i}) is very large, so assumed mean A = 150 and class interval is h = 20.
So, u_{i }= (x_{i} – A)/h = u_{i }= (x_{i} – 150)/20
Daily wages
(Class interval) 
Number of workers
frequency (f_{i}) 
Midpoint (x_{i}) 
u_{i }= (x_{i} – 150)/20 
f_{i}u_{i } 
100120 
12 
110 
2 
24 
120140 
14 
130 
1 
14 
140160 
8 
150 
0 
0 
160180 
6 
170 
1 
6 
180200 
10 
190 
2 
20 
Total 
Sum f_{i }= 50 


Sum f_{i}u_{i} = 12 
Mean = xÌ„ = A + hâˆ‘f_{i}u_{i} /âˆ‘f_{i }=150 + (20 Ã— 12/50) = 150 – 4.8 = 145.20
Thus, mean daily wage = Rs. 145.20
3. The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is Rs 18. Find the missing frequency f.
Answer:
Here, the value of midpoint (x_{i}) mean xÌ„ = 18
Class interval 
Number of children (f_{i}) 
Midpoint (x_{i}) 
f_{i}x_{i } 
1113 
7 
12 
84 
1315 
6 
14 
84 
1517 
9 
16 
144 
1719 
13 
18 = A 
234 
1921 
f 
20 
20f 
2123 
5 
22 
110 
2325 
4 
24 
96 
Total 
f_{i} = 44+f 

Sum f_{i}x_{i} = 752+20f 
Mean = xÌ„ = âˆ‘f_{i}x_{i} /âˆ‘f_{i }= (752+20f)/(44+f)
â‡’ 18 = (752+20f)/(44+f)
â‡’ 18(44+f) = (752+20f)
â‡’ 792+18f = 752+20f
â‡’ 792+18f = 752+20f
â‡’ 792 – 752 = 20f – 18f
â‡’ 40 = 2f
â‡’ f = 20
4. Thirty women were examined in a hospital by a doctor and the number of heart beats per minute were recorded and summarized as follows:
Answer:
x_{i }= (Upper limit + Lower limit)/2
Class size (h) = 3
Assumed mean (A) = 75.5
Class Interval 
Number of women (f_{i}) 
Midpoint (x_{i}) 
u_{i} = (x_{i} – 75.5)/h 
f_{i}u_{i} 
6568 
2 
66.5 
3 
6 
6871 
4 
69.5 
2 
8 
7174 
3 
72.5 
1 
3 
7477 
8 
75.5 
0 
0 
7780 
7 
78.5 
1 
7 
8083 
4 
81.5 
3 
8 
8386 
2 
84.5 
3 
6 

Sum f_{i}= 30 


Sum f_{i}u_{i }= 4 
Mean = xÌ„ = A + hâˆ‘f_{i}u_{i} /âˆ‘f_{i }= 75.5 + 3Ã—(4/30) = 75.5 + 4/10 = 75.5 + 0.4 = 75.9
The mean heart beats per minute for these women is 75.9
5. In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number mangoes. The following was the distribution of mangoes according to the number of boxes.
Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?
Answer:
Since, the given data is not continuous so we add 0.5 to the upper limit and subtract 0.45 from the lower limit.
Here, assumed mean (A) = 57
Class size (h) = 3
Class Interval 
Number of boxes (f_{i}) 
Midpoint (x_{i}) 
d_{i} = x_{i} – A 
f_{i}d_{i} 
49.552.5 
15 
51 
6 
90 
52.555.5 
110 
54 
3 
330 
55.558.5 
135 
57 = A 
0 
0 
58.561.5 
115 
60 
3 
345 
61.564.5 
25 
63 
6 
150 

Sum f_{i} = 400 


Sum f_{i}d_{i} = 75 
Mean = xÌ„ = A + âˆ‘f_{i}d_{i} /âˆ‘f_{i }= 57 + (75/400) = 57 + 0.1875 = 57.19
6. The table below shows the daily expenditure on food of 25 households in a locality:
Find the mean daily expenditure on food by a suitable method.
Answer:
Find the midpoint of the given interval using the formula.
Midpoint (x_{i}) = (upper limit + lower limit)/2
Let is assume the mean (A) = 225
Class size (h) = 50
= 225+50(7/25)
= 22514
= 211
Therefore, the mean daily expenditure on food is 211
7. To find out the concentration of SO^{2} in the air (in parts per million, i.e., ppm), the data was collected for 30 localities in a certain city and is presented below:
Find the mean concentration of SO_{2} in the air.
Answer:
The formula to find out the mean is
Mean = xÌ„ = âˆ‘f_{i}x_{i} /âˆ‘f_{i}
= 2.96/30
= 0.099 ppm
Therefore, the mean concentration of SO_{2} in air is 0.099 ppm.
8. A class teacher has the following absentee record of 40 students of a class for the whole
term. Find the mean number of days a student was absent.
Number of days

06 
610 
1014 
1420 
2028 
2838 
3840 
Number of students 
11 
10 
7 
4 
4 
3 
1 
Answer:
Find the midpoint of the given interval using the formula.
Midpoint (x_{i}) = (upper limit + lower limit)/2
The mean formula is,
Mean = xÌ„ = âˆ‘f_{i}x_{i} /âˆ‘f_{i}
= 499/40
= 12.48 days
Therefore, the mean number of days a student was absent = 12.48.
9. The following table gives the literacy rate (in percentage) of 35 cities. Find the mean
literacy rate.
Literacy rate (in %) 
4555 
5565 
6575 
7585 
8598 
Number of cities 
3 
10 
11 
8 
3 
Answer:
Class Interval 
Frequency (f_{i}) 
(x_{i}) 
d_{i} = x_{i} – a 
u_{i} = d_{i}/h 
f_{i}u_{i}

4555 
3 
50 
20 
2 
6 
5565 
10 
60 
10 
1 
10 
6575 
11 
70 
0 
0 
0 
7585 
8 
80 
10 
1 
8 
8595 
3 
90 
20 
2 
6 

Sum f_{i} = 35 



Sum f_{i}u_{i} = 2 
Mean = xÌ„ = a + (âˆ‘f_{i}u_{i} /âˆ‘f_{i}) Ñ… h
= 70 + (2/35) Ñ… 10 = 69.42
EXERCISE 14.2
1. The following table shows the ages of the patients admitted in a hospital during a year:
Age (in years) 
515 
1525 
2535 
3545 
4555 
5565 
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 out the modal class, let us the consider the class interval with high frequency
Here, the greatest frequency = 23, so the modal class = 35 â€“ 45,
l = 35,
class width (h) = 10,
f_{m} = 23,
f_{1} = 21 and f_{2} = 14
The formula to find the mode is
Mode = l+ [(f_{m}f_{1})/(2f_{m}f_{1}f_{2})]Ã—h
Substitute the values in the formula, we get
Mode = 35+[(2321)/(462114)]Ã—10
Mode = 35+(20/11) = 35+1.8
Mode = 36.8 year
So the mode of the given data = 36.8 year
Calculation of Mean:
First find the midpoint using the formula, x_{i }= (upper limit +lower limit)/2
The mean formula is
Mean = xÌ„ = âˆ‘f_{i}x_{i} /âˆ‘f_{i}
= 2830/80
= 35.37 years
Therefore, the mean of the given data = 35.37 years.
2. The following data gives the information on the observed lifetimes (in hours) of 225 electrical components :
Lifetime (in hours) 
020 
2040 
4060 
6080 
80100 
100120 
Frequency 
10 
35 
52 
61 
38 
29 
Determine the modal lifetimes of the components.
Answer:
Given:
From the given data the modal class is 60â€“80.
l = 60,
The frequencies are:
f_{m} = 61, f_{1} = 52, f_{2} = 38 and h = 20
The formula to find the mode is
Mode = l+ [(f_{m}f_{1})/(2f_{m}f_{1}f_{2})]Ã—h
Substitute the values in the formula, we get
Mode =60+[(6152)/(1225238)]Ã—20
Mode = 60+((9 x 20)/32)
Mode = 60+(45/8) = 60+ 5.625
Therefore, modal lifetime of the components = 65.625 hours.
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 
Number of families 
10001500 
24 
15002000 
40 
20002500 
33 
25003000 
28 
30003500 
30 
35004000 
22 
40004500 
16 
45005000 
7 
Answer:
For Mode:
Given data:
Modal class = 15002000,
l = 1500,
Frequencies:
f_{m} = 40 f_{1} = 24, f_{2} = 33 and
h = 500
Mode formula:
Mode = l+ [(f_{m}f_{1})/(2f_{m}f_{1}f_{2})]Ã—h
Substitute the values in the formula, we get
Mode =1500+[(4024)/(802433)]Ã—500
Mode = 1500+((16Ã—500)/23)
Mode = 1500+(8000/23) = 1500 + 347.83
Therefore, modal monthly expenditure of the families = Rupees 1847.83
Calculation for mean:
First find the midpoint using the formula, x_{i }=(upper limit +lower limit)/2
Let us assume a mean, A be 2750
The formula to calculate the mean,
Mean = xÌ„ = a +(âˆ‘f_{i}u_{i} /âˆ‘f_{i})Ã—h
Substitute the values in the given formula
= 2750+(35/200)Ã—500
= 275087.50
= 2662.50
So, the mean monthly expenditure of the families = Rupees 2662.50
4. The following distribution gives the statewise teacherstudent ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret the two measures.
Answer:
5. The given distribution shows the number of runs scored by some top batsmen of the world in oneday cricket matches:
Find mode of the data.
Answer:
6. A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarized it in the table given below:
Find the mode of the data.
Answer:
EXERCISE 14.3
1. The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them.
Answer:
2. If the median of the distribution given below is 28.5, then find the values of x and y.
Answer:
3. A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if policies are only given to persons having age 18 years onwards but less than 60 years.
Answer:
4. The lengths of 40 leaves of a plant are measured correct to the nearest millimeter and data obtained is represented in the following table. Find the median length of the leaves.
Answer:
EXERCISE 14.4
1. The following distribution gives the daily income of 50 workers of a factory:
Convert the distribution above to a less than type cumulative frequency distribution and draw its ogive.
Answer:
Now, by drawing the points on the graph,
i.e., (120, 12); (140, 26); 160, 34); (180, 40); (200, 50)
2. During the medical checkup of 35 students of a class, their weights were recorded as follows:
Draw a less than type ogive for the given data. Hence obtain the median weight from the graph and verify the result by using the formula.
Answer:
Hence, the points for graph are:
(38, 0), (40, 3), (42, 5), (44, 9), (46, 14), (48, 28), (50, 32), (52, 35)
3. The following table gives production yield per hectare of wheat of 100 farms of a village.
Change the distribution to a more than type distribution and draw its ogive.
Answer:
The points for the graph are:
(50, 100), (55, 98), (60, 90), (65, 78), (70, 54), (75, 16)