Class 10 NCERT Solutions – Chapter 14 Statistics – Exercise 14.1
Question 1. A survey was conducted by a group of students as a part of their environment awareness program, 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?
Solution:
Step 1: Let us find out the Class Mark (x_{i}) for the following class intervals by using the formula
Class Mark = (Upper class Limit + Lower Class Limit)/2
Step 2: Now, we will multiply the classmark with the number of times they have occurred, i.e, with the frequency.
Step 3: Now we will apply the general formula to calculate the mean
Now, Let’s see the detailed solution:
No.of Plants (Class Interval)  No. of Houses (Frequency) (f_{i})  Class Mark (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 
Now, after creating this table we will be able to find the mean very easily –
= 16
= 8.1
Hence, we come to the conclusion that the number of plants per house is 8.1. Since the numeral value of frequency(fi) and the class mark(xi) is small so we use DIRECT METHOD to find the mean number of plants per house.
Question 2. Consider the following distribution of daily wages of 50 workers of a factory.
Daily Wages (in ₹)  500520  520540  540560  560580  580600 
Number of Workers  12  14  8  6  10 
Find the mean daily wages of the workers of the factory by using an appropriate method.
Solution:
Step 1: Let us find out the Class Mark (xi) for the following class intervals by using the formula
Class Mark = (Upper class Limit + Lower Class Limit)/2
Step 2: In this case, the value of midpoint (xi) is very large, so let us assume the mean value, A = 150, and the class interval is h = 20.
u_{i} = (x_{i} – A)/h
=> u_{i }= (xi – 150)/20
Step 3: Now we will apply the Assumed Mean Formula to calculate the mean
Now, Let’s see the detailed solution:
Daily wages (Class interval)  Number of workers frequency (fi)  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 
So, the formula to find out the mean is:
Mean =
= 150 + (20 × 12/50)
= 150 – 4.8
= 145.20
Thus, mean daily wage of the workers = Rs. 145.20.
Question 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.
Daily pocket allowance (in ₹)  1113  1315  1517  1719  1921  2123  2325 
Number of children  7  6  9  13  f  5  4 
Solution:
Step 1: Let us find out the Class Mark (xi) for the following class intervals by using the formula
Class Mark = (Upper class Limit + Lower Class Limit)/2
Step 2: Now, we will multiply the classmark with the number of times they have occurred, i.e, with the frequency. As a certain frequency is missing and we have an odd number of class intervals hence, we will assume the middleClass Mark as our Assumed Mean(A).
Step 3: Now we will apply the general formula to calculate the mean
Now, Let’s see the detailed solution:
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 
The mean formula is
Mean =
= (752 + 20f)/(44 + f)
Now substitute the values and equate to find the missing frequency (f)
⇒ 18 = (752 + 20f)/(44 + f)
⇒ 18(44 + f) = (752 + 20f)
⇒ 792 + 18f = 752 + 20f
⇒ 792 + 18f = 752 + 20f
⇒ 792 – 752 = 20f – 18f
⇒ 40 = 2f
⇒ f = 20
So, the missing frequency, f = 20.
Question 4. Thirty women were examined in a hospital by a doctor and the number of heartbeats per minute was recorded and summarized as follows. Find the mean heartbeats per minute for these women, choosing a suitable method.
Number of heartbeats per minute  6568  6871  7174  7477  7780  8083  8386 
Number of Women  2  4  3  8  7  4  2 
Solution:
Step 1: Let us find out the Class Mark (xi) for the following class intervals by using the formula
Class Mark = (Upper class Limit + Lower Class Limit)/2
Step 2: In this case, the value of midpoint (xi) is very large, so let us assume the mean value, A = 75.5 and class size is h = 3.
d_{i} = (x_{i} – A)
=> d_{i} = (x_{i} – 75.5)
Step 3: Now we will apply the Assumed Mean Formula to calculate the mean
Now, Let’s see the detailed solution:
Class Interval  Number of women (f_{i})  Midpoint (x_{i})  d_{i} = (x_{i} – 75.5)  f_{i}d_{i} 

6568  2  66.5  9  18 
6871  4  69.5  6  24 
7174  3  72.5  3  9 
7477  8  75.5 = A  0  0 
7780  7  78.5  3  21 
8083  4  81.5  6  24 
8386  2  84.5  9  18 
 Sum ∑f_{i} = 30 

 Sum ∑f_{i}u_{i} = 12 
Mean =
= 75.5 + (12/30)
= 75.5 + 2/5
= 75.5 + 0.4
= 75.9
Therefore, the mean heartbeats per minute for these women is 75.9
Question 5. In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying numbers of mangoes. The following was the distribution of mangoes according to the number of boxes.
Number of Mangoes  5052  5355  5658  5961  6264 
Number of Boxes  15  110  135  115  25 
Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?
Solution:
Step 1: In the above table we find that the class intervals are not continuous and hence to make them a continuous set of data we add 0.5 to the upper limit and subtract 0.45 from the lower limit as the gap between two intervals is 1. Then find the Mid Point by using the formula
Class Mark = (Upper class Limit + Lower Class Limit)/2
Step 2: In this case, let us assume the mean value, A = 57 and class size is h = 3.
Step 3: Since the frequency values are big, hence we are using the STEPDEVIATION METHOD.
Now, Lets see the detailed solution:
Class Interval  Number of boxes (f_{i})  Midpoint (x_{i})  d_{i} = x_{i} – A  u_{i}=(x_{i} – A)/h  f_{i}u_{i} 

49.552.5  15  51  6  2  30 
52.555.5  110  54  3  1  110 
55.558.5  135  57 =A  0  0  0 
58.561.5  115  60  3  1  115 
61.564.5  25  63  6  2  50 
 Sum ∑f_{i} = 400 


 Sum ∑f_{i}u_{i} = 25 
Mean =
= 57 + 3 * (25/400)
= 57 + 0.1875
= 57.19
Therefore, the mean number of mangoes kept in a packing box is 57.19
Question 6. The table below shows the daily expenditure on the food of 25 households in a locality. Find the mean daily expenditure on food by a suitable method.
Daily Expenditure (in ₹)  100150  150200  200250  250300  300350 
Number of Households  4  5  12  2  2 
Solution:
Step 1: Let us find out the Class Mark (xi) for the following class intervals by using the formula
Class Mark = (Upper class Limit + Lower Class Limit)/2
Step 2: In this case, the value of midpoint (xi) is very large, so let us assume the mean value, A = 225 and class size is h = 50.
d_{i} = (x_{i} – A)
=> d_{i} = (x_{i} – 225)
u_{i} = (x_{i} – A)/h
=> u_{i} = (x_{i} – 225)/50
Step 3: Now we will apply the Step Deviation Formula to calculate the mean
Now, Let’s see the detailed solution:
Class Interval  Number of households (f_{i})  Midpoint (x_{i})  d_{i} = x_{i} – A  u_{i} = d_{i}/50  f_{i}u_{i} 

100150  4  125  100  2  8 
150200  5  175  50  1  5 
200250  12  225 = A  0  0  0 
250300  2  275  50  1  2 
300350  2  325  100  2  4 
 Sum ∑f_{i} = 25 


 Sum ∑f_{i}u_{i} = 7 
Mean =
= 225 + 50 (7/25)
= 225 – 14
= 211
Therefore, the mean daily expenditure on food is ₹211
Question 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:
Concentration of SO_{2} (in ppm)  Frequency 

0.000.04  4 
0.040.08  9 
0.080.12  9 
0.120.16  2 
0.160.20  4 
0.200.24  2 
Find the mean concentration of SO_{2} in the air.
Solution:
Step 1: Let us find out the Class Mark (xi) for the following class intervals by using the formula
Class Mark = (Upper class Limit + Lower Class Limit)/2
Step 2: Now, we will multiply the classmark with the number of times they have occurred, i.e, with the frequency.
Step 3: Now we will apply the general formula to calculate the mean
Now, Let’s see the detailed solution:
Concentration of SO_{2 }(in ppm)  Frequency (f_{i})  Midpoint (x_{i})  f_{i}x_{i} 

0.000.04  4  0.02  0.08 
0.040.08  9  0.06  0.54 
0.080.12  9  0.10  0.90 
0.120.16  2  0.14  0.28 
0.160.20  4  0.18  0.72 
0.200.24  2  0.22  0.44 
 Sum ∑f_{i} = 30 
 Sum ∑f_{i}x_{i} = 2.96 
The formula to find out the mean is
Mean =
= 2.96/30
= 0.099 ppm
Therefore, the mean concentration of SO_{2} in the air is 0.099 ppm.
Question 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 
Solution:
Step 1: Let us find out the Class Mark (xi) for the following class intervals by using the formula
Class Mark = (Upper class Limit + Lower Class Limit)/2
Step 2: Now, we will multiply the classmark with the number of times they have occurred, i.e, with the frequency.
Step 3: Now we will apply the general formula to calculate the mean
Now, Let’s see the detailed solution:
Class Interval  Frequency (f_{i})  Midpoint (x_{i})  f_{i}x_{i} 

06  11  3  33 
610  10  8  80 
1014  7  12  84 
1420  4  17  68 
2028  4  24  96 
2838  3  33  99 
3840  1  39  39 
 Sum ∑f_{i} = 40 
 Sum ∑f_{i}x_{i} = 499 
The mean formula is,
Mean =
= 499/40
= 12.48 days
Therefore, the mean number of days a student was absent = 12.48.
Question 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  8595 
Number of Cities  3  10  11  8  3 
Solution:
Step 1: Let us find out the Class Mark (xi) for the following class intervals by using the formula
Class Mark = (Upper class Limit + Lower Class Limit)/2
Step 2: In this case, the value of midpoint (xi) is very large, so let us assume the mean value, A = 70 and class size is h = 10.
d_{i} = (x_{i} – A)
=> d_{i} = (x_{i} – 70)
u_{i} = (x_{i} – A)/h
=> u_{i} = (x_{i} – 70)/10
Step 3: Now we will apply the Step Deviation Formula to calculate the mean
Now, Let’s see the detailed solution:
Class Interval  Frequency (f_{i})  Class Mark(x_{i})  di = 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 = A  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 
So,
Mean =
= 70 + (2/35) × 10
= 69.42
Therefore, the mean literacy rate = 69.42%.
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