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NCERT solutions for class 10 maths chapter 14 Statistics Excercise: 14.1
Which method did you use for finding the mean, and why?
Answer:
Number of plants  Number of houses
 Classmark


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 
=20 
=162 
Mean,
We used the direct method in this as the values of are small.
Q2 Consider the following distribution of daily wages of 50 workers of a factory. Find the mean daily wages of the workers of the factory by using an appropriate method.
Answer:
Let the assumed mean be a = 550
Daily Wages  Number of workers  Classmark



500520  12  510  40  480 
520540  14  530  20  280 
540560  8  550  0  0 
560580  6  570  20  120 
580600  10  590  40  400 
=50 
=240 
Mean,
Therefore, the mean daily wages of the workers of the factory is Rs. 545.20
Answer:
Daily pocket allowance  Number of children  Classmark


1113  7  12  84 
1315  6  14  84 
1517  9  16  144 
1719  13  18  234 
1921  f  20  20f 
2123  5  22  110 
2325  4  24  96 
=44+f 
=752+20f 
Mean,
Therefore the missing f = 20
Q4 Thirty women were examined in a hospital by a doctor and the number of heartbeats per minute was recorded and summarised as follows. Find the mean heartbeats per minute for these women, choosing a suitable method.
Answer:
Let the assumed mean be a = 75.5
No. of heartbeats per minute  Number of women  Classmark



6568  2  66.5  9  18 
6871  4  69.5  6  24 
7174  3  72.5  3  9 
7477  8  75.5  0  0 
7780  7  78.5  3  21 
8083  4  81.5  6  24 
8386  2  84.5  9  18 
=30 
=12 
Mean,
Therefore, the mean heartbeats per minute of these women are 75.9
Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?
Answer:
Let the assumed mean be a = 57
Number of mangoes  Number of boxes  Classmark



5052  15  51  6  90 
5355  110  54  3  330 
5658  135  57  0  0 
5961  115  60  3  345 
6264  25  63  6  150 
=400 
=75 
Mean,
Therefore, the mean number of mangoes kept in a packing box is approx 57.19
Q6 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 rupees  100150  150200  200250  250300  300350 
Number of households  4  5  12  2  2 
Answer:
Let the assumed mean be a = 225 and h = 50
Daily Expenditure  Number of households  Classmark




100150  4  125  100  2  8 
150200  5  175  50  1  5 
200250  12  225  0  0  0 
250300  2  275  50  1  2 
300350  2  325  100  2  4 
=25 
= 7 
Mean,
Therefore, the mean daily expenditure on food is Rs. 211
Find the mean concentration of in the air.
Answer:
Class Interval  Frequency
 Classmark


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 
=30 
=2.96 
Mean,
Therefore, the mean concentration of in the air is 0.099 ppm
Answer:
Number of days  Number of Students  Classmark


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 
=40 
=499 
Mean,
Therefore, the mean number of days a student was absent is 12.48 days.
Answer:
Let the assumed mean be a = 75 and h = 10
Literacy rates  Number of cities  Classmark




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 
= 35 
= 2 
Mean,
Therefore, the mean mean literacy rate is 69.43%
NCERT solutions for class 10 maths chapter 14 Statistics Excercise: 14.2
Q1 The following table shows the ages of the patients admitted in a hospital during a year:
Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.
Answer:
The class having maximum frequency is the modal class.
The maximum frequency is 23 and hence the modal class = 3545
Lower limit (l) of modal class = 35, class size (h) = 10
Frequency ( ) of the modal class = 23, frequency ( ) of class preceding the modal class = 21, frequency ( ) of class succeeding the modal class = 14.
Now,
Age  Number of patients  Classmark


515  6  10  60 
1525  11  20  220 
2535  21  30  630 
3545  23  40  920 
4555  14  50  700 
5565  5  60  300 
=80 
=2830 
Mean,
The maximum number of patients are in the age group of 36.8, whereas the average age of all the patients is 35.37.
Determine the modal lifetimes of the components.
Answer:
The class having maximum frequency is the modal class.
The maximum frequency is 61 and hence the modal class = 6080
Lower limit (l) of modal class = 60, class size (h) = 20
Frequency ( ) of the modal class = 61 frequency ( ) of class preceding the modal class = 52, frequency ( ) of class succeeding the modal class = 38.
Thus, the modal lifetime of 225 electrical components is 65.62 hours
Answer:
The class having maximum frequency is the modal class.
The maximum frequency is 40 and hence the modal class = 15002000
Lower limit (l) of modal class = 1500, class size (h) = 500
Frequency ( ) of the modal class = 40 frequency ( ) of class preceding the modal class = 24, frequency ( ) of class succeeding the modal class = 33.
Thus, the Mode of the data is Rs. 1847.82
Now,
Let the assumed mean be a = 2750 and h = 500
Expenditure  Number of families  Classmark




10001500  24  1250  1500  3  72 
15002000  40  1750  1000  2  80 
20002500  33  2250  500  1  33 
25003000  28  2750  0  0  0 
30003500  30  3250  500  1  30 
35004000  22  3750  1000  2  44 
40004500  16  4250  1500  3  48 
45005000  7  4750  2000  4  28 
=200 
= 35 
Mean,
Thus, the Mean monthly expenditure is Rs. 2662.50
Answer:
The class having maximum frequency is the modal class.
The maximum frequency is 10 and hence the modal class = 3035
Lower limit (l) of modal class = 30, class size (h) = 5
Frequency ( ) of the modal class = 10 frequency ( ) of class preceding the modal class = 9, frequency ( ) of class succeeding the modal class = 3
Thus, Mode of the data is 30.625
Now,
Let the assumed mean be a = 32.5 and h = 5
Class  Number of states  Classmark




1520  3  17.5  15  3  9 
2025  8  22.5  10  2  16 
2530  9  27.5  5  1  9 
3035  10  32.5  0  0  0 
3540  3  37.5  5  1  3 
4045  0  42.5  10  2  0 
4550  0  47.5  15  3  0 
5055  2  52.5  20  4  8 
=35 
= 23 
Mean,
Thus, the Mean of the data is 29.22
Answer:
The class having maximum frequency is the modal class.
The maximum frequency is 18 and hence the modal class = 40005000
Lower limit (l) of modal class = 4000, class size (h) = 1000
Frequency ( ) of the modal class = 18 frequency ( ) of class preceding the modal class = 4, frequency ( ) of class succeeding the modal class = 9
Thus, Mode of the data is 4608.70
Answer:
The class having maximum frequency is the modal class.
The maximum frequency is 20 and hence the modal class = 4050
Lower limit (l) of modal class = 40, class size (h) = 10
Frequency ( ) of the modal class = 20 frequency ( ) of class preceding the modal class = 12, frequency ( ) of class succeeding the modal class = 11
Thus, Mode of the data is 44.70
Statistics Class 10 Maths Chapter 14 Solution Excercise: 14.3
Answer:
Let the assumed mean be a = 130 and h = 20
Class  Number of consumers  Cumulative Frequency  Classmark




6585  4  4  70  60  3  12 
85105  5  9  90  40  2  10 
105125  13  22  110  20  1  13 
125145  20  42  130  0  0  0 
145165  14  56  150  20  1  14 
165185  8  64  170  40  2  16 
185205  4  68  190  60  3  12 
= 68 
= 7 
MEDIAN:
Median class = 125145; Cumulative Frequency = 42; Lower limit, l = 125;
c.f. = 22; f = 20; h = 20
Thus, the median of the data is 137
MODE:
The class having maximum frequency is the modal class.
The maximum frequency is 20 and hence the modal class = 125145
Lower limit (l) of modal class = 125, class size (h) = 20
Frequency ( ) of the modal class = 20; frequency ( ) of class preceding the modal class = 13, frequency ( ) of class succeeding the modal class = 14.
Thus, Mode of the data is 135.76
MEAN:
Mean,
Thus, the Mean of the data is 137.05
Q2 If the median of the distribution given below is 28.5, find the values of x and y.
Answer:
Class  Number of consumers  Cumulative Frequency 
010  5  5 
1020  x  5+x 
2030  20  25+x 
3040  15  40+x 
4050  y  40+x+y 
5060  5  45+x+y 
= 60 
Now,
Given median = 28.5 which lies in the class 2030
Therefore, Median class = 2030
Frequency corresponding to median class, f = 20
Cumulative frequency of the class preceding the median class, c.f. = 5 + x
Lower limit, l = 20; Class height, h = 10
Also,
Therefore, the required values are: x=8 and y=7
Answer:
Class  Frequency
 Cumulative Frequency 
1520  2  2 
2025  4  6 
2530  18  24 
3035  21  45 
3540  33  78 
4045  11  89 
4550  3  92 
5055  6  98 
5560  2  100 
Therefore, Median class = 3545
Frequency corresponding to median class, f = 21
Cumulative frequency of the class preceding the median class, c.f. = 24
Lower limit, l = 35; Class height, h = 10
Thus, the median age is 35.75 years.
Find the median length of the leaves.
(Hint: The data needs to be converted to continuous classes for finding the median since the formula assumes continuous classes. The classes then change to
117.5  126.5, 126.5  135.5, . . ., 171.5  180.5.)
Answer:
The data needs to be converted to continuous classes for finding the median since the formula assumes continuous classes.
Class  Frequency
 Cumulative Frequency 
117.5126.5  3  3 
126.5135.5  5  8 
135.5144.5  9  17 
144.5153.5  12  29 
153.5162.5  5  34 
162.5171.5  4  38 
171.5180.5  2  40 
Therefore, Median class = 144.5153.5
Lower limit, l = 144.5; Class height, h = 9
Frequency corresponding to median class, f = 12
Cumulative frequency of the class preceding the median class, c.f. = 17
Thus, the median length of the leaves is 146.75 mm
Q5 The following table gives the distribution of the lifetime of 400 neon lamps :
Find the median lifetime of a lamp.
Answer:
Class  Frequency
 Cumulative Frequency 
15002000  14  14 
20002500  56  70 
25003000  60  130 
30003500  86  216 
35004000  74  290 
40004500  62  352 
45005000  48  400 
Therefore, Median class = 30003500
Lower limit, l = 3000; Class height, h = 500
Frequency corresponding to median class, f = 86
Cumulative frequency of the class preceding the median class, c.f. = 130
Thus, the median lifetime of a lamp is 3406.97 hours
Thus, the median length of the leaves is 146.75 mm
Answer:
Class  Number of surnames  Cumulative Frequency  Classmark


14  6  6  2.5  15 
47  30  36  5.5  165 
710  40  76  8.5  340 
1013  16  92  11.5  184 
1316  4  96  14.5  51 
1619  4  100  17.5  70 
= 100 
= 825 
MEDIAN:
Median class = 710; Lower limit, l = 7;
Cumulative frequency of preceding class, c.f. = 36; f = 40; h = 3
Thus, the median of the data is 8.05
MODE:
The class having maximum frequency is the modal class.
The maximum frequency is 40 and hence the modal class = 710
Lower limit (l) of modal class = 7, class size (h) = 3
Frequency ( ) of the modal class = 40; frequency ( ) of class preceding the modal class = 30, frequency ( ) of class succeeding the modal class = 16
Thus, Mode of the data is 7.88
MEAN:
Mean,
Thus, the Mean of the data is 8.25
Answer:
Class  Number of students  Cumulative Frequency 
4045  2  2 
4550  3  5 
5055  8  13 
5560  6  19 
6065  6  25 
6570  3  28 
7075  2  30 
MEDIAN:
Median class = 5560; Lower limit, l = 55;
Cumulative frequency of preceding class, c.f. = 13; f = 6; h = 5
Thus, the median weight of the student is 56.67 kg
1
Q1.The following distribution gives the daily income of 50 workers of a factory.
Answer:
Daily Income (UpperClass Limit)  Cumulative Frequency 
Less than 120  12 
Less than 140  26 
Less than 160  34 
Less than 180  40 
Less than 200  50 
Now,
Taking upperclass interval on the xaxis and their respective frequencies on the yaxis,
Q2
Answer:
Taking upperclass interval on the xaxis and their respective frequencies on the yaxis,
Marking a point on the curve whose ordinate is 17.5 gives an xordinate= 46.5.
Hence, the Median of the data is 46.5
Now,
Weight (Class)  Frequency  Cumulative Frequency 
>38  0  0 
3840  3  3 
4042  2  5 
4244  4  9 
4446  5  14 
4648  14  28 
4850  4  32 
5052  3  35 
Median class = 4648; Lower limit, l = 46;
Cumulative frequency of preceding class, c.f. = 14; f = 14; h = 2
Thus, the median using formula is 46.5 which verifies the result.
Answer:
Production yield (UpperClass Limit)  Cumulative Frequency 
More than or equal to 50  100 
More than or equal to 55  98 
More than or equal to 60  90 
More than or equal to 65  78 
More than or equal to 70  54 
More than or equal to 75  16 
Now,
Taking lower class limit on the xaxis and their respective frequencies on the yaxis,