Class 10 RD Sharma Solutions – Chapter 7 Statistics – Exercise 7.3 | Set 2
Question 14. Find the mean of the following frequency distribution:
Class interval: | 25 – 29 | 30 – 34 | 35 – 39 | 40 – 44 | 45 – 49 | 50 – 54 | 55 – 59 |
Frequency: | 14 | 22 | 16 | 6 | 5 | 3 | 4 |
Solution:
Let’s consider the assumed mean (A) = 42
Class interval | Mid – value xi | di = xi – 42 | ui = (xi – 42)/5 | fi | fiui |
25 – 29 | 27 | -15 | -3 | 14 | -42 |
30 – 34 | 32 | -10 | -2 | 22 | -44 |
35 – 39 | 37 | -5 | -1 | 16 | -16 |
40 – 44 | 42 | 0 | 0 | 6 | 0 |
45 – 49 | 47 | 5 | 1 | 5 | 5 |
50 – 54 | 52 | 10 | 2 | 3 | 6 |
55 – 59 | 57 | 15 | 3 | 4 | 12 |
| | | | N = 70 | Σ fiui = -79 |
From the table it’s seen that,
A = 42 and h = 5
Mean = A + h x (Σfi ui/N)
= 42 + 5 x (-79/70)
= 42 – 79/14
= 42 – 5.643
= 36.357
Question 15. For the following distribution, calculate mean using all suitable methods:
Size of item: | 1 – 4 | 4 – 9 | 9 – 16 | 16 – 20 |
Frequency: | 6 | 12 | 26 | 20 |
Solution:
By direct method
Class interval | Mid value xi | Frequency fi | fixi |
1 – 4 | 2.5 | 6 | 15 |
4 – 9 | 6.5 | 12 | 18 |
9 – 16 | 12.5 | 26 | 325 |
16 – 27 | 21.5 | 20 | 430 |
| | N = 64 | Sum = 848 |
Mean = (sum/N) + A
= 848/64
= 13.25
By assuming mean method
Let the assumed mean (A) = 65
Class interval | Mid value xi | ui = (xi – A) = xi – 65 | Frequency fi | fiui |
1 – 4 | 2.5 | -4 | 6 | -25 |
4 – 9 | 6.5 | 0 | 12 | 0 |
9 – 16 | 12.5 | 6 | 26 | 196 |
16 – 27 | 21.5 | 15 | 20 | 300 |
| | | N = 64 | Sum = 432 |
Mean = A + sum/N
= 6.5 + 6.75
= 13.25
Question 16. The weekly observation on cost of living index in a certain city for the year 2004 – 2005 are given below. Compute the weekly cost if living index.
Cost of living index | Number of students | Cost of living index | Number of students |
1400 – 1500 | 5 | 1700 – 1800 | 9 |
1500 – 1600 | 10 | 1800 – 1900 | 6 |
1600 – 1700 | 20 | 1900 – 2000 | 2 |
Solution:
Let the assumed mean (A) = 1650
Class interval | Mid value xi | di = xi – A = xi – 1650 | | Frequency fi | fiui |
1400 – 1500 | 1450 | -200 | -2 | 5 | -10 |
1500 – 1600 | 1550 | -100 | -1 | 10 | -10 |
1600 – 1700 | 1650 | 0 | 0 | 20 | 0 |
1700 – 1800 | 1750 | 100 | 1 | 9 | 9 |
1800 – 1900 | 1850 | 200 | 2 | 6 | 12 |
1900 – 2000 | 1950 | 300 | 3 | 2 | 6 |
| | | | N = 52 | Sum = 7 |
We have
A = 16, h = 100
Mean = A + h (sum/N)
= 1650 + (175/13)
= 21625/13
= 1663.46
Question 17. The following table shows the marks scored by 140 students in an examination of a certain paper:
Marks: | 0 – 10 | 10 – 20 | 20 – 30 | 30 – 40 | 40 – 50 |
Number of students: | 20 | 24 | 40 | 36 | 20 |
Calculate the average marks by using all the three methods: direct method, assumed mean deviation and shortcut method.
Solution:
(i) Direct method:
Class interval | Mid value xi | Frequency fi | fixi |
0 – 10 | 5 | 20 | 100 |
10 – 20 | 15 | 24 | 360 |
20 – 30 | 25 | 40 | 1000 |
30 – 40 | 35 | 36 | 1260 |
40 – 50 | 45 | 20 | 900 |
| | N = 140 | Sum = 3620 |
Mean = sum/N
= 3620/140
= 25.857
(ii) Assumed mean method:
Let the assumed mean = 25
Mean = A + (sum/N)
Class interval | Mid value xi | ui = (xi – A) | Frequency fi | fiui |
0 – 10 | 5 | -20 | 20 | -400 |
10 – 20 | 15 | -10 | 24 | -240 |
20 – 30 | 25 | 0 | 40 | 0 |
30 – 40 | 35 | 10 | 36 | 360 |
40 – 50 | 45 | 20 | 20 | 400 |
| | | N = 140 | Sum = 120 |
Mean = A + (sum/N)
= 25 + (120/140)
= 25 + 0.857
= 25.857
(iii) Step deviation method:
Let the assumed mean (A) = 25
Class interval | Mid value xi | di = xi – A = xi – 25 | | Frequency fi | fiui |
0 – 10 | 5 | -20 | -2 | 20 | -40 |
10 – 20 | 15 | -10 | -1 | 24 | -24 |
20 – 30 | 25 | 0 | 0 | 40 | 0 |
30 – 40 | 35 | 10 | 1 | 36 | 36 |
40 – 50 | 45 | 20 | 2 | 20 | 40 |
| | | | N = 140 | Sum = 12 |
Mean = A + h(sum/N)
= 25 + 10(12/140)
= 25 + 0.857
= 25.857
Question 18. The mean of the following frequency distribution is 62.8 and the sum of all the frequencies is 50. Compute the miss frequency f1 and f2.
Class: | 0 – 20 | 20 – 40 | 40 – 60 | 60 – 80 | 80 – 100 | 100 – 120 |
Frequency: | 5 | f1 | 10 | f2 | 7 | 8 |
Solution:
Class interval | Mid value xi | Frequency fi | fixi |
0 – 20 | 10 | 5 | 50 |
20 – 40 | 30 | fi | 30fi |
40 – 60 | 50 | 10 | 500 |
60 – 80 | 70 | f2 | 70f2 |
80 – 100 | 90 | 7 | 630 |
100 – 120 | 110 | 8 | 880 |
| | N = 50 | Sum = 30f1 + 70f2 + 2060 |
Given,
Sum of frequency = 50
5 + f1 + 10 + f2 + 7 + 8 = 50
f1 + f2 = 20
3f1 + 3f2 = 60 —(1) [Multiply both side by 3]
and mean = 62.8
Sum/N = 62.8
(30f1 + 70f2 + 2060)/50 = 62.8
30f1 + 70f2 = 3140 – 2060
30f1 + 70f2 = 1080
3f1 + 7f2 = 108 —(2) [divide it by 10]
Subtract equation (1) from equation (2)
3f1 + 7f2 – 3f1 – 3f2 = 108 – 60
4f2 = 48
f2 = 12
Put value of f2 in equation (1)
3f1 + 3(12) = 60
f1 = 24/3 = 8
f1 = 8, f2 = 12
Question 19. The following distribution shows the daily pocket allowance given to the children of a multistory building. The average pocket allowance is Rs 18.00. Find out the missing frequency.
Class interval: | 11 – 13 | 13 – 15 | 15 – 17 | 17 – 19 | 19 – 21 | 21 – 23 | 23 – 25 |
Frequency: | 7 | 6 | 9 | 13 | – | 5 | 4 |
Solution:
Given mean = 18,
Let the missing frequency be v
Class interval | Mid interval xi | Frequency fi | fixi |
11 – 13 | 12 | 7 | 84 |
13 – 15 | 14 | 6 | 88 |
15 – 17 | 16 | 9 | 144 |
17 – 19 | 18 | 13 | 234 |
19 – 21 | 20 | x | 20x |
21 – 23 | 22 | 5 | 110 |
23 – 25 | 14 | 4 | 56 |
| | N = 44 + x | Sum = 752 + 20x |
Mean = sum/N
18 = 752 + 20×44 + x
792 + 18x = 752 + 20x
2x = 40
x = 20
Question 20. If the mean of the following distribution is 27. Find the value of p.
Class: | 0 – 10 | 10 – 20 | 20 – 30 | 30 – 40 | 40 – 50 |
Frequency: | 8 | P | 12 | 13 | 10 |
Solution:
Class interval | Mid value xi | Frequency fi | fixi |
0 – 10 | 5 | 8 | 40 |
10 – 20 | 15 | P | 152 |
20 – 30 | 25 | 12 | 300 |
30 – 40 | 35 | 13 | 455 |
40 – 50 | 45 | 16 | 450 |
| | N = 43 + P | Sum = 1245 + 15p |
Given mean = 27
Mean = sum/N
1245 + 15p43 + p = 27
1245 + 15p = 1161 + 27p
12p = 84
P =7
Question 21. In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contain varying number of mangoes. The following was the distribution of mangoes according to the number of boxes.
Number of mangoes: | 50 – 52 | 53 – 55 | 56 – 58 | 59 – 61 | 62 – 64 |
Number of boxes: | 15 | 110 | 135 | 115 | 25 |
Find the mean number of mangoes kept in packing box. Which method of finding the mean did you choose?
Solution:
Number of mangoes | Number of boxes |
50 – 52 | 15 |
53 – 55 | 110 |
56 – 58 | 135 |
59 – 61 | 115 |
62 – 64 | 25 |
We may observe that class internals are not continuous
There is a gap between two class intervals. So we have to add ½ from lower class limit to each interval and class mark (xi) may be obtained by using the relation
xi = upperlimit + lowerclasslimit2
Class size (h) of this data = 3
Now taking 57 as assumed mean (a) we may calculated di, ui, fiui as follows
Class interval | Frequency fi | Mid values xi | di = xi – A = xi – 25 | | fiui |
49.5 – 52.5 | 15 | 51 | -6 | -2 | -30 |
52.5 – 55.5 | 110 | 54 | -3 | -1 | -110 |
55.5 – 58.5 | 135 | 57 | 0 | 0 | 0 |
58.5 – 61.5 | 115 | 60 | 3 | 1 | 115 |
61.5 – 64.5 | 25 | 63 | 6 | 2 | 50 |
Total | N = 400 | | | | Sum = 25 |
Now we have N
Sum = 25
Mean = A +h (sum/N)
= 57 + 3 (45/400)
= 57 + 3/16
= 57 + 0.1875
= 57.19
Clearly mean number of mangoes kept in packing box is 57.19
Question 22. The table below shows the daily expenditure on food of 25 households in a locality
Daily expenditure (In Rs): | 100 – 150 | 150 – 200 | 200 – 250 | 250 – 300 | 300 – 350 |
Number of households: | 4 | 5 | 12 | 2 | 2 |
Find the mean daily expenditure on food by a suitable method.
Solution:
We may calculate class mark (xi) for each interval by using the relation
xi = upperlimit + lowerclasslimit2
Class size = 50
Now, Taking 225 as assumed mean (xi) we may calculate di, ui, fiui as follows
Daily expenditure | Frequency f1 | Mid value xi | di = xi – 225 | | fiui |
100 – 150 | 4 | 125 | -100 | -2 | -8 |
150 – 200 | 5 | 175 | -50 | -1 | -5 |
200 – 250 | 12 | 225 | 0 | 0 | 0 |
250 – 300 | 2 | 275 | 50 | 1 | 2 |
300 – 350 | 2 | 325 | 100 | 2 | 4 |
| N = 25 | | | | Sum = -7 |
Now we may observe that
N = 25
Sum = -7
225 + 50 (-7/25)
225 – 14 = 211
So, mean daily expenditure on food is Rs 211
Question 23. To find out the concentration of SO2 in the air (in parts per million i.e ppm) the data was collected for localities for 30 localities in a certain city and is presented below:
Concentration of SO2 (in ppm) | Frequency |
0.00 – 0.04 | 4 |
0.04 – 0.08 | 9 |
0.08 – 0.12 | 9 |
0.12 – 0.16 | 2 |
0.16 – 0.20 | 4 |
0.20 – 0.24 | 2 |
Find the mean concentration of SO2 in the air
Solution:
We may find class marks for each interval by using the relation
x = upperlimit + lowerclasslimit2x =
Class size of this data = 0.04
Now taking 0.04 assumed mean (xi) we may calculate di, ui, fiui as follows
Concentration of SO2 | Frequency f1 | Class interval xi` | di = xi – 0.14 | ui | fiui |
0.00 – 0.04 | 4 | 0.02 | -0.12 | -3 | -12 |
0.04 – 0.08 | 9 | 0.06 | -0.08 | -2 | -18 |
0.08 – 0.12 | 9 | 0.10 | -0.04 | -1 | -9 |
0.12 – 0.16 | 2 | 0.14 | 0 | 0 | 0 |
0.16 – 0.20 | 4 | 0.18 | 0.04 | 1 | 4 |
0.20 – 0.24 | 2 | 0.22 | 0.08 | 2 | 4 |
Total | N = 30 | | | | Sum = -31 |
From the table we may observe that
N = 30
Sum = -31
= 0.14 + (0.04)(-31/30)
= 0.099 ppm
So mean concentration of SO2 in the air is 0.099 ppm.
Question 24. 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: | 0 – 6 | 6 – 10 | 10 – 14 | 14 – 20 | 20 – 28 | 28 – 38 | 38 – 40 |
Number of students: | 11 | 10 | 7 | 4 | 4 | 3 | 1 |
Solution:
We may find class mark of each interval by using the relation
x = upperlimit + lowerclasslimit2x =
Now, taking 16 as assumed mean (a) we may
Calculate di and fidi as follows
Number of days | Number of students fi | Xi | d = xi + 10 | fidi |
0 – 6 | 11 | 3 | -13 | -143 |
6 – 10 | 10 | 8 | -8 | -280 |
10 – 14 | 7 | 12 | -4 | -28 |
14 – 20 | 7 | 16 | 0 | 0 |
20 – 28 | 8 | 24 | 8 | 32 |
28 – 36 | 3 | 33 | 17 | 51 |
30 – 40 | 1 | 39 | 23 | 23 |
Total | N = 40 | | | Sum = -145 |
Now we may observe that
N = 40
Sum = -145
= 16 + (-145/40)
= 16 – 3.625
= 12.38
So mean number of days is 12.38 days, for which student was absent.
Question 25. The following table gives the literacy rate (in percentage) of 35 cities. find the mean literacy rate.
Literacy rate (in %): | 45 – 55 | 55 – 65 | 65 – 75 | 75 – 85 | 85 – 95 |
Number of cities: | 3 | 10 | 11 | 8 | 3 |
Solution:
We may find class marks by using the relation
x = upperlimit + lowerclasslimit2x =
Class size (h) for this data = 10
Now taking 70 as assumed mean (a) wrong
Calculate di, ui, fiui as follows
Literacy rate (in %) | Number of cities (fi) | Mid value xi | di = xi – 70 | ui = di – 50 | fiui |
45 – 55 | 3 | 50 | -20 | -20 | -6 |
55 – 65 | 10 | 60 | -10 | -1 | -10 |
65 – 75 | 11 | 70 | 0 | 0 | 0 |
75 – 85 | 8 | 80 | 10 | 1 | 8 |
85 – 95 | 3 | 90 | 20 | 2 | 6 |
Total | N = 35 | | | | Sum = -2 |
Now we may observe that
N = 35
Sum = -2
= 70 + (-2/35)
= 70 – 4/7
= 70 – 0.57
= 69.43
So, mean literacy rate is 69.43%
Question 21. In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contain varying number of mangoes. The following was the distribution of mangoes according to the number of boxes.
Number of mangoes: | 50 – 52 | 53 – 55 | 56 – 58 | 59 – 61 | 62 – 64 |
Number of boxes: | 15 | 110 | 135 | 115 | 25 |
Find the mean number of mangoes kept in packing box. Which method of finding the mean did you choose?
Solution:
Number of mangoes | Number of boxes |
50 – 52 | 15 |
53 – 55 | 110 |
56 – 58 | 135 |
59 – 61 | 115 |
62 – 64 | 25 |
We may observe that class internals are not continuous
There is a gap between two class intervals. So we have to add ½ from lower class limit to each interval and class mark (xi) may be obtained by using the relation
xi = upperlimit + lowerclasslimit2
Class size (h) of this data = 3
Now taking 57 as assumed mean (a) we may calculated di, ui, fiui as follows
Class interval | Frequency fi | Mid values xi | di = xi – A = xi – 25 | | fiui |
49.5 – 52.5 | 15 | 51 | -6 | -2 | -30 |
52.5 – 55.5 | 110 | 54 | -3 | -1 | -110 |
55.5 – 58.5 | 135 | 57 | 0 | 0 | 0 |
58.5 – 61.5 | 115 | 60 | 3 | 1 | 115 |
61.5 – 64.5 | 25 | 63 | 6 | 2 | 50 |
Total | N = 400 | | | | Sum = 25 |
Now we have N
Sum = 25
Mean = A +h (sum/N)
= 57 + 3 (45/400)
= 57 + 3/16
= 57 + 0.1875
= 57.19
Clearly mean number of mangoes kept in packing box is 57.19
Question 22. The table below shows the daily expenditure on food of 25 households in a locality
Daily expenditure (In Rs): | 100 – 150 | 150 – 200 | 200 – 250 | 250 – 300 | 300 – 350 |
Number of households: | 4 | 5 | 12 | 2 | 2 |
Find the mean daily expenditure on food by a suitable method.
Solution:
We may calculate class mark (xi) for each interval by using the relation
xi = upperlimit + lowerclasslimit2
Class size = 50
Now, Taking 225 as assumed mean (xi) we may calculate di, ui, fiui as follows
Daily expenditure | Frequency f1 | Mid value xi | di = xi – 225 | | fiui |
100 – 150 | 4 | 125 | -100 | -2 | -8 |
150 – 200 | 5 | 175 | -50 | -1 | -5 |
200 – 250 | 12 | 225 | 0 | 0 | 0 |
250 – 300 | 2 | 275 | 50 | 1 | 2 |
300 – 350 | 2 | 325 | 100 | 2 | 4 |
| N = 25 | | | | Sum = -7 |
Now we may observe that
N = 25
Sum = -7
225 + 50 (-7/25)
225 – 14 = 211
So, mean daily expenditure on food is Rs 211
Question 23. To find out the concentration of SO2 in the air (in parts per million i.e ppm) the data was collected for localities for 30 localities in a certain city and is presented below:
Concentration of SO2 (in ppm) | Frequency |
0.00 – 0.04 | 4 |
0.04 – 0.08 | 9 |
0.08 – 0.12 | 9 |
0.12 – 0.16 | 2 |
0.16 – 0.20 | 4 |
0.20 – 0.24 | 2 |
Find the mean concentration of SO2 in the air
Solution:
We may find class marks for each interval by using the relation
x = upperlimit + lowerclasslimit2x =
Class size of this data = 0.04
Now taking 0.04 assumed mean (xi) we may calculate di, ui, fiui as follows
Concentration of SO2 | Frequency f1 | Class interval xi` | di = xi – 0.14 | ui | fiui |
0.00 – 0.04 | 4 | 0.02 | -0.12 | -3 | -12 |
0.04 – 0.08 | 9 | 0.06 | -0.08 | -2 | -18 |
0.08 – 0.12 | 9 | 0.10 | -0.04 | -1 | -9 |
0.12 – 0.16 | 2 | 0.14 | 0 | 0 | 0 |
0.16 – 0.20 | 4 | 0.18 | 0.04 | 1 | 4 |
0.20 – 0.24 | 2 | 0.22 | 0.08 | 2 | 4 |
Total | N = 30 | | | | Sum = -31 |
From the table we may observe that
N = 30
Sum = -31
= 0.14 + (0.04)(-31/30)
= 0.099 ppm
So mean concentration of SO2 in the air is 0.099 ppm.
Question 24. 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: | 0 – 6 | 6 – 10 | 10 – 14 | 14 – 20 | 20 – 28 | 28 – 38 | 38 – 40 |
Number of students: | 11 | 10 | 7 | 4 | 4 | 3 | 1 |
Solution:
We may find class mark of each interval by using the relation
x = upperlimit + lowerclasslimit2x =
Now, taking 16 as assumed mean (a) we may
Calculate di and fidi as follows
Number of days | Number of students fi | Xi | d = xi + 10 | fidi |
0 – 6 | 11 | 3 | -13 | -143 |
6 – 10 | 10 | 8 | -8 | -280 |
10 – 14 | 7 | 12 | -4 | -28 |
14 – 20 | 7 | 16 | 0 | 0 |
20 – 28 | 8 | 24 | 8 | 32 |
28 – 36 | 3 | 33 | 17 | 51 |
30 – 40 | 1 | 39 | 23 | 23 |
Total | N = 40 | | | Sum = -145 |
Now we may observe that
N = 40
Sum = -145
= 16 + (-145/40)
= 16 – 3.625
= 12.38
So mean number of days is 12.38 days, for which student was absent.
Question 25. The following table gives the literacy rate (in percentage) of 35 cities. find the mean literacy rate.
Literacy rate (in %): | 45 – 55 | 55 – 65 | 65 – 75 | 75 – 85 | 85 – 95 |
Number of cities: | 3 | 10 | 11 | 8 | 3 |
Solution:
We may find class marks by using the relation
x = upperlimit + lowerclasslimit2x =
Class size (h) for this data = 10
Now taking 70 as assumed mean (a) wrong
Calculate di, ui, fiui as follows
Literacy rate (in %) | Number of cities (fi) | Mid value xi | di = xi – 70 | ui = di – 50 | fiui |
45 – 55 | 3 | 50 | -20 | -20 | -6 |
55 – 65 | 10 | 60 | -10 | -1 | -10 |
65 – 75 | 11 | 70 | 0 | 0 | 0 |
75 – 85 | 8 | 80 | 10 | 1 | 8 |
85 – 95 | 3 | 90 | 20 | 2 | 6 |
Total | N = 35 | | | | Sum = -2 |
Now we may observe that
N = 35
Sum = -2
= 70 + (-2/35)
= 70 – 4/7
= 70 – 0.57
= 69.43
So, mean literacy rate is 69.43%
Last Updated :
03 Mar, 2021
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