Rs Aggarwal 2018 Solutions for Class 10 Math Chapter 9 Mean, Median, Mode Of Grouped Data, Cumulative Frequency Graph And Ogive are provided here with simple step-by-step explanations. These solutions for Mean, Median, Mode Of Grouped Data, Cumulative Frequency Graph And Ogive are extremely popular among Class 10 students for Math Mean, Median, Mode Of Grouped Data, Cumulative Frequency Graph And Ogive Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2018 Book of Class 10 Math Chapter 9 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2018 Solutions. All Rs Aggarwal 2018 Solutions for class Class 10 Math are prepared by experts and are 100% accurate.
Page No 370:
Question 1:
If the mean of 5 observations x, x + 2, x + 4, x + 6 and x + 8 is 11, find the value of x. [CBSE 2014]
Answer:
Mean of given observations =
Hence, the value of x is 7.
Page No 370:
Question 2:
If the mean of 25 observations is 27 and each observation is decreased by 7, what will be the new mean?
Answer:
​Mean of given observations =
Mean of 25 observations = 27
∴ Sum of 25 observations = 27 × 25 = 675
If 7 is subtracted from every number, then the sum = 675 − (25 × 7)
= 675 − 175
= 500
Then, new mean =
Thus, the new mean will be 20.
Page No 370:
Question 3:
Compute the mean of following data: [CBSE 2013]
Class | 1−3 | 3−5 | 5−7 | 7−9 |
Frequency | 12 | 22 | 27 | 19 |
Answer:
The given data is shown as follows:
Class | Frequency (fi) | Class mark (xi) | fixi |
1−3 | 12 | 2 | 24 |
3−5 | 22 | 4 | 88 |
5−7 | 27 | 6 | 162 |
7−9 | 19 | 8 | 152 |
Total | ∑ fi = 80 | ∑ fixi = 426 |
The mean of given data is given by
Thus, the mean of the following data is 5.325.
Page No 370:
Question 4:
Find the mean, using direct method:
Class | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 |
Frequency | 7 | 5 | 6 | 12 | 8 | 2 |
Answer:
Class |
Frequency |
Mid Values |
|
0-10 |
7 |
5 |
35 |
10-20 |
5 |
15 |
75 |
20-30 |
6 |
25 |
150 |
30-40 |
12 |
35 |
420 |
40-50 |
8 |
45 |
360 |
50-60 |
2 |
55 |
110 |
|
|
|
|
Page No 370:
Question 5:
Find the mean, using direct method:
Class | 25-35 | 35-45 | 45-55 | 55-65 | 65-75 |
Frequency | 6 | 10 | 8 | 12 | 4 |
Answer:
Class |
Frequency |
Mid values |
|
25-35 |
6 |
30 |
180 |
35-45 |
10 |
40 |
400 |
45-55 |
8 |
50 |
400 |
55-65 |
12 |
60 |
720 |
65-75 |
4 |
70 |
280 |
|
|
|
|
Page No 370:
Question 6:
Find the mean, using direct method:
Class | 0-100 | 100-200 | 200-300 | 300-400 | 400-500 |
Frequency | 6 | 9 | 15 | 12 | 8 |
Answer:
Class |
Frequency |
Mid values |
|
0-100 |
6 |
50 |
300 |
100-200 |
9 |
150 |
1350 |
200-300 |
15 |
250 |
3750 |
300-400 |
12 |
350 |
4200 |
400-500 |
8 |
450 |
3600 |
|
|
|
|
Page No 370:
Question 7:
Using an appropriate method, find the mean of following frequency distribution:
Class interval | 84−90 | 90−96 | 96−102 | 102−108 | 108−114 | 114−120 |
Frequency | 8 | 10 | 16 | 23 | 12 | 11 |
Which method did you use, and why?
Answer:
The given data is shown as follows:
Class interval | Frequency (fi) | Class mark (xi) | fixi |
84−90 | 8 | 87 | 696 |
90−96 | 10 | 93 | 930 |
96−102 | 16 | 99 | 1584 |
102−108 | 23 | 105 | 2415 |
108−114 | 12 | 111 | 1332 |
114−120 | 11 | 117 | 1287 |
Total | ∑ fi = 80 | ∑ fixi = 8244 |
The mean of given data is given by
​Thus, the mean of the following data is 103.05 .
Page No 370:
Question 8:
If the mean of the following frequency distribution is 24, find the value of p. [CBSE 2013]
Class | 0−10 | 10−20 | 20−30 | 30−40 | 40−50 |
Frequency | 3 | 4 | p | 3 | 2 |
Answer:
The given data is shown as follows:
Class | Frequency (fi) | Class mark (xi) | fixi |
0−10 | 3 | 5 | 15 |
10−20 | 4 | 15 | 60 |
20−30 | p | 25 | 25p |
30−40 | 3 | 35 | 105 |
40−50 | 2 | 45 | 90 |
Total | ∑ fi = 12 + p | ∑ fixi = 270 + 25p |
The mean of given data is given by
Hence, the value of p is 18.
Page No 371:
Question 9:
The following distribution shows the daily pocket allowance of children of a locality. If the mean pocket allowance is ₹18, find the missing frequency f.
Daily pocket allowance (in ₹) | 11−13 | 13−15 | 15−17 | 17−19 | 19−21 | 21−23 | 23−25 |
Number of children | 7 | 6 | 9 | 13 | f | 5 | 4 |
Answer:
The given data is shown as follows:
Daily pocket allowance (in ₹) | Number of children (fi) | Class mark (xi) | fixi |
11−13 | 7 | 12 | 84 |
13−15 | 6 | 14 | 84 |
15−17 | 9 | 16 | 144 |
17−19 | 13 | 18 | 234 |
19−21 | f | 20 | 20f |
21−23 | 5 | 22 | 110 |
23−25 | 4 | 24 | 96 |
Total | ∑ fi = 44 + f | ∑ fixi = 752 + 20f |
The mean of given data is given by
​Hence, the value of f is 20.
Page No 371:
Question 10:
If the mean of the following frequency distribution is 54, find the value of p. [CBSE 2006C]
Class | 0−20 | 20−40 | 40−60 | 60−80 | 80−100 |
Frequency | 7 | p | 10 | 9 | 13 |
Answer:
The given data is shown as follows:
Class | Frequency (fi) | Class mark (xi) | fixi |
0−20 | 7 | 10 | 70 |
20−40 | p | 30 | 30p |
40−60 | 10 | 50 | 500 |
60−80 | 9 | 70 | 630 |
80−100 | 13 | 90 | 1170 |
Total | ∑ fi = 39 + p | ∑ fixi = 2370 + 30p |
The mean of given data is given by
​Thus, the value of p is 11.
Page No 371:
Question 11:
The mean of the following data is 42, find the missing frequencies x and y if the sum of the frequencies is 100.
Class interval | 0−10 | 10−20 | 20−30 | 30−40 | 40−50 | 50−60 | 60−70 | 70−80 |
Frequency | 7 | 10 | x | 13 | y | 10 | 14 | 9 |
Answer:
The given data is shown as follows:
Class interval | Frequency (fi) | Class mark (xi) | fixi |
0−10 | 7 | 5 | 35 |
10−20 | 10 | 15 | 150 |
20−30 | x | 25 | 25x |
30−40 | 13 | 35 | 455 |
40−50 | y | 45 | 45y |
50−60 | 10 | 55 | 550 |
60−70 | 14 | 65 | 910 |
70−80 | 9 | 75 | 675 |
Total | ∑ fi = 63 + x + y | ∑ fixi = 2775 + 25x + 45y |
Sum of the frequencies = 100
Now, The mean of given data is given by
If x = 12, then y = 37 − 12 = 25
​Thus, the value of x is 12 and y is 25.
Page No 371:
Question 12:
The daily expenditure of 100 families are given below. Calculate f1 and f2 if the mean daily expenditure is ₹188.
Expenditure (in ₹) |
140−160 | 160−180 | 180−200 | 200−220 | 220−240 |
Number of families | 5 | 25 | f1 | f2 | 5 |
Answer:
The given data is shown as follows:
Expenditure (in ₹) |
Number of families (fi) |
Class mark (xi) | fixi |
140−160 | 5 | 150 | 750 |
160−180 | 25 | 170 | 4250 |
180−200 | f1 | 190 | 190f1 |
200−220 | f2 | 210 | 210f2 |
220−240 | 5 | 230 | 1150 |
Total | ∑ fi = 35 + f1 + f2 | ∑ fixi = 6150 + 190f1 + 210f2 |
Sum of the frequencies = 100
Now, The mean of given data is given by
If f1 = 50, then f2 = 65 − 50 = 15
​Thus, the value of f1 is 50 and f2 is 15.
Page No 371:
Question 13:
The mean of the following frequency distribution is 57.6 and the sum of the observations is 50
Class | 0-20 | 20-40 | 40-60 | 60-80 | 80-100 | 100-120 |
Frequency | 7 | f1 | 12 | f2 | 8 | 5 |
Answer:
Class |
Frequency |
Mid values |
|
0-20 |
7 |
10 |
70 |
20-40 |
f |
30 |
30 f |
40-60 |
12 |
50 |
600 |
60-80 |
18- f |
70 |
1260-70 f1 |
80-100 |
8 |
90 |
720 |
100-120 |
5 |
110 |
550 |
|
|
|
|
Page No 371:
Question 14:
During a medical check-up, the number of heartbeats per minute of 30 patients were recorded and summarised as follows:
Number of heartbeats per minute | 65−68 | 68−71 | 71−74 | 74−77 | 77−80 | 80−83 | 83−86 |
Number of patients | 2 | 4 | 3 | 8 | 7 | 4 | 2 |
Find the mean heartbeats per minute for these patients, choosing a suitable method.
Answer:
Using Direct method, the given data is shown as follows:
Number of heartbeats per minute | Number of patients (fi) |
Class mark (xi) | fixi |
65−68 | 2 | 66.5 | 133 |
68−71 | 4 | 69.5 | 278 |
71−74 | 3 | 72.5 | 217.5 |
74−77 | 8 | 75.5 | 604 |
77−80 | 7 | 78.5 | 549.5 |
80−83 | 4 | 81.5 | 326 |
83−86 | 2 | 84.5 | 169 |
Total | ∑ fi = 30 | ∑ fixi = 2277 |
The mean of given data is given by
​Thus, the mean heartbeats per minute for these patients is 75.9.
Page No 372:
Question 15:
Find the mean, using assumed-mean method:
Marks | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 |
Number of students | 12 | 18 | 27 | 20 | 17 | 6 |
Answer:
Class |
Frequency |
Mid values |
Deviation |
|
0-10 |
12 |
5 |
-20 |
-240 |
10-20 |
18 |
15 |
-10 |
-180 |
20-30 |
27 |
25=A |
0 |
0 |
30-40 |
20 |
35 |
10 |
200 |
40-50 |
17 |
45 |
20 |
340 |
50-60 |
6 |
55 |
30 |
180 |
|
|
|
|
|
Page No 372:
Question 16:
Find the mean, using assumed-mean method:
Class | 100-120 | 120-140 | 140-160 | 160-180 | 180-200 |
Frequency | 10 | 20 | 30 | 15 | 5 |
Answer:
Class |
Frequency |
Mid values |
Deviation |
|
100-120 |
10 |
110 |
-40 |
-400 |
120-140 |
20 |
130 |
-20 |
-400 |
140-160 |
30 |
150=A |
0 |
0 |
160-180 |
15 |
170 |
20 |
300 |
180-200 |
5 |
190 |
40 |
200 |
|
|
|
|
|
Page No 372:
Question 17:
Find the mean, using assumed-mean method:
Class | 0-20 | 20-40 | 40-60 | 60-80 | 80-100 | 100-120 |
Frequency | 20 | 35 | 52 | 44 | 38 | 31 |
Answer:
Class |
Frequency |
Mid Values |
Deviation |
|
0-20 |
20 |
10 |
-40 |
-800 |
20-40 |
35 |
30 |
-20 |
-700 |
40-60 |
52 |
50=A |
0 |
0 |
60-80 |
44 |
70 |
20 |
880 |
80-100 |
38 |
90 |
40 |
1520 |
100-120 |
31 |
110 |
60 |
1860 |
|
|
|
|
|
Page No 372:
Question 18:
Literacy rate (%) | 45−55 | 55−65 | 65−75 | 75−85 | 85−95 |
Number of cities | 4 | 11 | 12 | 9 | 4 |
Answer:
Using Direct method, the given data is shown as follows:
Literacy rate (%) | Number of cities (fi) |
Class mark (xi) | fixi |
45−55 | 4 | 50 | 200 |
55−65 | 11 | 60 | 660 |
65−75 | 12 | 70 | 840 |
75−85 | 9 | 80 | 720 |
85−95 | 4 | 90 | 360 |
Total | ∑ fi = 40 | ∑ fixi = 2780 |
The mean of given data is given by
​Thus, the mean literacy rate is 69.5%.
Page No 372:
Question 19:
Class | 0−10 | 10−20 | 20−30 | 30−40 | 40−50 |
Frequency | 7 | 10 | 15 | 8 | 10 |
Answer:
Let us choose a = 25, h = 10, then di = xi − 25 and ui = .
Using Step-deviation method, the given data is shown as follows:
Class | Frequency (fi) |
Class mark (xi) | di = xi − 25 | ui = | fiui |
0−10 | 7 | 5 | −20 | −2 | −14 |
10−20 | 10 | 15 | −10 | −1 | −10 |
20−30 | 15 | 25 | 0 | 0 | 0 |
30−40 | 8 | 35 | 10 | 1 | 8 |
40−50 | 10 | 45 | 20 | 2 | 20 |
Total | ∑ fi = 50 | ∑ fiui = 4 |
The mean of given data is given by
​Thus, the mean is 25.8.
Page No 372:
Question 20:
Class | 5−15 | 15−25 | 25−35 | 35−45 | 45−55 | 55−65 | 65−75 |
Frequency | 6 | 10 | 16 | 15 | 24 | 8 | 7 |
Answer:
Let us choose a = 40, h = 10, then di = xi − 40 and ui = .
Using Step-deviation method, the given data is shown as follows:
Class | Frequency (fi) |
Class mark (xi) | di = xi − 40 | ui = | fiui |
5−15 | 6 | 10 | −30 | −3 | −18 |
15−25 | 10 | 20 | −20 | −2 | −20 |
25−35 | 16 | 30 | −10 | −1 | −16 |
35−45 | 15 | 40 | 0 | 0 | 0 |
45−55 | 24 | 50 | 10 | 1 | 24 |
55−65 | 8 | 60 | 20 | 2 | 16 |
65−75 | 7 | 70 | 30 | 3 | 21 |
Total | ∑ fi = 86 | ∑ fiui = 7 |
The mean of given data is given by
​Thus, the mean is 40.81.
Page No 372:
Question 21:
​The weights of tea in 70 packets are shown in the following table:
Weight (in grams) |
200−201 | 201−202 | 202−203 | 203−204 | 204−205 | 205−206 |
Number of packets | 13 | 27 | 18 | 10 | 1 | 1 |
Find the mean weight of packets using step-deviation method. [CBSE 2013]
Answer:
​Let us choose a = 202.5, h = 1, then di = xi − 202.5 and ui = .
Using Step-deviation method, the given data is shown as follows:
Weight (in grams) |
Number of packets (fi) | Class mark (xi) | di = xi − 202.5 | ui = | fiui |
200−201 | 13 | 200.5 | −2 | −2 | −26 |
201−202 | 27 | 201.5 | −1 | −1 | −27 |
202−203 | 18 | 202.5 | 0 | 0 | 0 |
203−204 | 10 | 203.5 | 1 | 1 | 10 |
204−205 | 1 | 204.5 | 2 | 2 | 2 |
205−206 | 1 | 205.5 | 3 | 3 | 3 |
Total | ∑ fi = 70 | ∑ fiui = −38 |
The mean of given data is given by
​Hence, the mean is 201.96 g.
Page No 373:
Question 22:
Class | 20−30 | 30−40 | 40−50 | 50−60 | 60−70 |
Frequency | 25 | 40 | 42 | 33 | 10 |
Answer:
Let us choose a = 45, h = 10, then di = xi − 45 and ui = .
Using Step-deviation method, the given data is shown as follows:
Class | Frequency (fi) |
Class mark (xi) | di = xi − 45 | ui = | fiui |
20−30 | 25 | 25 | −20 | −2 | −50 |
30−40 | 40 | 35 | −10 | −1 | −40 |
40−50 | 42 | 45 | 0 | 0 | 0 |
50−60 | 33 | 55 | 10 | 1 | 33 |
60−70 | 10 | 65 | 20 | 2 | 20 |
Total | ∑ fi = 150 | ∑ fiui = −37 |
The mean of given data is given by
​Thus, the mean is 42.534.
Page No 373:
Question 23:
In an annual examination, marks (out of 90) obtained by students of class X in mathematics are given below:
Marks obtained | 0−15 | 15−30 | 30−45 | 45−60 | 60−75 | 75−90 |
Number of students | 2 | 4 | 5 | 20 | 9 | 10 |
Find the mean marks.
Answer:
​Let us choose a = 52.5, h = 15, then di = xi − 52.5 and ui = .
Using Step-deviation method, the given data is shown as follows:
Marks obtained | Number of students (fi) | Class mark (xi) | di = xi − 37.5 | ui = | fiui |
0−15 | 2 | 7.5 | −45 | −3 | −6 |
15−30 | 4 | 22.5 | −30 | −2 | −8 |
30−45 | 5 | 37.5 | −15 | −1 | −5 |
45−60 | 20 | 52.5 | 0 | 0 | 0 |
60−75 | 9 | 67.5 | 15 | 1 | 9 |
75−90 | 10 | 82.5 | 30 | 2 | 20 |
Total | ∑ fi = 50 | ∑ fiui = 10 |
The mean of given data is given by
​Thus, the mean is 55.5.
Page No 373:
Question 24:
Find the arithmetic mean of each of the following frequency distributions using step-deviation method:
Age (in years) | 18-24 | 24-30 | 30-36 | 36-42 | 42-48 | 48-54 |
Number of workers | 6 | 8 | 12 | 8 | 4 | 2 |
Answer:
Class |
Frequency |
Mid values |
|
|
18-24 |
6 |
21 |
−2 |
−12 |
24-30 |
8 |
27 |
−1 |
−8 |
30-36 |
12 |
33 = A |
0 |
0 |
36-42 |
8 |
39 |
1 |
8 |
42-48 |
4 |
45 |
2 |
8 |
48-54 |
2 |
51 |
3 |
6 |
|
|
|
|
|
Page No 373:
Question 25:
Find the arithmetic mean of each of the following frequency distributions using step-deviation method:
Class | 500-520 | 520-540 | 540-560 | 560-580 | 580-600 | 600-620 |
Frequency | 14 | 9 | 5 | 4 | 3 | 5 |
Answer:
Class |
Frequency |
Mid values |
|
|
500-520 |
14 |
510 |
−2 |
−28 |
520-540 |
9 |
530 |
−1 |
−9 |
540-560 |
5 |
550 = A |
0 |
0 |
560-580 |
4 |
570 |
1 |
4 |
580-600 |
3 |
590 |
2 |
6 |
600-620 |
5 |
610 |
3 |
15 |
|
|
|
|
|
Page No 373:
Question 26:
Find the mean age from the following frequency distribution:
Age (in years) | 25-29 | 30-34 | 35-39 | 40-44 | 45-49 | 50-54 | 55-59 |
No. of person | 4 | 14 | 22 | 16 | 6 | 5 | 3 |
Answer:
Converting the series into exclusive form, we get:
Class |
Frequency |
Mid values |
|
|
24.5-29.5 |
4 |
27 |
−3 |
−12 |
29.5-34.5 |
14 |
32 |
−2 |
−28 |
34.5-39.5 |
22 |
37 |
−1 |
−22 |
39.5-44.5 |
16 |
42 = A |
0 |
0 |
44.5-49.5 |
6 |
47 |
1 |
6 |
49.5-54.5 |
5 |
52 |
2 |
10 |
54.5-59.5 |
3 |
57 |
3 |
9 |
|
|
|
|
|
Page No 373:
Question 27:
The following table shows the age distribution of patients of malaria in a village during a particular month.
Age (in years) | 5-14 | 15-24 | 25-34 | 35-44 | 45-54 | 55-64 |
No. of cases | 6 | 11 | 21 | 23 | 14 | 5 |
Find the average age of the patients.
Answer:
Converting the series into exclusive form, we get:
Class |
Frequency |
Mid values |
|
|
4.5-14.5 |
6 |
9.5 |
−2 |
−12 |
14.5-24.5 |
11 |
19.5 |
−1 |
−11 |
24.5-34.5 |
21 |
29.5 = A |
0 |
0 |
34.5-44.5 |
23 |
39.5 |
1 |
23 |
44.5-54.5 |
14 |
49.5 |
2 |
28 |
54.5-64.5 |
5 |
59.5 |
3 |
15 |
|
|
|
|
|
Page No 373:
Question 28:
Weight of 60 eggs were recorded as given below:
Weight (in grams) |
75−79 | 80−84 | 85−89 | 90−94 | 95−99 | 100−104 | 105−109 |
Number of eggs | 4 | 9 | 13 | 17 | 12 | 3 | 2 |
Calculate their mean weight to the nearest gram.
Answer:
Let us choose a = 92, h = 5, then di = xi − 92 and ui = .
Using Step-deviation method, the given data is shown as follows:
Weight (in grams) |
Number of eggs (fi) | Class mark (xi) | di = xi − 92 | ui = | fiui |
74.5−79.5 | 4 | 77 | −15 | −3 | −12 |
79.5−84.5 | 9 | 82 | −10 | −2 | −18 |
84.5−89.5 | 13 | 87 | −5 | −1 | −13 |
89.5−94.5 | 17 | 92 | 0 | 0 | 0 |
94.5−99.5 | 12 | 97 | 5 | 1 | 12 |
99.5−104.5 | 3 | 102 | 10 | 2 | 6 |
104.5−109.5 | 2 | 107 | 15 | 3 | 6 |
Total | ∑ fi = 60 | ∑ fiui = −19 |
The mean of given data is given by
​
​Thus, the mean weight to the nearest gram is 90 g.
Page No 374:
Question 29:
​The following table shows the marks scored by 80 students in an examination:
Marks | Less than 5 | Less than 10 | Less than 15 | Less than 20 | Less than 25 | Less than 30 | Less than 35 | Less than 40 |
Number of students | 3 | 10 | 25 | 49 | 65 | 73 | 78 | 80 |
Calculate the mean marks correct to 2 decimal places.
Answer:
Let us choose a = 17.5, h = 5, then di = xi − 17.5 and ui = .
Using Step-deviation method, the given data is shown as follows:
Marks | Number of students (cf) | Frequency (fi) | Class mark (xi) | di = xi − 17.5 | ui = | fiui |
0−5 | 3 | 3 | 2.5 | −15 | −3 | −9 |
5−10 | 10 | 7 | 7.5 | −10 | −2 | −14 |
10−15 | 25 | 15 | 12.5 | −5 | −1 | −15 |
15−20 | 49 | 24 | 17.5 | 0 | 0 | 0 |
20−25 | 65 | 16 | 22.5 | 5 | 1 | 16 |
25−30 | 73 | 8 | 27.5 | 10 | 2 | 16 |
30−35 | 78 | 5 | 32.5 | 15 | 3 | 15 |
35−40 | 80 | 2 | 37.5 | 20 | 4 | 8 |
Total | ∑ fi = 80 | ∑ fiui = 17 |
The mean of given data is given by
​
​Thus, the mean marks correct to 2 decimal places is 18.56.
Page No 380:
Question 1:
In a hospital, the ages of diabetic patients were recorded as follows. Find the median age. [CBSE 2014]
Age (in years) | 0−15 | 15−30 | 30−45 | 45−60 | 60−75 |
Number of patients | 5 | 20 | 40 | 50 | 25 |
Answer:
We prepare the cumulative frequency table, as shown below:
Age (in years) | Number of patients (fi) | Cumulative Frequency (cf) |
0−15 | 5 | 5 |
15−30 | 20 | 25 |
30−45 | 40 | 65 |
45−60 | 50 | 115 |
60−75 | 25 | 140 |
Total | N = ∑ fi = 140 |
Now, N = 140 .
The cumulative frequency just greater than 70 is 115 and the corresponding class is 45−60.
Thus, the median class is 45−60.
∴ l = 45, h = 15, f = 50, N = 140 and cf = 65.
Now,
Hence, the median age is 46.5 years.
Page No 380:
Question 2:
Compute the median from the following data:
Marks | 0-7 | 7-14 | 14-21 | 21-28 | 28-35 | 35-42 | 42-49 |
Number of students | 3 | 4 | 7 | 11 | 0 | 16 | 9 |
Answer:
Class |
Frequency (f) |
Cumulative frequency |
0-7 |
3 |
3 |
7-14 |
4 |
7 |
14-21 |
7 |
14 |
21-28 |
11 |
25 |
28-35 |
0 |
25 |
35-42 |
16 |
41 |
42-49 |
9 |
50 |
|
N=∑f=50 |
|
Page No 380:
Question 3:
The following table shows the daily wages of workers in a factory:
Daily wages (in Rs) | 0-100 | 100-200 | 200-300 | 300-400 | 400-500 |
Number of workers | 40 | 32 | 48 | 22 | 8 |
Find the median daily wage income of the workers.
Answer:
Class |
Frequency(f) |
Cumulative frequency |
0-100 |
40 |
40 |
100-200 |
32 |
72 |
200-300 |
48 |
120 |
300-400 |
22 |
142 |
400-500 |
8 |
150 |
|
N=∑f=150 |
|
Page No 381:
Question 4:
Calculate the median from the following frequency distribution:
Class | 5-10 | 10-15 | 15-20 | 20-25 | 20-30 | 30-35 | 35-40 | 40-45 |
Frequency | 5 | 6 | 15 | 10 | 5 | 4 | 2 | 2 |
Answer:
Class |
Frequency(f) |
Cumulative frequency |
5-10 |
5 |
5 |
10-15 |
6 |
11 |
15-20 |
15 |
26 |
20-25 |
10 |
36 |
25-30 |
5 |
41 |
30-35 |
4 |
45 |
35-40 |
2 |
47 |
40-45 |
2 |
49 |
|
N=∑f=49 |
|
Page No 381:
Question 5:
Given below is the number of units of electricity consumed in a week in a certain locality:
Consumption (in units) |
65-85 | 85-105 | 105-125 | 125-145 | 145-165 | 165-185 | 195-205 |
Number of consumers |
4 | 5 | 13 | 20 | 14 | 7 | 4 |
Calculate the median
Answer:
Class |
Frequency(f) |
Cumulative frequency |
65-85 |
4 |
4 |
85-105 |
5 |
9 |
105-125 |
13 |
22 |
125-145 |
20 |
42 |
145-165 |
14 |
56 |
165-185 |
7 |
63 |
185-205 |
4 |
67 |
|
N=∑f=67 |
|
Page No 381:
Question 6:
Calculate the median from the following data:
Height (in cm) |
135-140 | 140-145 | 145-150 | 150-155 | 155-160 | 160-165 | 165-170 | 170-175 |
No. of boys |
6 | 10 | 18 | 22 | 20 | 15 | 6 | 3 |
Answer:
Class |
Frequency(f) |
Cumulative frequency |
135-140 |
6 |
6 |
140=145 |
10 |
16 |
145-150 |
18 |
34 |
150-155 |
22 |
56 |
155-160 |
20 |
76 |
160-165 |
15 |
91 |
165-170 |
6 |
97 |
170-175 |
3 |
100 |
|
N=∑f=100 |
|
Page No 381:
Question 7:
Calculate the missing frequency from the following distribution, it being given that the median of the distribution is 24.
Class | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 |
Frequency | 5 | 25 | ? | 18 | 7 |
Answer:
Class | Frequency (fi) | c.f |
0-10 | 5 | 5 |
10-20 | 25 | 30 |
20-30 | x | x+30 |
30-40 | 18 | x+48 |
40-50 | 7 | x+55 |
Page No 381:
Question 8:
The median of the following data is 16. Find the missing frequencies a and b if the total of frequencies is 70.
Class | 0−5 | 5−10 | 10−15 | 15−20 | 20−25 | 25−30 | 30−35 | 35−40 |
Frequency | 12 | a | 12 | 15 | b | 6 | 6 | 4 |
Answer:
We prepare the cumulative frequency table, as shown below:
Class | Frequency (fi) | Cumulative frequency (cf) |
0−5 | 12 | 12 |
5−10 | a | 12 + a |
10−15 | 12 | 24 + a |
15−20 | 15 | 39 + a |
20−25 | b | 39 + a + b |
25−30 | 6 | 45 + a + b |
30−35 | 6 | 51 + a + b |
35−40 | 4 | 55 + a + b |
Total | N = ∑fi = 70 |
55 + a + b = 70 ⇒ a + b = 15 ...(1)
Median is 16, which lies in 15−20. So, the median class is 15−20.
∴ l = 15, h = 5, N = 70, f = 15 and cf = 24 + a
Now,
∴ b = 15 − a [From (1)]
⇒ b = 15 − 8
⇒ b = 7
Hence, a = 8 and b = 7.
Page No 381:
Question 9:
In the following data the median of the runs scored by 60 top batsmen of the world in one-day international cricket matches is 5000. Find the missing frequencies x and y.
Runs scored | 2500−3500 | 3500−4500 | 4500−5500 | 5500−6500 | 6500−7500 | 7500−8500 |
Number of batsmen | 5 | x | y | 12 | 6 | 2 |
Answer:
We prepare the cumulative frequency table, as shown below:
Runs scored | Number of batsmen (fi) | Cumulative frequency (cf) |
2500−3500 | 5 | 5 |
3500−4500 | x | 5 + x |
4500−5500 | y | 5 + x + y |
5500−6500 | 12 | 17 + x + y |
6500−7500 | 6 | 23 + x + y |
7500−8500 | 2 | 25 + x + y |
Total | N = ∑fi = 60 |
25 + x + y = 60 ⇒ x + y = 35 ...(1)
Median is 5000, which lies in 4500−5500. So, the median class is 4500−5500.
∴ l = 4500, h = 1000, N = 60, f = y and cf = 5 + x
Now,
∴ y = 35 − x [From (1)]
⇒ y = 35 − 15
⇒ y = 20
Hence, x = 15 and y = 20.
Page No 381:
Question 10:
If the median of the following frequency distribution is 32.5, find the values of f1 and f2.
Class interval |
0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50-60 | 60-70 | Total |
Frequency | f1 | 5 | 9 | 12 | f2 | 3 | 2 | 40 |
Answer:
Class |
Frequency(f) |
Cumulative frequency |
0-10 |
|
|
10-20 |
5 |
+5 |
20-30 |
9 |
+14 |
30-40 |
12 |
+26 |
40-50 |
|
+26 |
50-60 |
3 |
+29 |
60-70 |
2 |
+31 |
|
N=∑f=40 |
|
Page No 382:
Question 11:
Calculate the median for the following data:
Age (in years) | 19-25 | 26-32 | 33-39 | 40-46 | 47-53 | 54-60 |
Frequency | 35 | 96 | 68 | 102 | 35 | 4 |
Answer:
First, we will convert the data into exclusive form.
Class |
Frequency(f) |
Cumulative frequency |
18.5-25.5 |
35 |
35 |
25.5-32.5 |
96 |
131 |
32.5-39.5 |
68 |
199 |
39.5-46.5 |
102 |
301 |
46.5-53.5 |
35 |
336 |
53.5-60.5 |
4 |
340 |
|
N=∑f=340 |
|
Page No 382:
Question 12:
Find the median wages for the following frequencies distribution:
Wages per day (in Rs) |
61-70 | 71-80 | 81-90 | 91-100 | 101-110 | 111-120 |
No. of women workers |
5 | 15 | 20 | 30 | 20 | 8 |
Answer:
Converting the given data into exclusive form, we get:
Class |
Frequency(f) |
Cumulative frequency |
60.5-70.5 |
5 |
5 |
70.5-80.5 |
15 |
20 |
80.5-90.5 |
20 |
40 |
90.5-100.5 |
30 |
70 |
100.5-110.5 |
20 |
90 |
110.5-120.5 |
8 |
98 |
|
N=∑f=98 |
|
Page No 382:
Question 13:
Find the median from the following data:
Class | 1-5 | 6-10 | 11-15 | 16-20 | 21-25 | 26-30 | 31-35 | 35-40 | 41-45 |
Frequency | 7 | 10 | 16 | 32 | 24 | 16 | 11 | 5 | 2 |
Answer:
Converting into exclusive form, we get: