RD Sharma Solutions Class 10 Statistics Exercise 7.4

RD Sharma Solutions Class 10 Chapter 7 Exercise 7.4

RD Sharma Class 10 Solutions Chapter 7 Ex 7.4 PDF Free Download

Exercise 7.4

 

1. Following are the lives in hours of 15 pieces of the components of aircraft engine. Find the median: 715, 724, 725, 710, 729, 745, 694, 699, 696, 712, 734, 728, 716, 705, 719

 

Soln:

Lives in hours of is pieces are = 715, 724, 725, 710, 729, 745, 694, 699, 696, 712, 734, 728, 716, 705, 719

Arrange the above data in ascending order = 694, 696, 699, 705, 710, 712, 715, 716, 719, 721, 725, 728, 729, 734, 745

N = 15 (odd)

Median = \(\left ( \frac{N + 1}{2} \right )^{th}\) terms

= \(\left ( \frac{15 + 1}{2} \right )^{th}\) terms = 8th terms = 716

 

2.The following is the distribution of height of students of a certain class in a certain city:

 

Height (in cm): 160-162                163-165                166-168                169-171                172-174
No of students: 15                           118                         142                         127                         18

 

Find the median height.

 

Soln:

 

Class interval (exclusive) Class interval  (inclusive) Class interval frequency Cumulative frequency
160 – 162 159.5 – 162.5 15 15
163 – 165 162.5 – 165.5 118 133 (F)
166 – 168 165.5 – 168.5 142 (f) 275
169 – 171 168.5 – 171.5 127 402
172 – 174 171.5 – 174.5 18 420
N = 420

 

We have

N = 420

N/2 = 420/ 2 = 120

The cumulative frequency just greater than N/2 is 275 then 165.5 – 168.5 is the median class such, that

L = 165.5, f = 142, F = 133 and h = 168.5 – 165.5 = 3

Mean = \(L + \frac{\frac{N}{2}-F}{f}\times h\)

= \(165.5 + \frac{\frac{10\times 2}{142}=10\) = 166.5 + 1.63 = 168.13

 

3. Following is the distribution of I.Q of 100 students. Find the median I.Q.

 

I.Q: 55-64     65-74     75-84     85-94     95-104     105-114     115-124    125-134      135-144
No of students: 1              2              9              22           33              22                 8                     2              1

 

Soln:

 

Class interval (exclusive) Class interval  (inclusive) Class interval frequency Cumulative frequency
55 – 64 54.5 – 64-5 1 1
65 – 74 64.5 – 74.5 2 3
75 – 84 74.5 – 84.5 9 12
85 – 94 84.5 – 94.5 22 34 (f)
95 – 104 94.5 – 104.5 33 (f) 67
105 – 114 104.5 – 114.5 22 89
115 – 124 114.5 – 124.5 8 97
125 – 134 124.5 – 134.5 2 99
135 – 144 134.5 – 144.5 1 100
N = 100

 

We have N = 100

N/ 2 = 100/ 2 = 50

The cumulative frequency just greater than N/ 2 is 67 then the median class is 94.5-104.5 such that

L=94.5,F = 33 h =104.5 – 94.5=10

Mean = \(L + \frac{\frac{N}{2}-F}{f}\times h\)

= \(94.5 + \frac{50-34}{33}\times 10\) = 94.5 + 4.88 = 99.35

 

4.Calculate the median from the following data:

 

Rent (in Rs): 15-25     25-35     35-45     45-55     55-65     65-75     75-85     85-95
No of houses: 8              10           15           25           40           20           15           7

 

Soln:

Class interval Frequency Cumulative frequency
15 – 25 8 8
25 – 35 10 18
35 – 45 15 33(f)
45 – 55 25 58
55 – 65 40(f) 28
65 – 75 20 38
75 – 85 15 183
85 – 95 7 140
N = 140

 

We have N = 140

N/ 2 = 140/ 2 = 70

The cumulative frequency just greater than N/ 2 is 98 then median class is 55 – 65 such that

L = 55, f = 40, F = 58, h = 65 – 55 = 10

Mean = \(L + \frac{\frac{N}{2}-F}{f}\times h\)

= \(55 + \frac{70 – 58}{40}\times 10\) = 55 + 3 = 58

 

5.Calculate the median from the following data:

 

Marks below:    10           20           30           40           50           60           70           80
No of students:                15           35           60           84           96           127         198         250

 

Soln:

 

Marks below No of students Class interval Frequency Cumulative frequency
10 15 0 – 10 15 15
20 35 10 – 20 20 35
30 60 20 – 30 25 60
40 84 30 – 40 24 84
50 96 40 – 50 12 96(F)
60 127 50 – 60 31 (f) 127
70 198 60 – 70 71 198
80 250 70 – 80 52 250
N = 250

 

We have N = 250

N/ 2 = 250/ 2 = 125

The cumulative frequency just greater than N/ 2 is 127 then median class is 50 – 60 such that

L = 50, f = 31, F = 96, h = 60 -50 = 10

Mean = \(L + \frac{\frac{N}{2}-F}{f}\times h\)

= \(50 + \frac{125-96}{31}\times 10\) = 50 + 9.35 = 59.35

 

6.Calculate the missing frequency from the following distribution, it being given that the median of the distribution is 24.

Age in years: 0-10       10-20     20-30     30-40     40-50
No of persons: 5              25           ?              18           7

Soln:

Class interval Frequency Cumulative frequency
0 – 10 5 5
10 – 20 25 30 (F)
20 – 30 x (f) 30 + x
30 – 40 18 48 + x
40 – 50 7 55 + x
N = 170

 

Given

Median = 24

Then, median class = 20 – 30

L = 20,   h = 30 -20 = 10, f = x,      F = 30

Median = \(L + \frac{\frac{N}{2}-F}{f}\times h\)

24 = \(20 + \frac{\frac{55+x}{2}-30}{x}\times 10\)

24 – 20 = \(\frac{\frac{55+x}{2}-30}{x}\times 10\)

4x = \(\left(\frac{55+x}{2}-30\right)\times 10\)

4x = 275 + 5x – 300

4x – 5x = -25

-x = -25

x = 25

Missing frequency = 25

 

 

7.The following table gives the frequency distribution of married women by age at marriage.

Age (in years) Frequency Age (in years) Frequency
15 – 19 53 40 – 44 9
20 – 24 140 45 – 49 5
25 – 29 98 50 – 54 3
30 – 34 32 55 – 59 3
35 – 39 12 60 and above 2

Calculate the median and interpret the results

Soln:

Class interval (exclusive) Class interval (inclusive) Frequency Cumulative frequency
15 – 19 14.5 – 19.5 53 53 (F)
20 – 24 19.5 – 24.5 140 (f) 193
25 – 29 24.5 – 29.5 98 291
30 – 34 29.5 – 34.5 32 323
35 – 39 34.5 – 39.5 12 335
40 – 44 39.5 – 44.5 9 344
45 – 49 44.5 – 49.5 5 349
50 – 54 49.5 – 54.5 3 352
55 – 54 54.5 – 59.5 3 355
60 and above 59.5 and above 2 357
N = 357

 

N = 357

N/ 2 = 357/ 2 = 178.5

The cumulative frequency just greater than N/ 2 is 193,

Then the median class is 19.5 – 24.5 such that l = 19.5, f = 140, F = 53, h = 25.5 – 19.5 = 5

Median = \(l + \frac{\frac{N}{2}-F}{f}\times h\)

Median = \(19.5 + \frac{178.5-53}{140}\times 5\)

Median = 23.98

Nearly half the women were married between the ages of 15 and 25

 

 

8.The following table gives the distribution of the life time of 400 neon lamps:

Life time: Number of lamps
1500 – 2000 14
2000 – 2500 56
2500 – 3000 60
3000 – 3500 86
3500 – 4000 74
4000 – 4500 62
4500 – 5000 48

Find the median life.

Soln: We can find cumulative frequencies with their respective class intervals as below

Life time Number of lamps fi Cumulative frequency (cf)
1500 – 2000 14 14
2000 – 2500 56 70
2500 – 3000 60 130
3000 – 3500 86 216
3500 – 4000 74 290
4000 – 4500 62 352
4500 – 5000 48 400
Total (n) 400

Now we may observe that cumulative frequency just greater than n/ 2 (400/ 2 = 200) is 216 belongs to class interval 3000 – 3500

Median class = 3000 – 3500

Lower limits (l) of median class = 3000

Frequency (f) of median class = 86

Cumulative frequency (cf) of class preceding median class = 130

Class size (h) = 500

Median = \(l + \left ( \frac{\frac{n}{2}-cf}{f}\right )\times h \)

= \(3000 + \left (\frac{200 -130}{86}\right) \times 500 )\)

= 3000 + (35000/ 86)

= 3406.98

So, median life time of lamps is 3406.98 hours

 

 

9.The distribution below gives the weight of 30 students in a class. Find the median weight of students:

Weight (in kg): 40-45     45-50     50-55     55-60     60-65     65-70     70-75
No of students: 2              3              8              6              6              3              2

Soln: We may find cumulative frequency with their respective class intervals as below

Weight (in kg) Number of students fi Cumulative frequency (cf)
40 – 45 2 2
45 – 50 3 5
50 – 55 8 13
55 – 60 6 19
60 – 65 6 25
65 – 70 3 28
70 – 75 2 30

Cumulative frequency just greater than n/ 2 (i.e. 30/ 2 = 15) is 19, belonging to class interval 55 – 60

Median class = 55 – 60

Lower limit (l) of median class = 55

Frequency (f) of median class = 6

Cumulative frequency (cf) = 13

Class size (h) = 5

Median = \(l + \left ( \frac{\frac{n}{2}-cf}{f}\right )\times h \)

= \(55 + \left (\frac{15 -13}{6}\right) \times 5 )\)

= 55 +10/ 6

= 56.666

So, median weight is 56.67 kg

 

 

10.Find the missing frequencies and the median for the following distribution if the mean is 1.46

No of accidents:                               0              1              2              3              4              5 Total
Frequencies (no of days): 46           ?              ?              25           10           5 200

                               

Soln:

No of accidents (x) No of days (f) fx
0 46 0
1 x x
2 y 2y
3 25 75
4 10 40
5 5 25
N = 200 Sum = x + 2y + 140

 

Given

N = 200

46 + x + y + 25 + 10 + 5 = 200

x + y = 200 – 46 – 25 – 10 – 5

x + y = 114 —- (1)

And, Mean = 1.46

Sum/ N = 1.46

(x + 2y + 140)/ 200 = 1.46

x + 2y = 292 – 140

x + 2y = 152 —- (2)

Subtract equation (1) from equation (2)

x + 2y – x – y = 152 – 114

y = 38

Putting the value of y in equation (1), we have x = 114 – 38 = 76

No of accidents No of days Cumulative frequency
0 46 46
1 76 122
2 38 160
3 25 185
4 10 195
5 5 200
N = 200

We have,

N = 200

N/ 2 = 200/ 2 = 100

The cumulative frequency just more than N/ 2 is 122 then the median is 1

 

 

11.An incomplete distribution is given below:

Variable: 10-20     20-30     30-40     40-50     50-60     60-70     70-80
Frequency: 12           30           ?              65           ?              25           18

You are given that the median value is 46 and the total number of items is 230.

(i) Using the median formula fill up the missing frequencies.

(ii) Calculate the AM of the completed distribution.

Soln:

(i)

Class interval Frequency Cumulative frequency
10 – 20 12 12
20 – 30 30 42
30 – 40 x 42+ x (F)
40 – 50 65 (f) 107 + x
50 – 60 Y 107 + x + y
60 – 70 25 132 + x + y
70 – 80 18 150 + x + y
N = 150

 

Given

Median = 46

Then, median class = 40 – 50

L = 40, h = 50 – 40 = 10, f = 65, F = 42 + x

Median = \(L+\frac{\frac{N}{2}-F}{f}\times h\)

46 = \(40+\frac{115-\left(42+x\right)}{65}\times 10\)

46 – 40 = \(\frac{115-42-x}{65}\times 10\)

6 (65/ 10) = 73 –x

39 = 73 – x

x = 73 – 39 = 34

Given

N = 230

12 + 30 + 34 + 65 + y + 25 + 18 = 230

184 + y = 230

Y = 230 – 184

Y = 46

(ii)

Class interval Mid value x Frequency f Fx
10 – 20 15 12 180
20 – 30 25 30 750
30 – 40 35 34 1190
40 – 50 45 65 2925
50 – 60 55 46 2530
60 – 70 65 25 1625
70 – 80 75 18 1350
N = 230 \(\Sigma fx = 10550\)

 

Mean = \(\frac{\Sigma fx}{N}\)

= 10550/ 230 = 45.87

 

 

12.If the median of the following frequency distribution is 28.5 find the missing frequencies:

Class interval: 0-10       10-20     20-30     30-40     40-50     50-60     Total
Frequency: 5              f1             20           15           f2             5              60

Soln:

Class interval Frequency Cumulative frequency
0 – 10 5 5
10 – 20 f1 5 + f1 (F)
20 – 30 20 (f) 25 + f1
30 – 40 15 40 + f1
40 – 50 f2 40 + f1 + f2
N = 60

Given

Median = 28.5

Then, median class = 20 – 30

Median = \(l+\frac{\frac{N}{2}-F}{f}\times h\)

28.5 = \(20\frac{30- \left( 5+ f_{1}\right)}{20}\times 10\)

28.5 -20 = \(\frac{30- 5_{1}}{20}\times 10\)

8.5 = \(\frac{25 –f_{1}}{2}\)

17 = 25 – f_{1}

f1 = 25 – 17 = 8

Given

Sum of frequencies = 60

5 + f1 + 20 + 15 + f2 + 5 = 60

5 + 8 + 20 + 15 + f2 + 5 = 60

f2 = 7

f1 = 8 and f2 = 7

 

 

13.The median of the following data is 525. Find the missing frequency, if it is given that there are 100 observations in the data.

Class interval Frequency Class interval Frequency
0 – 100 2 500 – 600 20
100 – 200 5 600 – 700 f2
200 – 300 f1 700 – 800 9
300 – 400 12 800 – 900 7
400 – 500 17 900 – 1000 4

Soln:

Class interval Frequency Cumulative frequency
0 – 100 2 2
100 – 200 5 7
200 – 300 f1 7 + f1
300 – 400 12 19 + f1
400 – 500 17 36 + f­1 (F)
500 – 600 20 (f) 56 + f1
600 – 700 f2 56 + f1 + f2
700 – 800 9 65 + f1 + f2
800 – 900 7 72 + f1 + f2
900 – 1000 4 76 + f1 + f2
N = 100

 

Given

Median                = 525

Then, median class = 500 – 600

L = 500, f = 20, F = 36 + f1, h = 600 – 500 = 100

Median = \(L+\frac{\frac{N}{2}-F}{f}\times h\)

525 = \(500+\frac{50-\left( 36+ f_{1}\right)}{20}\times 100\)

525 = 500 +\(\frac{50- 36- f_{1}}{20}\times 100\)

25 = (14 – f1) x 5

5 = 14 – f1

f1 = 14 – 5 = 9

Given

Sum of frequencies = 100

2 + 5 + f1 + 12 + 17 + 20 + f2 + 9 + 7 + 4 = 100

2 + 5 + 9 + 12 + 17 + 20 + f2 + 9 + 7 + 4 = 100

85 + f2 = 100

f2 = 100 – 85 = 15

f1 = 9 and f2 = 15

 

 

14.If the median of the following data is 32.5, find the missing frequencies.

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

Soln:

Class interval Frequency Cumulative frequency
0 – 10 f1 f1
10 – 20 5 5 + f1
20 – 30 9 14 + f1
30 – 40 12 (f) 26 + f1
40 – 50 f2 26 + f1 + f2
50 – 60 3 29 + f1 + f2
60 – 70 2 31 + f1 + f2
N = 40

Given

Median = 32.5

The median class = 90 – 40

L = 30, h = 40 – 30 = 10, f = 12, F = 14 + f1

Median = \(L + \frac{\frac{N}{2}-F}{f}\times h\)

32.5 = \(30 + \frac{20- \left( 14+ f_{1}\right)}{12}\times 10\)

32.5 – 30 = \(\frac{20- \left( 14+ f_{1}\right)}{12}\times 10\)

2.5 (12) = (6 – f1) * 10

30 = (6 – f1) * 10

3 = 6 – f1

f1 = 6 – 3 = 3

Given

Sum of frequencies = 40

f1 + 5 + 9 + 12 + f2 + 3 + 2 = 40

3 + 5 + 9 + 12 + f2 + 3 + 2 = 40

34 + f2 = 40

f2 = 40 – 34 = 6

f1 = 3 and f2 = 6

 

 

15.Compute the median for each of the following data

(i)   (ii)  
Marks No of students Marks No of students
Less than 10 0 More than 80 150
Less than 30 10 More than 90 141
Less than 50 25 More than 100 124
Less than 70 43 More than 110 105
Less than 90 65 More than 120 60
Less than 110 87 More than 130 27
Less than 130 96 More than 140 12
Less than 150 100 More than 150 0

 

Soln:(i)

Marks No of students Class interval Frequency Cumulative frequency
Less than 10 0 0 – 10 0 0
Less than 30 10 10 – 30 10 10
Less than 50 25 30 ­– 50 15 25
Less than 70 43 50 – 70 18 43 (F)
Less than 90 65 70 – 90 22 (f) 65
Less than 110 87 90 – 110 22 87
Less than 130 96 110 – 130 9 96
Less than 150 100 130 – 150 4 100
N = 100

We have

N = 100

N/ 2 = 100/ 2 = 50

The cumulative frequency just greater than N/ 2 is 65 then median class is 70 – 90 such that

L = 70, h = 90 – 70 = 20, f = 22, F = 43

Median = \(L + \frac{\frac{N}{2}-F}{f}\times h\)

= \(70 + \frac{50- 43}{22}\times 20\)

= \(70 + \frac{7\times 20}{22}\)

= 70 + 6.36

= 76.36

(ii)

Marks No of students Class interval Frequency Cumulative frequency
More than 80 150 80 – 90 9 9
More than 90 141 90 – 100 17 26
More than 100 124 100 – 110 19 45 (F)
More than 110 105 110 – 120 45 (f) 90
More than 120 60 120 – 130 33 123
More than 130 27 130 – 140 15 138
More than 140 12 140 – 150 12 150
More than 150 0 150 – 160 0 150
N  = 150

We have

N = 150

N/ 2 = 150/ 2 = 75

The cumulative frequency just more than N/ 2 is 90 then the median class is 110 – 120 such that

L = 70, h = 120 – 110 = 10, f = 45, F = 45

Median = \(L + \frac{\frac{N}{2}-F}{f}\times h\)

= \(110 + \frac{75- 45}{45}\times 10\)

= \(110 + \frac{30\times 10}{45}\)

= 110 + 6.67

= 116.67

 

 

16.A survey regarding the height (in cm) of 51 girls of class X of a school was conducted and the following data was obtained:

Height in cm number of girls

 

Less than 140

Less than 145

Less than 150

Less than 155

Less than 160

Less than 165

 

4

11

29

40

46

51

Find the median height.

Soln: To calculate the median height, we need to find the class intervals and their corresponding frequencies.

The given distribution being of the less than type, 140, 145, 150, 155, 160, 165 give the upper limits of the corresponding class intervals. So, the classes should be below 140, 140-145, 145-150, 150-155, 155-160, 160-165. Observe that from the given distribution, we find that there are 4 girls with height less than 140, i.e. the frequency of class interval below 140 is 4. Now, there are 11 girls with heights less than 145 and 4 girls with height less than 140. Therefore, the number of girls with height in the interval 140 – 145 is 11 – 4 = 7. Similarly, the frequency of 145 – 150 is 29 – 11 = 18, for 150 – 155, it is 40 – 29 = 11, and so on. So, our frequency distribution table with given cumulative frequencies becomes:

Class interval Frequency Cumulative frequency
Below 140

140 – 145

145 – 150

150 – 155

155 – 160

160 – 165

4

7

18

11

6

5

4

11

29

40

46

51

 

Now n = 51. S0, n/ 2 = 51/ 2 =25.5 this observation lies in the class 145 – 150

Then,

L (the lower limit) = 145

cf (the cumulative frequency of the class preceding 145 – 150) = 11

f (the frequency of the median class 145 – 150) = 18

h (the class size) = 5

Using the formula, median = \(l + \left ( \frac{\frac{n}{2}-cf}{f} \right ) \times h \), we have

Median = \(145 + \left ( \frac{\frac{25.5-11}{18}\right )\times 5 \)

= 145 + 72.5/ 18 = 149.03

So, the median height of the girls is 149.03 cm

This means that the height of about 50% of the girls in less than this height, and 50% are taller than this height.

 

 

17.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.

Ages in years Number of policy holders
Below 20

Below 25

Below 30

Below 35

Below 40

Below 45

Below 50

Below 55

Below 60

 

2

6

24

45

78

89

92

98

100

Soln: Here class width is not same. There is no need to adjust the frequencies according to class interval. Now given frequencies table is less type represented with upper class limits. As policies were given only to persons having age 18 years onwards but less than 60 years we can define class intervals with their respective cumulative frequency as below.

Age (in years) Number of policy holders fi Cumulative frequency (cf)
18 – 20 2 2
20 – 25 6 – 2 = 4 6
25 – 30 24 – 6 = 18 24
30 – 35 45 – 24 = 21 45
35 – 40 78 – 45 = 33 78
40 – 45 89 – 78 = 11 89
45 – 50 92 – 89 = 3 92
50 – 55 98 – 92 = 6 98
55 – 60 100 – 98 = 2 100
Total

Now from table we may observe that n = 100

Cumulative frequency (cf) just greater than n/ 2 (i.e. 100/ 2 = 50) is 78 belonging to interval 35 – 40

So median class                = 35 – 40

Lower limit (l) of median class = 35

Class size (h) = 5

Frequency (f) of median class = 33

Cumulative frequency (cf) of class preceding median class = 45

Median = \(l + \left ( \frac{\frac{n}{2}-cf}{f}\right )\times h \)

= \(35 + \left ( \frac{50 – 45}{33}\right )\times 5 \)

= \( 35 + \frac {25}{33}\)

= 35.76

So median age is 35.76 years

 

 

18.The lengths of 40 leaves of a plant are measured correct to the nearest millimeter, and the data obtained is represented in the following table:

Length (in mm) No of leaves
118 – 126 3
127 – 135 5
136 – 144 9
145 – 153 12
154 – 162 5
163 – 171 4
172 – 180 2

 Find the mean length of life

Soln: The given data is not having continuous class intervals. We can observe the difference between two class intervals is 1. So we have to add and subtract

1/ 2 = 0.5 to upper class limits and lower class limits

Now continuous class intervals with respective cumulative frequencies can be represented as below:

Length (in mm) Number of leaves fi Cumulative frequency (cf)
117.5 – 126.5 3 3
126.5 – 135.5 5 8
135.5 – 144.5 9 17
144.5 – 153.5 12 29
153.5 – 162.5 5 34
162.5 – 171.5 4 38
171.5 – 180.5 2 40

From the table we may observe that cumulative frequency just greater then n/ 2 (i.e. 40/ 2 =20) is 29, belongs to class interval 144.5 – 153.5

Median class = 144.5 – 153.5

Lower limit (l) = 144.5

Class size (h) = 9

Frequency (f) of median class = 12

Cumulative frequency (cf) of class preceding median class = 17

Median = \(l + \left ( \frac{\frac{n}{2}-cf}{f}\right )\times h \)

= \(144.5 + \left (\frac{20 -17}{12}\right) \times 9 )\)

= 144.5 + 9/ 4 = 146.75

So median length of leaves is 146.75mm

 

 

19.An incomplete distribution is given as follows:

Variable:             0-10       10-20     20-30     30-40     40-50     50-60     60-70
Frequency:         10           20           ?              40           ?              25           15

You are given that the median value is 35 and sum is all the frequencies are 170. Using the median formula, fill up the missing frequencies

Soln:

Class interval Frequency Cumulative frequency
0 – 10 10 10
10 – 20 20 30
20 – 30 f1 30 + f1 (F)
30 – 40 40 (F) 70 + f1
40 – 50 f2 70 + f1 + f2
50 – 60 25 95 + f1 + f2
60 – 70 15 110 + f1 + f2
N = 170

 

Given

Median = 35

Then median class = 30 – 40

L = 30, h = 40 – 30 = 10, f = 40, F = 30 + f1

Median = \(L+\frac{\frac{N}{2}-F}{f}\times h\)

35 = \(30+\frac{85-( 30+ f_{1})}{40}\times 10\)

35 – 30 = \(\frac{85–30–f_{1}}{40}\times 10\)

5 = \(\frac{55 – f_{1}}{4}\)

20 = 55 – f1

f1 = 55 – 20 = 35

Given

Sum of frequencies = 170

10 + 20 + fi + 40 + f2 + 25 + 15 = 170

10 + 20 + 35 + 40 + f2 + 25 + 15 = 170

f2 = 25

f1 = 35 and f2 =25

 


Practise This Question

Find the zeroes of the following quadratic polynomial:
6x2+x5