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Class 10 RD Sharma Solutions – Chapter 7 Statistics – Exercise 7.3 | Set 2

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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 d= x– 42 u= (x– 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 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 <strong>u_i=(\frac{x_i-1650}{100})</strong> 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 -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 <strong>u_i=(\frac{x_i-25}{10})</strong> 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 \frac{752+20x}{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: 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\frac{1245+15p}{43+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\frac{upperlimit+lowerclasslimit}{2}   

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 <strong>u_i=(\frac{x_i-25}{10})</strong> 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\frac{upperlimit+lowerclasslimit}{2}

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 <strong>u_i=(\frac{x_i-225}{50})</strong> 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

Mean(\overline{x})=a+(sumN)\times h\overline{x}=a+(\frac{sum}{N})\times h

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 = \frac{upperlimit+lowerclasslimit}{2}

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

Mean(\overline{x})=a+(sumN)\times h\overline{x}=a+(\frac{sum}{N})\times h

= 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 = \frac{upperlimit+lowerclasslimit}{2}

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

Mean(\overline{x})=a+(sumN)\times \overline{x}=a+(\frac{sum}{N})

= 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 = \frac{upperlimit+lowerclasslimit}{2}

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 

Mean(\overline{x})=a+(sumN)\times h\overline{x}=a+(\frac{sum}{N})\times h

= 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\frac{upperlimit+lowerclasslimit}{2}   

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 <strong>u_i=(\frac{x_i-25}{10})</strong> 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\frac{upperlimit+lowerclasslimit}{2}

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 <strong>u_i=(\frac{x_i-225}{50})</strong> 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

Mean(\overline{x})=a+(sumN)\times h\overline{x}=a+(\frac{sum}{N})\times h

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 = \frac{upperlimit+lowerclasslimit}{2}

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

Mean(\overline{x})=a+(sumN)\times h\overline{x}=a+(\frac{sum}{N})\times h

= 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 = \frac{upperlimit+lowerclasslimit}{2}

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

Mean(\overline{x})=a+(sumN)\times \overline{x}=a+(\frac{sum}{N})

= 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 = \frac{upperlimit+lowerclasslimit}{2}

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 

Mean(\overline{x})=a+(sumN)\times h\overline{x}=a+(\frac{sum}{N})\times h

= 70 + (-2/35)

= 70 – 4/7

= 70 – 0.57

= 69.43

So, mean literacy rate is 69.43%


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Last Updated : 03 Mar, 2021
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