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

  • Last Updated : 03 Mar, 2021
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Question 14. Find the mean of the following frequency distribution:

Class interval:25 – 2930 – 3435 – 3940 – 4445 – 4950 – 5455 – 59
Frequency:1422166534

Solution:

Let’s consider the assumed mean (A) = 42

Class intervalMid – value xid= x– 42u= (x– 42)/5fifiui
25 – 2927-15-314-42
30 – 3432-10-222-44
35 – 3937-5-116-16
40 – 44420060
45 – 49475155
50 – 545210236
55 – 5957153412
    N = 70Σ 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 – 44 – 99 – 1616 – 20
Frequency:6122620

Solution:

By direct method

Class intervalMid value xiFrequency fifixi
1 – 42.5615
4 – 96.51218
9 – 1612.526325
16 – 2721.520430
  N = 64Sum = 848

Mean = (sum/N) + A

= 848/64



= 13.25

By assuming mean method

Let the assumed mean (A) = 65

Class intervalMid value xiui = (xi – A) = xi – 65Frequency fifiui
1 – 42.5-46-25
4 – 96.50120
9 – 1612.5626196
16 – 2721.51520300
   N = 64Sum = 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 indexNumber of studentsCost of living indexNumber of students
1400 – 15001700 – 18009
1500 – 1600101800 – 19006
1600 – 1700201900 – 20002

Solution:

Let the assumed mean (A) = 1650

Class intervalMid value xidi = xi – A = xi – 1650<strong>u_i=(\frac{x_i-1650}{100})</strong>Frequency fifiui
1400 – 15001450-200-25-10
1500  – 16001550-100-110-10
1600 – 1700165000200
1700 – 18001750100199
1800 – 190018502002612
1900 – 20001950300326
    N = 52Sum = 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 – 1010 – 2020 – 3030 – 4040 – 50
Number of students:2024403620

Calculate the average marks by using all the three methods: direct method, assumed mean deviation and shortcut method.

Solution:

(i) Direct method:

Class intervalMid value xiFrequency fifixi
0 – 10520100
10 – 201524360
20 – 3025401000
30 – 40 35361260
40 – 504520900
  N = 140Sum = 3620

Mean = sum/N

= 3620/140

= 25.857

(ii) Assumed mean method:



Let the assumed mean = 25

Mean = A + (sum/N)

Class intervalMid value xiui = (xi – A)Frequency fifiui
0 – 10-2020-400
10 – 2015-1024-240
20 – 30250400
30 – 40351036360
40 – 50452020400
   N = 140Sum = 120

Mean = A + (sum/N)

= 25 + (120/140)

= 25 + 0.857

= 25.857

(iii) Step deviation method:

Let the assumed mean (A) = 25

Class intervalMid value xidi = xi – A = xi – 25<strong>u_i=(\frac{x_i-25}{10})</strong>Frequency fifiui
0 – 10 5-20-220-40
10 – 2015-10-124-24
20 – 302500400
30 – 40351013636
40 – 50452022040
    N = 140Sum = 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 – 2020 – 4040 – 6060 – 80 80 – 100100 – 120
Frequency:5f110f278

Solution:

Class intervalMid value xiFrequency fifixi
0 – 20 10550
20 – 4030fi30fi
40 – 605010500
60 – 8070f270f2
80 – 100907630
100 – 1201108880
  N = 50Sum = 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 – 1313 – 1515 – 1717 – 1919 – 2121 – 2323 – 25
Frequency:76913    –54

Solution:

Given mean = 18,

Let the missing frequency be v

Class intervalMid interval xiFrequency fifixi
11 – 1312784
13 – 1514688
15 – 17169144
17 – 191813234
19 – 2120x20x
21 – 23225110
23 – 2514456
  N = 44 + xSum = 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 – 1010 – 2020 – 3030 – 4040 – 50
Frequency:P121310

Solution:

Class intervalMid value xiFrequency fifixi
0 – 105840
10 – 2015P152
20 – 302512300
30 – 403513455
40 – 504516450
  N = 43 + PSum = 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 – 5253 – 5556 – 5859 – 6162 – 64
Number of boxes: 1511013511525

Find the mean number of mangoes kept in packing box. Which method of finding the mean did you choose?

Solution:

Number of mangoesNumber of boxes
50 – 5215
53 – 55110
56 – 58135
59 – 61115
62 – 6425

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 intervalFrequency fiMid values xidi = xi – A = xi – 25<strong>u_i=(\frac{x_i-25}{10})</strong>fiui
49.5 – 52.51551-6-2-30
52.5 – 55.511054-3-1-110
55.5 – 58.513557000
58.5 – 61.51156031115
61.5 – 64.525636250
TotalN = 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 – 150150 – 200200 – 250250 – 300300 – 350
Number of households:451222

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 f1Mid value xidi = xi – 225<strong>u_i=(\frac{x_i-225}{50})</strong>fiui
100 – 1504125-100-2-8
150 – 2005175-50-1-5
200 – 25012225000
250 – 30022755012
300 – 350232510024
 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 – 0.044
0.04 – 0.089
0.08 – 0.129
0.12 – 0.162
0.16 – 0.204
0.20 – 0.242

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 SO2Frequency f1Class interval xi`di = xi – 0.14uifiui
0.00 – 0.0440.02-0.12-3-12
0.04 – 0.0890.06-0.08-2-18
0.08 – 0.1290.10-0.04-1-9
0.12 – 0.1620.14000
0.16 – 0.2040.180.0414
0.20 – 0.2420.220.0824
TotalN = 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 – 66 – 1010 – 1414 – 2020 – 2828 – 3838 – 40
Number of students:111074431

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 daysNumber of students fiXid = xi + 10fidi
0 – 6113-13-143
6 – 10108-8-280
10 – 14712-4-28
14 – 2071600
20 – 28824832
28 – 363331751
30 – 401392323
TotalN = 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 – 5555 – 6565 – 7575 – 8585 – 95
Number of cities:3101183

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 xidi = xi – 70ui = di – 50fiui
45 – 55350-20-20-6
55 – 651060-10-1-10
65 – 751170000
75 – 858801018
85 – 953902026
TotalN = 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 – 5253 – 5556 – 5859 – 6162 – 64
Number of boxes: 1511013511525

Find the mean number of mangoes kept in packing box. Which method of finding the mean did you choose?

Solution:

Number of mangoesNumber of boxes
50 – 5215
53 – 55110
56 – 58135
59 – 61115
62 – 6425

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 intervalFrequency fiMid values xidi = xi – A = xi – 25<strong>u_i=(\frac{x_i-25}{10})</strong>fiui
49.5 – 52.51551-6-2-30
52.5 – 55.511054-3-1-110
55.5 – 58.513557000
58.5 – 61.51156031115
61.5 – 64.525636250
TotalN = 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 – 150150 – 200200 – 250250 – 300300 – 350
Number of households:451222

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 f1Mid value xidi = xi – 225<strong>u_i=(\frac{x_i-225}{50})</strong>fiui
100 – 1504125-100-2-8
150 – 2005175-50-1-5
200 – 25012225000
250 – 30022755012
300 – 350232510024
 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 – 0.044
0.04 – 0.089
0.08 – 0.129
0.12 – 0.162
0.16 – 0.204
0.20 – 0.242

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 SO2Frequency f1Class interval xi`di = xi – 0.14uifiui
0.00 – 0.0440.02-0.12-3-12
0.04 – 0.0890.06-0.08-2-18
0.08 – 0.1290.10-0.04-1-9
0.12 – 0.1620.14000
0.16 – 0.2040.180.0414
0.20 – 0.2420.220.0824
TotalN = 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 – 66 – 1010 – 1414 – 2020 – 2828 – 3838 – 40
Number of students:111074431

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 daysNumber of students fiXid = xi + 10fidi
0 – 6113-13-143
6 – 10108-8-280
10 – 14712-4-28
14 – 2071600
20 – 28824832
28 – 363331751
30 – 401392323
TotalN = 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 – 5555 – 6565 – 7575 – 8585 – 95
Number of cities:3101183

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 xidi = xi – 70ui = di – 50fiui
45 – 55350-20-20-6
55 – 651060-10-1-10
65 – 751170000
75 – 858801018
85 – 953902026
TotalN = 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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