In this chapter, we provide RD Sharma Class 10 Ex 7.3 Solutions Chapter 7 Statistics for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Class 10 Ex 7.3 Solutions Chapter 7 Statistics pdf, Now you will get step by step solution to each question.

Textbook | NCERT |

Class | Class 10 |

Subject | Maths |

Chapter | Chapter 7 |

Chapter Name | Statistics |

Exercise | 7.3 |

Category | RD Sharma Solutions |

**RD Sharma Solutions for Class 10 Chapter** **7**** Statistics** Ex 7.3 Download PDF

**Statistics**Ex 7.3 Download PDF

**Chapter 7: Statistics Exercise – 7.3**

**Question: 1**

The following table gives the distribution of total household expenditure (in rupees) of manual workers in a city.

Expenditure (in rupees) (x) | Frequency (f_{i}) | Expenditure (in rupees) (x_{i}) | Frequency (f_{i}) |

100 – 150 | 24 | 300 – 350 | 30 |

150 – 200 | 40 | 350 – 400 | 22 |

200 – 250 | 33 | 400 – 450 | 16 |

250 – 300 | 28 | 450 – 500 | 7 |

Find the average expenditure (in rupees) per household

**Solution:**

Let the assumed mean (A) = 275

Class interval | Mid value (x_{i}) | d_{i }= x_{i} – 275 | u_{i }= (x_{i }– 275)/50 | Frequency f_{i} | f_{i}u_{i} |

100 – 150 | 125 | -150 | -3 | 24 | -12 |

150 – 200 | 175 | -100 | -2 | 40 | -80 |

200 – 250 | 225 | -50 | -1 | 33 | -33 |

250 – 300 | 275 | 0 | 0 | 28 | 0 |

300 – 350 | 325 | 50 | 1 | 30 | 30 |

350 – 400 | 375 | 100 | 2 | 22 | 44 |

400 – 450 | 425 | 150 | 3 | 16 | 48 |

450 – 500 | 475 | 200 | 4 | 7 | 28 |

N = 200 | Sum = -35 |

We have A = 275, h = 50

Mean = A + h * sum/N

= 275 + 50 * – 35/200

= 275 – 8.75

= 266.25

**Question: 2**

A survey was conducted by a group of students as a part of their environmental awareness program, in which they collected the following data regarding the number of plants in 200 houses in a locality. Find the mean number of plants per house.

Number of Number of plants: | 0 – 2 | 2 – 4 | 4 – 6 | 6 – 8 | 8 – 10 | 10 – 12 | 12 – 14 |

Number of houses: | 1 | 2 | 5 | 6 | 2 | 3 |

Which method did you use for finding the mean, and why?

**Solution:**

Let us find class marks (x_{i}) = (upper class limit + lower class limit)/2. Now we may compute x_{i} and f_{i}x_{i} as following.

Number of plants | Number of house (f_{i}) | x_{i} | F_{i}x_{i} |

0 – 2 | 1 | 1 | 1 |

2 – 4 | 2 | 3 | 6 |

4 – 6 | 1 | 5 | 5 |

6 – 8 | 5 | 7 | 35 |

8 – 10 | 6 | 9 | 54 |

10 – 12 | 2 | 11 | 22 |

12 – 14 | 3 | 13 | 39 |

Total | N = 20 | Sum = 162 |

From the table we may observe that N = 20, Sum = 162

So mean number of plants per house is 8.1. We have used for the direct method values Xi and fi are very small

**Question: 3**

Consider the following distribution of daily wages of workers of a factory

Daily wages (in Rs) | 100 – 120 | 120 – 140 | 140 – 160 | 160 – 180 | 180 – 200 |

Number of workers: | 12 | 16 | 8 | 6 | 10 |

Find the mean daily wages of the workers of the factory by using an appropriate method.

**Solution:**

Let the assume mean (A) = 150

Class interval | Mid value x_{i} | d_{i }= x_{i} – 150 | u_{i} = (x_{i }– 150)/20 | Frequency f_{i} | F_{i}u_{i} |

100 – 120 | 110 | -40 | -2 | 12 | -24 |

120 – 140 | 130 | -20 | -1 | 14 | -14 |

140 – 160 | 150 | 0 | 0 | 8 | 0 |

160 – 180 | 170 | 20 | 1 | 6 | 6 |

180 – 200 | 190 | 40 | 2 | 10 | 20 |

N = 50 | Sum = -12 |

We have N = 50, h = 20

= 150 – 4.8

= 145.2

**Question: 4**

Thirty women were examined in a hospital by a doctor and the number of heart beats per minute recorded and summarized as follows. Find the mean heart beats per minute for these women, choosing a suitable method. Number of heart

Beats Per minute: | 65 – 68 | 68 – 71 | 71 – 74 | 74 – 77 | 77 – 80 | 80 – 83 | 83 – 86 |

Number of women: | 2 | 4 | 3 | 8 | 7 | 4 | 2 |

**Solution:**

We may find marks of each interval (xi) by using the relation (xi¬) = (upper class limit + lower class limit)/2 Class size of this data = 3 Now talking 75.5 as assumed mean (a) We may calculate di, ui, fiui as following

Number of heart beats per minute | Number of women (x_{i}) | x_{i} | d_{i} = x_{i} – 75.5 | u_{i} = (x_{i }– 755)/h | f_{i}u_{i} |

65-68 | 2 | 66.5 | -9 | -3 | -6 |

68-71 | 9 | 69.5 | -6 | -2 | -8 |

71-74 | 3 | 72.5 | -3 | -1 | -3 |

74-77 | 8 | 75.5 | 0 | 0 | 0 |

77-80 | 7 | 78.5 | 3 | 1 | 7 |

80-83 | 4 | 81.5 | 6 | 2 | 8 |

83-86 | 2 | 84.5 | 9 | 3 | 6 |

N = 30 | Sum = 4 |

Now we may observe from table that N = 30, sum = 4

= 75.5 + 0.4

= 75.9

So mean heart beats per minute for those women are 75.9 beats per minute

**Question: 5**

Find the mean of each of the following frequency distributions: (5 – 14)

Class interval: | 0 – 6 | 6 – 12 | 12 – 18 | 18 – 24 | 24 – 30 |

Frequency: | 6 | 8 | 10 | 9 | 7 |

**Solution:**

Let us assume mean be 15

Class interval | Mid – value | d_{i }= x_{i }– 15 | u_{i }= (x_{i }– 15)/6 | f_{i} | f_{i}u_{i} |

0 – 6 | 3 | -12 | -2 | 6 | -12 |

12-Jun | 9 | -6 | -1 | 8 | -8 |

18-Dec | 15 | 0 | 0 | 10 | 0 |

18 – 24 | 21 | 6 | 1 | 9 | 9 |

24 – 30 | 27 | 18 | 2 | 7 | 14 |

N = 40 | Sum = 3 |

A = 15, h = 6, N = 40

= 15 + 0.45

= 15.47

**Question: 6**

Class interval: | 50 – 70 | 70 – 90 | 90 – 110 | 110 – 130 | 130 – 150 | 150 – 170 |

Frequency: | 18 | 12 | 13 | 27 | 8 | 22 |

**Solution:**

Let us assumed mean be 100

Class interval | Mid-value x_{i} | d_{i }= x_{i} – 100 | u_{i }= (x_{i} – 100)/20 | f_{i} | f_{i}u_{i} |

50 – 70 | 60 | -40 | -2 | 18 | -36 |

70 – 90 | 80 | -20 | -1 | 12 | -12 |

90 – 110 | 100 | 0 | 0 | 13 | 0 |

110 – 130 | 120 | 20 | 1 | 27 | 27 |

130 – 150 | 140 | 40 | 2 | 8 | 16 |

150 – 170 | 160 | 60 | 3 | 22 | 66 |

N = 100 | Sum = 61 |

A = 100, h = 20

= 100 + 12.2

= 122.2

**Question: 7**

Class interval: | 0- 8 | 8 – 16 | 16 – 24 | 24 – 32 | 32 – 40 |

Frequency: | 6 | 7 | 10 | 8 | 9 |

**Solution:**

Let the assumed mean (A) = 20

Class interval | Mid- value | d_{i}= x_{i }– 20 | u_{i} = (x_{i }– 20)/8 | f_{i} | f_{i}u_{i} |

0-8 | 4 | -16 | -2 | 6 | -12 |

16-Aug | 12 | -8 | -1 | 7 | -7 |

16-24 | 20 | 0 | 0 | 10 | 0 |

24-32 | 28 | 8 | 1 | 8 | 8 |

32-40 | 36 | 16 | 2 | 9 | 18 |

N = 40 | Sum = 7 |

We have A = 20, h = 8

Mean = A + h (sum/N)

= 20 + 8 (7/40)

= 20 + 1.4

= 21.4

**Question: 8**

Class interval: | 0 – 6 | 6 – 12 | 12 – 18 | 18 – 24 | 24 – 30 |

Frequency: | 7 | 5 | 10 | 12 | 6 |

**Solution:**

Let the assumed mean be (A) = 15

Class interval | Mid – value | d_{i }= x_{i }-15 | u_{i }= (x_{i }-15)/6 | Frequency f_{i} | f_{i}u_{i} |

0 – 6 | 3 | -12 | -2 | -1 | -14 |

6 – 12 | 9 | -6 | -1 | 5 | -5 |

12 – 18 | 15 | 0 | 0 | 10 | 0 |

18 – 24 | 21 | 6 | 1 | 12 | 12 |

24 – 30 | 27 | 12 | 2 | 6 | 12 |

N = 40 | Sum = 5 |

We have A = 15, h = 6

Mean = A + h(sum/N)

= 15 + 6 (5/40)

= 15 + 0.75

= 15.75

**Question: 9**

Class interval: | 0 – 10 | 20 – 30 | 20 – 30 | 30 – 40 | 40 – 50 |

Frequency: | 6 | 7 | 10 | 8 | 9 |

**Solution:**

Let the assumed mean (A) = 25

Class interval | Mid – value | d_{i }= x_{i }-25 | u_{i }= (x_{i }– 25)/10 | Frequency f_{i} | f_{i}u_{i} |

0 – 10 | 5 | -20 | -2 | 9 | -18 |

20-Oct | 15 | -10 | -1 | 10 | -12 |

20 – 30 | 25 | 0 | 0 | 15 | 0 |

30 – 40 | 35 | 10 | 1 | 10 | 10 |

40 – 50 | 45 | 20 | 2 | 14 | 28 |

N = 60 | Sum = 8 |

We have A = 25, h = 10

Mean = A + h (sum/N)

= 25 + 19 (8/60)

= 25 + (4/3)

= 26.333

**Question: 10**

Class interval: | 0- 8 | 8 – 16 | 16 – 24 | 24 – 32 | 32 – 40 |

Frequency: | 5 | 9 | 10 | 8 | 8 |

**Solution:**

Let the assumed mean (A) = 20

Class interval | Mid value x_{i} | d_{i}= x_{i} – 20 | u_{i }= (x_{i }-20)/8 | Frequency f_{i} | f_{i}u_{i} |

0 – 8 | 4 | -16 | -2 | 5 | -10 |

8 – 16 | 12 | -8 | -1 | 9 | -9 |

16 – 24 | 20 | 0 | 0 | 10 | 0 |

24 – 32 | 28 | 8 | 1 | 8 | 8 |

32 – 40 | 36 | 16 | 2 | 8 | 16 |

N = 40 | Sum = 5 |

We have, A = 20, h = 8

Mean = A + h (sum/N)

= 20 + 8 (5/ 40)

= 20 + 1

= 21

**Question: 11**

Class interval: | 0- 8 | 8 – 16 | 16 – 24 | 24 – 32 | 32 – 40 |

Frequency: | 5 | 6 | 4 | 3 | 2 |

**Solution:**

Let the assumed mean (A) = 20

Class interval | Mid value x_{i} | d_{i}= x_{i} – 20 | u_{i }= (x_{i }-20)/8 | Frequency f_{i} | f_{i}u_{i} |

0 – 8 | 4 | -16 | -2 | -2 | -10 |

8 – 16 | 12 | -8 | -1 | -1 | -6 |

16 – 24 | 20 | 0 | 0 | 0 | 0 |

24 – 32 | 28 | 8 | 1 | 1 | 3 |

32 – 40 | 36 | 16 | 2 | 2 | 4 |

N = 20 | Sum = -9 |

We have, A = 20, h = 8

Mean = A + h (sum/N)

= 20 + 8 (-9/ 20)

= 20 – (72/20)

= 20 – 3.6

= 16.4

**Question: 12**

Class interval: | 20 – 30 | 30 – 50 | 50 – 70 | 70 – 90 | 90- 110 | 110 – 130 |

Frequency: | 5 | 8 | 12 | 20 | 3 | 2 |

**Solution:**

Let the assumed mean (A) = 60

Class interval | Mid value x_{i} | d_{i}= x_{i} – 60 | u_{i }= (x_{i }-60)/20 | Frequency f_{i} | f_{i}u_{i} |

10 – 30 | 20 | -40 | -2 | 5 | -10 |

30 – 50 | 40 | -20 | -1 | 8 | -8 |

50 – 70 | 60 | 0 | 0 | 12 | 0 |

70 – 90 | 80 | 20 | 1 | 20 | 20 |

90 – 110 | 100 | 40 | 2 | 3 | 6 |

110 – 130 | 120 | 60 | 3 | 2 | 6 |

N = 50 | Sum = 14 |

We have A = 60, h = 20

Mean = A + h (sum/N)

= 60 + 20 (14/ 5)

= 60 + 5.6

= 65.6

**Question: 13**

Class interval: | 25 – 35 | 35 – 45 | 45 – 55 | 55 – 65 | 65 – 75 |

Frequency: | 6 | 10 | 8 | 10 | 4 |

**Solution:**

Let the assumed mean (A) = 50

Class interval | Mid value x_{i} | d_{i}= x_{i} – 50 | u_{i }= (x_{i }– 50)/ 10 | Frequency f_{i} | f_{i}u_{i} |

25 – 35 | 30 | -20 | -2 | 6 | -12 |

35 – 45 | 40 | -10 | -1 | 10 | -10 |

45 – 55 | 50 | 0 | 0 | 8 | 0 |

55 – 65 | 60 | 10 | 1 | 12 | 12 |

65 – 75 | 70 | 20 | 2 | 4 | 8 |

N = 40 | Sum = -2 |

We have A = 50, h = 10

Mean = A + h (sum/N)

= 50 + 10 (-2/ 40)

= 50 – 0.5

= 49.5

**Question: 14**

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 the assumed mean (A) = 42

Class interval | Mid value x_{i} | d_{i }= x_{i} – 42 | u_{i }= (x_{i }– 42)/ 5 | Frequency f_{i} | f_{i}u_{i} |

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 | Sum = -79 |

We have A = 42, h = 5

Mean = A + h (sum/N)

= 42 + 5 (-79/70)

= 42 – 79/14

= 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

Mean = (sum/N)

= 848/ 64= 13.25

By assuming mean method Let the assumed mean (A) = 65

Class interval | Mid value x_{i} | u_{i }= (x_{i }– A) = x_{i }– 65 | Frequency f_{i} | f_{i}u_{i} |

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 of 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 x_{i} | d_{i }= x_{i} – A = x_{i}– 1650 | u_{i }= (x_{i }– 1650)100 | Frequency f_{i} | f_{i}u_{i} |

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 + 100 (7/52)

= 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 |

**Solution:**

(i) Direct method:

Class interval | Mid value x_{i} | Frequency f_{i} | f_{i}x_{i} |

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 x_{i} | u_{i }= (x_{i }– A) | Frequency f_{i} | f_{i}u_{i} |

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 x_{i} | d_{i}= x_{i} – A = x_{i}– 25 | u_{i }= (x_{i }– 25)10 | Frequency f_{i} | f_{i}u_{i} |

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 f_{1 }and f_{2}.

Class: | 0 – 20 | 20 – 40 | 40 – 60 | 60 – 80 | 80 – 100 | 100 – 120 |

Frequency: | 5 | f_{1} | 10 | f_{2} | 7 | 8 |

**Solution:**

Class interval | Mid value x_{i} | Frequency f_{i} | f_{i}x_{i} |

0 – 20 | 10 | 5 | 50 |

20 – 40 | 30 | f_{1} | 30f_{1} |

40 – 60 | 50 | 10 | 500 |

60 – 80 | 70 | f_{2} | 70f_{2} |

80 – 100 | 90 | 7 | 630 |

100 – 120 | 110 | 8 | 880 |

N = 30 +fi + f2 | Sum = 30f1+70 f_{2}+ 2060 |

Given, sum of frequency

50 = f_{1} + f_{2} + 30

f_{1} + f_{2} = 50 – 30

f_{1} +f_{2} = 20

3f_{1} + 3f_{2} = 60 —- (1) [multiply both side by 3]

And mean = 62.8 Sum/ N

= 62.8 = (30f_{1}+ 70f_{2}+ 2060)/50

= 3140 = 30f_{1} + 70f_{2} + 2060

30f_{1}+ 70f_{2} = 3140 – 2060

30f_{1} + 70f_{2} = 1080

3f_{1} + 7f_{2} = 108 —- (2) [divide it by 10]

Subtract equation (1) from equation (2)

3f_{1} + 7f_{2} – 3f_{1} – 3f_{2} = 108 – 60

4f_{2} = 48 = f_{2} = 12

Put value of f_{2} in equation (1)

3f_{1} + 3(12) = 60

f_{1} = 24/3 = 8

f_{1} = 8, f_{2} = 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 value x_{i} | Frequency f_{i} | f_{i}x_{i} |

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 |

792 + 18x = 752 + 20x

20x – 18x = 792 – 752

x = 40/2

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 x_{i} | Frequency f_{i} | f_{i}x_{i} |

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

1161 + 27p = 1245 + 15p

27p – 15p = 1245 – 1161

12p = 84

p = 84/12

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 (f_{i}) |

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 1/2 from lower class limit of each interval and class mark (xi) may be obtained by using the relation xi = (upper limit + lower class limit)/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 f_{i} | Mid value x_{i} | d_{i }= x_{i} – A | u_{i }= (x_{i }– A)/3 | f_{i}u_{i} |

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 = 400, Sum = 25,

Mean = A + h (sum/ N)

= 57 + 3 (25/400)

= 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 = (upper limit + lower class limit)/2. Class size = 50 Now, talking 225 as assumed mean (xi) we may calculate di ,ui, fiui as follows:

Daily expenditure | Frequency f_{i} | Mid value x_{i} | d_{i }= x_{i }– 225 | u_{i }= (x_{i }– 225)50 | f_{i}u_{i} |

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

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 SO_{2 }(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 SO_{2 }in the air

**Solution:**

We may find class marks for each interval by using the relation xi = (upper limit + lower class limit)/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 SO_{2} | Frequency f_{i} | Class interval x_{i} | d_{i} = x_{i }– 0.14 | u_{i} | f_{i}u_{i} |

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 | 0.1 | -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

= 0.04 + (-31/30) × (0.04)

= 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 xi = (upper limit + lower class limit)/2 Now, taking 16 as assumed mean (a) we may Calculate di and fi di as follows

Number of days | Number of students f_{i} | X_{i} | d = x_{i }– 16 | f_{i}d_{i} |

0 – 6 | 11 | 3 | -13 | -143 |

6 – 10 | 10 | 8 | -8 | -80 |

10 – 14 | 7 | 12 | -4 | -28 |

14 – 20 | 7 | 16 | 0 | 0 |

20 – 28 | 8 | 24 | 8 | 64 |

28 – 36 | 3 | 33 | 17 | 51 |

30 – 40 | 1 | 39 | 23 | 23 |

Total | N = 40 | Sum = -113 |

Now we may observe that N = 40, Sum = -113

16 + (-113/ 40)

= 16 – 2.825

= 13.75

So mean number of days is 13.75 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 cites: | 3 | 10 | 11 | 8 | 3 |

**Solution:**

We may find class marks by using the relation x_{i} = (upper limit + lower class limit)/2 Class size (h) for this data = 10 Now taking 70 as assumed mean (a) wrong Calculate d_{i },u_{i}, f_{i}u_{i} as follows

Literacy rate (in %) | Number of cities (f_{i}) | Mid value x_{i} | d_{i}= x_{i }– 70 | u_{i }= (d_{i} –A)10 | f_{i}u_{i} |

45 – 55 | 3 | 50 | -20 | -2 | – 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) × 10

= 70 – 4/7

= 70 – 0.57

= 69.43

So, mean literacy rate is 69.43 %

**All Chapter RD Sharma Solutions For Class10 Maths**

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