## Chapter 7 Statistics R.D. Sharma Solutions for Class 10th Math Exercise 7.5

1. Find the mode of the following data:(i) 3,5,7,4,5,3,5,6,8,9,5,3,5,3,6,9,7,4

(ii) 3, 3, 7, 4, 5, 3, 5, 6, 8, 9, 5, 3, 5, 3, 6, 9, 7,4

(iii) 15, 8, 26, 25, 24, 15, 18, 20, 24, 15, 19, 15Â Â Â

**Solution**

(i)

2. The shirt sizes worn by a group of 200 persons, who bought the shirt from a store, are as follows:

Shirt size: 37Â 38Â 39Â 40Â 41Â 42Â 43Â 44

Number of persons: 15Â 25Â 39Â 41Â 36Â 17Â 15Â 12

Find the model shirt size worn by the group.Â

(ii)Â

Â

(iii)Â

2. The shirt sizes worn by a group of 200 persons, who bought the shirt from a store, are as follows:

Shirt size: 37Â 38Â 39Â 40Â 41Â 42Â 43Â 44

Number of persons: 15Â 25Â 39Â 41Â 36Â 17Â 15Â 12

Find the model shirt size worn by the group.Â

**SolutionÂ Â**

3. Find the mode of the following distribution.

(i) Class-interval: 0-10Â 10-20Â 20-30Â 30-40Â 40-50Â 50-60Â 60-70Â 70-80

Frequency: 5Â Â 8Â 7Â 12Â 28Â 20Â 10Â 10

(ii) Class-interval: 10-15Â 15-20Â 20-25Â 25-30Â 30-35Â 35-40

Frequency: 30Â 45Â 75Â 35Â 25Â 15

(iii) Class-interval: 25-30Â 30-35Â 35-40Â 40-45Â 45-50Â 50-60

Frequency: 25Â 34Â 50Â 42Â 38Â 14

**Solution**

(i)

(ii)Â

(iii)Â

Age (in years): 16-18Â 18-20Â 20-22Â 22-24Â 24-26

Group A: 50Â 78Â 46Â 28Â 23

Group B: 54Â 89Â 40Â 25Â 17

**Solution**Â

marks obtained by the students in science.

Marks: 0-10Â 10-20Â 20-30Â 30-40Â 40-50Â 50-60Â 60-70Â 70-80Â 80-90Â 90-100

Frequency: 3Â 5Â 16Â 12Â 13Â 20Â 5Â 4Â 1Â 1

**Solution**

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

Height (in cm): 160-162Â 163-165Â 166-168Â 169-171Â 172-174

No. of students: 15Â 118Â 142Â 127Â 18

Find the average height of maximum number of students.Â Â

**Solution**

7.Â The following table shows the ages of the patients admitted in a hospital during a year:

Age (in years): 5-15Â 15-25Â 25-35Â 35-45Â 45-55Â 55-65

No. of students: 6Â 11Â 21Â 23Â 14Â 5

Find the mode and the mean of the data given above. Compare and interpret the twoÂ measures of central tendency.**Solution**

8. The following data gives the information on the observed lifetimes (in hours) of 225Â electrical components:

Lifetimes (in hours): 0-20Â 20-40Â 40-60Â 60-80Â 80-100Â 100-120

No. of components: 10Â 35Â 52Â 61Â 38Â 29

Determine the modal lifetimes of the components.

**Solution**

9. The following table gives the daily income of 50 workers of a factory:

Daily income (in Rs) : 100-120Â 120-140Â 140-160Â 160-180Â 180â€“200

Number of workers:Â 12Â Â 14Â Â 8Â Â 6Â 10

Find the mean, mode and median of the above data

**Solution**

10. The following distribution gives the state-wise teacher-student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret, the two measures:

**Solution**

11. Find the mean, median and mode of the following data:

Classes: 0-50Â 50-100Â 100-150Â 150-200Â 200-250Â 250-300Â 300-350

Frequency:Â 2Â Â 3Â Â 5Â Â 6Â 5Â Â 3Â 1

**Solution**Â Â

12.Â Â A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised in the table given below. Find the mode of the data :

Number of cars : 0-10Â 10-20Â 20-30Â 30-40Â 40-50Â 50-60Â 60-70Â 70-80Â

Frequency : 7Â 14Â 13Â 12Â Â 20Â Â 11Â Â 15Â Â 8Â Â

**Solution**

13. The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them.

Monthly consumption : 65-85Â 85-105Â 105-125Â 125-145Â 145-165Â 165-185Â 185-205

(in units)

No. of consumers:Â 4Â 5Â 13Â 20Â 14Â 8Â 4

**Solution**Â

14. 100 surnames were randomly picked up from a local telephone directly and the frequency

distribution of the number of letters in the English alphabets in the surnames was obtained

as follows:

Number of letters: l-4Â 4-7Â 7-10Â 10-13Â 13-16Â 16-19

Number surnames: 6Â 30Â 40Â 16Â 4Â 4

Determine the median number of letters in the surnames. Find the mean number of letters in the surnames. Also, find the modal size of the surnames.

**Solution**Â

Classes:Â 0-20Â 20-40Â 40-60Â 60-80Â 80-100Â 100-120Â 120-140

Frequency: 6Â 8Â 10Â 12Â Â 6Â Â 5Â Â 3

**Solution**

16. The following data gives the distribution of total monthly houshold expenditure of 200 families of a village. Find the modal monthly expenditure of the families. Also, find the mean monthly expenditure:

**Solution**

17. The given distribution shows the number of runs scored by some top batsmen of the world in one-day international cricket matches.Â

**Solution**

18. The frequency distribution table of agriculture holdings in a village is given below :

Area of land (in hectares) : 1-3Â 3-5 5-7 7-9 9-11 11-13

Number of families : 20Â 45Â 80Â 55Â 40Â 12

Find the modal agriculture holdings of the village .

**Solution**

The maximum class frequency is 80. The class corresponding to this frequency is 5-7.

So, the modal class is 5-7.

l(the lower limit of modal class) = 55

f

_{1}Â (frequency of the modal class) = 80
f

_{o}Â (frequency of the class preceding the modal class) = 45
f

_{2}Â (frequency of the class succeeding the modal class) = 55
h (class size) = 2

Mode = l+(f

_{1}-f_{o}/2f_{1}-f_{o}-f_{2}) Ã— h
= 5 + (80-45/2Ã—80-45-55) Ã— 2

= 5 + 35/60 Ã— 2

= 5 + 35/60

= 6.2

_{}
19. The monthly income of 100 families are given as below :

Income in | Number of families |

0-5000 | 8 |

5000 - 10000 | 26 |

10000 - 15000 | 41 |

15000 - 20000 | 16 |

20000 - 25000 | 3 |

25000 - 30000 | 3 |

30000 - 35000 | 2 |

35000 - 40000 | 1 |

Calculate the modal income .

_{}

**Solution**

The maximum class frequency is 41. The class corresponding to this frequency is 5-7.

So, the modal class is 10000-15000.

l (the lower limit of modal class) = 10000

f

_{1}Â (frequency of the modal class) = 41
f

_{o}Â (frequency of the class preceding the modal class) = 26
f

_{2}Â (frequency of the class succeeding the modal class) = 16
h (class size) = 5000

Mode = l+(f

_{1}-f_{o}/2f_{1}-f_{o}-f_{2}) Ã— h
= 10000 + (41-26/2 Ã— 41-26 - 16) Ã— 5000

= 10000 + 15/40 Ã— 5000

= 10000 +1875

= 11875