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Q1. The median of following observations, arranged in ascending order, is 25. Find x. 11, 13, 15, 19, x + 2, x + 4, 30, 35, 39, 46.

Solution

Here, n = 10 Median = average of 5^th and 6^th observations 25 = 25 2 = 2x + 6 50 - 6 = 2x 44 = 2x Thus, x = 22

Q2. Construct a frequency polygon for the following data: Age 0 - 2 2 - 4 4 - 6 6 - 8 8 - 10 10 - 12 12 - 14 14 - 16 16 - 18 Frequency 2 4 6 8 9 6 5 3 1

Solution

First we obtain the class marks as given in the following table. Age Class-mark Frequency 0 -2 1 2 2 - 4 3 4 4 - 6 5 6 6 - 8 7 8 8 - 10 9 9 10 - 12 11 6 12 - 14 13 5 14 - 16 15 3 16 - 18 17 1 Now we plot the points on the graph.

Q3. If the class marks in a frequency distribution are 19.5, 26.5, 33.5, 40.5 then the class corresponding to the class mark 33.5 is:

• 1) 16 - 23
• 2) 23 - 30
• 3) 30 - 37
• 4) 37 - 41

Solution

30 - 37 as 33.5 lies in between 30 and 37.

Q4. The median of the following observations arranged in ascending order is 27. Find x. 13, 15, 17, 21, x + 2, x + 4, 32, 37, 41 and 48

Solution

Number of observations = 10 54 = 2x + 6 2x = 48 x = 24

Q5. If the mean of 10, 12, 18, 11, p and 19 is 15, find the value of p.

Solution

_{90 = 70 + p} Then p = 90 - 70 Hence, p = 20

Q6. Find the median of the following data 19, 25, 59, 48, 35, 37, 30, 32, 51. If 25 is replaced by 52 what will be the new median?

Solution

Arranging data in ascending order, we have 19, 25, 30, 32, 35, 37, 48, 51, 59 So, n = 9 Thus, Median = 5^th obs = 35 Now, if 25 is replaced by 52, we have 19, 30, 32, 35, 37, 48, 51, 52, 59 Median = 37

Q7. Given below are the marks obtained by 50 students of class IX in a test of Hindi. 42, 21, 51, 37, 42, 37, 38, 42, 49, 52, 38, 53, 57, 47, 29, 59, 61, 33, 17, 71, 39, 43, 42, 39, 14, 7, 27, 19, 54, 51, 39, 43, 42, 16, 37, 67, 62, 39, 51, 53, 41, 53, 59, 37, 27, 33, 34, 42, 22, 31. With these data, form a less than type cumulative frequency table.(i) What is the range of marks. (ii) What can you conclude from the above data about the number of students who scored more than 35.

Solution

(i) Range of marks (ii) From cumulative frequency, we find that 14 students scored less than 35. Thus, 50 - 14 = 36 students i.e. 72% students scored more than 35 marks.

Q8. The mean of 10 numbers is 55. If one number is excluded, their mean becomes 50, the excluded number is:

• 1) 100
• 2) 60
• 3) 80
• 4) 70

Solution

The sum of 10 numbers = 10 55 = 550 The sum of 9 numbers = 9 50 = 450 Therefore, excluded number = 550 - 450 = 100

Q9. For what value of x, is the mode of the following data 17?15,16,17,13,17,16,14,x,17,16,15,15

Solution

Q10. Following table gives the distribution of the marks obtained by the students of a class. Marks 0 - 15 15 - 30 30 - 45 45 - 60 60 - 75 75 - 90 Number of students 5 12 28 30 35 13 Represent the data by Frequency polygon.

Solution

Class Class mark Frequency 0 - 15 7.5 5 15 - 30 22.5 12 30 - 45 37.5 28 45 - 60 52.5 30 60 - 75 67.5 35 75 - 90 82.5 13

Q11.

Solution

The number of students who practiced for over 20 hours are 1 + 1 + 2 = 4

Q12. Find the mean of first 10 prime numbers.

Solution

First 10 prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29 Sum = 129 n = 10

Q13. Median of the data 5, 9, 8, 6, 3, 5, 7, 12, 15 is

• 1) 7
• 2) 3
• 3) 6
• 4) 5

Solution

Arranging data in ascending order: 3, 5, 5, 6, 7, 8, 9, 12, 15 Number of observations = n = 9

Q14. For what value of y is the mode of the following data equal to 7? 3, 5, 6, 7, 3, 4, 7, 3, 6, y, 8, 7.

Solution

Q15. The relative humidity in Delhi, for the month of July, this year as reported by the Meteorological Department is given below : 90, 97, 92, 95, 93, 95, 93, 85, 83, 85, 83, 77, 83, 77, 74, 60, 71, 65, 80, 87, 82, 81, 76, 61, 63, 58, 58, 56, 57, 54, 98 Present the data in the form of frequency distribution.

Solution

Q16. If median of 29, 32, 48, 50, x, x + 2, 72, 78, 84, 95 (arranged in ascending order) is 63, find x.

Solution

Number of terms = 10 (even) Median = i.e. 63 =  2x + 2 = 126 2x = 124 x = 62

Q17. The mean weight per student in a group of 7 students is 55 kg. The individual weights of 6 of them in kg are 52, 54, 55, 53, 56, 54. Find the weight of the seventh student.

Solution

Q18.

Solution

Q19.

Solution

Q20. Find the mean of factors of 24.

Solution

Factors of 24 are 1, 2, 3, 4, 6, 8, 12 and 24 Sum = 60 n = 8 Thus, Mean = 

Q21. Find the mean of the following frequency distribution showing the marks obtained in a class test by 20 students. Marks 10 15 20 25 No. of students 4 5 8 3  

Solution

Marks (xi) 10 15 20 25 No. of students (fi) 4 5 8 3 fixi 40 75 160 75  

Q22. Draw a histogram for the following data:   Class interval Frequency 10 - 14 300 15 - 19 980 20 - 24 800 25 - 29 600 30 - 34 300 35 - 39 430 40 - 44 530

Solution

Make intervals continuous. Class interval Frequency 9.5 - 14.5 300 14.5 - 19.5 980 19.5 - 24.5 800 24.5 - 29.5 600 29.5 - 34.5 300 34.5 - 39.5 430 39.5 - 44.5 530 The required histogram is as follows:

Q23.

Solution

Q24. The class mark of class interval 60 - 70 will be

• 1) 60
• 2) 75
• 3) 65
• 4) 70

Solution

Class mark =

Q25. If the mean of the following distribution is 6, find the value of p. x 2 4 6 10 p + 5 f 3 2 3 1 2  

Solution

2p + 52 = 66 2p = 66 - 52 Thus, p = 7

Q26. Find the median of the following data: 15, 35, 18, 27, 19, 23, 29, 20, 28

Solution

Data can be arranged in ascending order as follows: 15, 18, 19, 20, 23, 27, 28, 29, 35 Here, n = 9 (odd) Thus, Median = 

Q27. Class Interval Frequency 0 - 5 2 5 - 10 3 10 - 15 4 15 - 20 1 20 - 25 5 25 - 30 3 Make a frequency polygon for above given frequency table.

Solution

Class Interval Class Mark Frequency 0 - 5 2.5 2 5 - 10 7.5 3 10 - 15 12.5 4 15 - 20 17.5 1 20 - 25 22.5 5 25 - 30 27.5 3 ABCDEFGH is the frequency polygon.

Q28. The marks obtained by 40 students of class IX in Mathematics are given below: 81, 55, 68, 79, 85, 43, 29, 68, 54, 73, 47, 35, 72, 64, 95, 44, 50, 77, 64, 35, 79, 52, 45, 54, 70, 83, 62, 64, 72, 92, 84, 76, 63, 43, 54, 38, 73, 68, 52, 54 Prepare a grouped continuous frequency distribution table with class-size of 10 marks.

Solution

Lowest = 29, Highest = 95 Marks Tally Frequency 20-30 | 1 30-40 ||| 3 40-50   5 50-60 ||| 8 60-70 ||| 8 70-80 |||| 9 80-90 |||| 4 90-100 || 2

Q29. The range of the data 25.7, 16.3, 2.8, 21.7, 24.3, 22.7, 24.9 is:

• 1) 20.5
• 2) 22
• 3) 22.9
• 4) 21.7

Solution

Range = Maximum value - Minimum value = 25.7 - 2.8 = 22.9

Q30. Mean of 25 observations was found to be 78.4. But later on, it was discovered that one observation 96 was misread as 69. Find the correct mean.

Solution

Corrected Mean = 

Q31. Find the missing frequencies f₁ and f₂ in the following frequency distribution, if it is known that the mean of the distribution is 1.46 x₁ 0 1 2 3 4 5 Total f₁ 46 f₁ f₂ 25 10 5 200  

Solution

x_i  f_i f_ix_i­ 0 46 0 1 f₁ f₁ 2 f₂ 2f₂ 3 25 75 4 10 40 5 5 25 N = 86 + f₁ + f₂ We have N = 200 200 = 86 + f₁­ + f₂ Thus, f₁ = 114 - f₂    .....(i) Mean = 1.46 292 = 140 + f₁ + 2f₂ f₁ + 2f₂ = 152  .....(ii) Solving (i) and (ii), we get 114 - f₂ + 2f₂ = 152 Thus, f₂ = 38 Hence, f₁ = 114 - 38 = 76

Q32. The mean of prime numbers between 20 and 30 is:

• 1) 27
• 2) 26
• 3) 25
• 4) 21

Solution

The only two prime numbers between 20 and 30 are 23 and 29. = 26

Q33.

Solution

(i) A child spent 7 hours in a school. (ii) A child spent his maximum time (8 hours) in sleeping and minimum time (1 hour) in chores and homework. (iii) Time spent by a child in watching TV = 3 hours and time spent in playing is 2 hours. So, a child spent 3 - 2 = 1 hour extra in watching tv.

Q34. The mean of 25 observations is 36. If the mean of the first 13 observations is 32 and that of last 13 observations is 39. Find the 13^th observation.

Solution

Sum of first 13 observations = 13 32 = 416 Sum of last 13 observations = 13 39 = 507 Sum of all 25 observations = 25 36 = 900 The 13^th observation = 416 + 507 - 900 = 23 13^th observation = 23

Q35. The following observations have been arranged in ascending order: 29, 32, 48, 58, x, x + 2, 64, 78, 84, 90. If median of the data is 63, find the mode of the data.

Solution

29, 32, 48, 58, x, x + 2, 64, 78, 84, 90. 126 = 2x + 2 2x = 124 x = 62 observations are 29, 32, 48, 58, 62, 64, 64, 78, 84 90 As highest frequency = 2(Since 64 occurs 2 times) Mode = 64

Q36. The range of a data is 45. The maximum value in the data is thrice the minimum value. So what are the minimum and maximum values in the data?

Solution

Q37.

Solution

Q38. In a study of diabetic patients in a village, the following observations were noted. Age in years 10-20 20-30 30-40 40-50 50-60 60-70 Number of patients 2 5 12 19 9 4 Represent the above data by a frequency polygon.

Solution

The frequency distribution table with class marks is given below: Class- intervals Class marks Frequency 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 5 15 25 35 45 55 65 75 0 2 5 12 19 9 4 0 Plot points (5,0), (15,2)………(65, 4) and (75, 0) to get the frequency polygon as follows:

Q39.

Solution

Q40. The mean of the following distribution is 50.  x_i 10 30 50 70 90 f_i 17 5a + 3 32 7a - 11 19 Find the value of a.

Solution

x_i  10 30 50 70 90 f_i 17 5a + 3 32 7a - 11 19 f_ix_i 170 150a+90 1600 490a-770 1710 Mean = 

Q41. The following are the runs made by 22 players in a one day cricket series played between India and Pakistan. 79, 28, 45, 99, 3, 46, 8, 0, 3, 7, 75, 24, 73, 122, 46, 27, 16, 7, 100, 3, 67, 53 Construct a frequency distribution table for the above data with equal class intervals, one of these being 0 - 20 (20 not included).

Solution

Frequency distribution table: Class Interval Tally Frequency 0-20   8 20-40   3 40-60 4 60-80 4 80-100 1 100-120 1 120-140 1 Sum 22

Q42. Obtain the mean of the following data: Variable (x_i­) 4 6 8 10 12 Frequency (f_i) 4 8 14 11 3

Solution

Q43. The blood groups of 30 students are recorded as follows : A, B, O, A, AB, O, A, O, B, A, O, B, A, AB, B, A, AB, B, A, A, O, A, AB, B, A, O, B, A, B, A Prepare a frequency distribution table for the data.

Solution

Q44.

Solution

Q45. Find the median of the following data: 23, 31, 47, 46, 27, 37, 30, 24.

Solution

The data can be arranged in ascending order as 23, 24, 27, 30, 31, 37, 46, 47 Here n = 8 (even) Median = Average of  and  observations = Average of 4^th and 5^th observations

Q46. Find the median of the given data: 26, 56, 32, 33, 60, 17, 34, 29 and 45. If 26 is replaced by 62, find the new median.

Solution

Q47. The points scored by a basket ball team in a series of match are as follows: 17, 2, 7, 27, 25, 5, 14, 18, 10, 24, 28, 10, 8, 7, 10, 28 Find the median and mode of the above data.

Solution

Arranging the data in ascending order, we get 2, 5, 7, 7, 8, 10, 10, 10, 14, 17, 18, 24, 25, 27, 28, 28 Since number of observations = 16, which is even Median is the mean of  and  observations As 10 occurs most frequently i.e. 3 times, mode of the data is 10.

Q48. Following are the marks obtained by 30 students in an examination: Taking class intervals 1-10, 11-20, ... make a frequency table for the above distribution.

Solution

Frequency table is as follows:

Q49. Find mean, mode and median for the following data: 10, 15, 18, 10, 10, 20, 10, 20, 15, 21, 15 and 25

Solution

(i) Mea (ii) Frequency of 10 is 4 which is maximum Mode = 10 (iii) Arrange the data in ascending order 10, 10, 10, 10, 15, 15, 15, 18, 20, 20, 21, 25 Here n = 12 (an even number)

Q50. Given below are the seats won by different political parties in the following outcomes of state assembly elections. Political party A B C D E F Seats won 75 55 41 32 11 42 (i) Draw a bar graph to represent the polling results. (ii) Which political party won the maximum number of seats?

Solution

(i) Bar graph: (ii) Party A won the maximum number of seats.

Q51. The electricity bills (in Rs.) of 25 houses in a certain locality are given below: 500, 440 ,100, 180, 150, 560, 300, 220, 300, 240, 150, 180, 120, 270, 420, 450, 250, 240, 200, 220, 140, 160, 250, 360, 470 Construct a grouped frequency table for this data with one of the class intervals as 100 - 200 (200 not included)

Solution

The frequency distribution table of the given data is as follows:

Q52. The following table gives the distribution of students of two sections according to the marks obtained by them. Section A Section B Mark Frequency Marks Frequency 0 - 10 2 0 - 10 5 10 - 20 12 10 - 20 11 20 - 30 18 20 - 30 15 30 - 40 13 30 - 40 12 40 - 50 5 40 - 50 7 Represent the marks of the students of both the sections on the same graph by two frequency polygons.

Solution

  Section A Section B Marks Class marks Frequency Marks Class marks Frequency 0 - 10 5 2 0 - 10 5 5 10 - 20 15 12 10 - 20 15 11 20 - 30 25 18 20 - 30 25 15 30 - 40 35 13 30 - 40 35 12 40 - 50 45 5 40 - 50 45 7 ABCDEFG is the frequency polygon for section A and AQRSTUG is the frequency polygon for section B.

Q53. The mean of the following distribution is 50. x_i 10 30 50 70 90 y_i 17 5p + 3 32 7p - 11 19 Find the value of p.

Solution

x_i f_i f_ix_i 10 17 170 30 5p + 3 30(5p + 3) 50 32 1600 70 7p - 11 70 (7p - 11) 90 19 1710 Mean =   50(60 + 12p) = 2800 + 640p 3000 - 2800 = 640p - 600p 200 = 40p Hence, p = 5

Q54. The following data shows IQs of some children. Represent it using the continuous frequency distribution.

Solution

Making the class intervals continuous, we get

Q55. The distribution of weight (in kg) of 100 people is given below. Weight in kg Frequency 40 - 45 13 45 - 50 25 50 - 55 28 55 - 60 15 60 - 65 12 65 - 70 5 70 - 75 2 Construct a histogram for the above distribution.

Solution

Steps of Construction: (i) We represent the weights on the horizontal axis. We choose the scale on horizontal axis as 1 cm = 5 kg. Also, since the first class interval is starting from 35 and not zero, we show it on the graph by marking a kink or a break on the axis. (ii) We represent the number of people (frequency) on the vertical axis. Since the maximum frequency is 28, we choose the scale as 1 cm = 5 people. (iii) We now draw rectangles (or rectangular bars) of width equal to the class-size and lengths according to the frequencies of the corresponding class intervals.

Q56. The following data has been arranged in ascending order: 12, 14, 17, 20, 22, x, 26, 28, 32, 36. If the median of the data is 23, find x. In the above data, if 32 is replaced with 23, find the new median.

Solution

Q57.

Solution

(i) Total exports = 1750 + 2000 + 1250 + 1750 = 6750 (All figures in lakhs) (ii) To Maldives, exports are below 1500 Lakhs.  (iii) To Mayanmar, maximum goods are exported.

Q58. Find the mean of the first ten prime numbers.

Solution

The first ten prime numbers are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29 Thus, Mean is given by

Q59. The value of  upto 50 decimal places is given below: 3.14159265358979323846264338327950288 (i)  Make a frequency distribution of the digits from 0 to 9 after the decimal point. (ii) What are the most and the least appearing digits?  

Solution

Q60.

Solution

Q61. The daily income of 50 doctors is given below: Daily income (in Rs.) No. of Doctors 0 - 1000 1000 - 2000 2000 - 3000 3000 - 4000 4000 - 5000 5000 - 6000 8 7 12 6 11 6 Draw a histogram for the above data.

Solution

Steps of Construction: (i) We represent the daily income on the horizontal axis. We choose the scale on horizontal axis as 1 cm = Rs. 1000. (ii) We represent the number of doctors (frequency) on the vertical axis. Since the maximum frequency is 12, we choose the scale as 1 cm = 2 doctors. (iii) We now draw rectangles (or rectangular bars) of width equal to the class-size and lengths according to the frequencies of the corresponding class intervals.

Q62. Draw histogram to represent the following distribution. Class Interval 5 - 10 10 - 15 15 - 25 25 - 45 45 - 75 Frequency 6 12 10 8 15

Solution

C.I F Width of the class Adjusted Frequency 5 - 10 6 5 10 -15 12 5 15 - 25 10 10 25 - 45 8 20 45 - 75 15 30

Q63.

Solution

Q64.

Solution