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Q1. Find the mode of the following distribution: Class 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 Frequency 15 18 16 5 6    

Solution

Modal class = 10 - 20   So, l = 10, f₁ = 18, f₀­ = 15, f₂ = 16, h = 10 = 10 + = 10 + 6 = 16

Q2. The mean of a data set with 10 observations is calculated as 17.85. If one more value is included in the data, then for the new data with 11 observations, mean becomes 18.50. Value of this 11^th observation is

• 1) 25
• 2) 24
• 3) 26
• 4) 28

Solution

Let x₁, x₂, x₃ …, x₁₀ be the 10 values of the given data. Let the 11^th observation be x₁₁. x₁ + x₂ + x₃ + ……..+ x₁₀ = 10 17.85 = 178.5 x₁ + x₂ + x₃ + ……..+ x₁₀ + x₁₁= 11 18.50 = 203.50 (x₁ + x₂ + x₃ +……..+ x₁₀) + x₁₁ = 203.50 x₁₁ = 203.50 - 178.50 = 25

Q3. The mean of the following data is: 45, 35, 20, 15, 25, 40

• 1) 25
• 2) 15
• 3) 30
• 4) 35

Solution

Q4. The mean of the following distribution is 22, find the missing frequency f: Class 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 Frequency 12 16 6 f 9

Solution

C.I f x_i f_i x_i 0 - 10 12 5 60 10 - 20 16 15 240 20 - 30 6 25 150 30 - 40 f 35 35f 40 - 50 9 45 405 43 + f 855 + 35f 946 + 22f = 855 + 35f 91 = 13f f= 7

Q5. For a symmetrical distribution, which is correct

• 1) Mean = Median = Mode
• 2) Mode =
• 3) Mean > Mode > Median
• 4) Mean < Mode < Median

Solution

In symmetrical distribution, Mean=Mode=Median

Q6. From the following data, draw the two types of cumulative frequency curves and hence determine the median. Height (in cm) Frequency 140 - 144 144 - 148 148 - 152 152 - 156 156 - 160 160 - 164 164 - 168 168 - 172 172 - 176 176 - 180 3 9 24 31 42 64 75 82 86 34  

Solution

We prepare less than series and more than series.   (i)Less than series Height in (cm) c.f. Less than 140 Less than 144 Less than 148 Less than 152 Less than 156 Less than 160 Less than 164 Less than 168 Less than 172 Less than 176 Less than 180 0 3 12 36 67 109 173 248 330 416 450 Now on graph paper plot the points (140, 0), (144, 3), (148, 12), (152, 36), (156, 67), (160, 109), (164, 173), (168, 248), (172, 330), (176, 416), (180, 450).   (ii)More than series Height in cm c.f. More than 140 More than 144 More than 148 More than 152 More than 156 More than 160 More than 164 More than 168 More than 172 More than 176 More than 180 450 447 438 414 383 341 277 202 120 34 0 Now on the same graph plot the points (140, 450), (144, 447), (148, 438), (152, 414), (156, 383), (160, 341), (164, 277), (168, 202), (172, 120), (176, 34), (180, 0)   From the graph, it is clear that 168 is the median.

Q7. The median of the observations given in order 16, 18, 20, 24 - x, 22 + 2x, 28, 30, 32 is 24. Find the value of x.

Solution

Total no. of terms = 8 Median =

Q8.

Solution

Q9. The marks obtained in a class test by 30 students of a class are as follows. Marks obtained Number of students More than or equal to 5 30 More than or equal to 10 28 More than or equal to 15 16 More than or equal to 20 14 More than or equal to 25 10 More than or equal to 30 7 More than or equal to 35 3 Draw a less than type and more than type ogive curves for the given data and hence find the median.

Solution

From the given data, we can get the following table. C I f cf 5 - 10 2 2 10 - 15 12 14 15 - 20 2 16 20 - 25 4 20 25 - 30 3 23 30 -35 4 27 35 - 40 3 30                         We plot points (5, 30) (10, 28) (15, 16) (20, 14) (25, 10) (30, 7) (35, 3) to get "more than" curve. We plot points (10, 2) (15, 14) (20, 16) (25, 20) (30, 23) (35, 27) (40, 30) to get "less than" curve.   Median=17. 5 (x-coordinate of intersection of both curves) 

Q10. Find the missing frequency f if the mode of the given data is 154.   Class 120 - 130 130 - 140 140 - 150 150 - 160 160 - 170 170 - 180 Frequency 2 8 12 f 8 7  

Solution

Class 120 - 130 130 - 140 140 - 150 150 - 160 160 - 170 170 - 180 Frequency 2 8 12 f 8 7 Mode = 154 Modal class = 150 - 160 So, l = 150, f₁ = f, f₀­ = 12, f₂ = 8, h = 10 We know, Mode = l + 154 = 150 +

Q11. Change the following frequency distribution to less than type distribution and draw its ogive. Hence, obtain the median value. Classes 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60 60 - 70 70 - 80 Frequency (f) 5 15 18 25 11 9 8

Solution

Classes Frequency Less than 20 5 Less than 30 20 Less than 40 38 Less than 50 63 Less than 60 74 Less than 70 83 Less than 80 91 Plot the points (20,5), (30,20), (40,38), (50,63), (60,74), (70,83), (80,91) and join them by free hand to obtain the required ogive.

Q12. For the data given below draw less than ogive curve. Marks 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60 Number of students 7 10 23 51 6 3  

Solution

We first prepare the cumulative frequency distribution table as given below: Marks No. of students Marks less than Cumulative frequency 0-10 7 10 7 10-20 10 20 17 20-30 23 30 40 30-40 51 40 91 40-50 6 50 97 50-60 3 60 100   Now, we mark the upper class limits along x-axis bt taking a suitable scale and the cumulative frequencies along y-axis by taking a suitable scale. Thus, we plot the points (10,7),(20,17),(30,40),(40,91),(50,97)and(60,100). Join the plotted points by a free hand to obtain the required ogive.   

Q13. For the following distribution the difference in the upper limit of median and modal class is CI 0-10 10-20 20-30 30-40 40-50 f 2 5 7 5 2

• 1) 40
• 2) 20
• 3) 10
• 4) 0

Solution

CI 0-10 10-20 20-30 30-40 40-50 f 2 5 7 5 2 cf 2 7 14 19 21 Highest frequency 7 is corresponding to the class 20-30 so it is the modal class. Upper limit of modal class = 30. =10.5 Class interval with cf just greater than 10.5 is 20-30. So the median class is 20-30. Upper limit of median class = 30 Difference between the upper limit of the modal and median class is 30-30=0.

Q14. The mean of the following frequency distribution is 8.4. Find the value of p. Classes 0 - 4 4 - 8 8 - 12 12 - 16 16 - 20 Frequency (f) 7 10 p 8 2

Solution

Classes 0 - 4 4 - 8 8 - 12 12 - 16 16 - 20 Frequency (f) 7 10 p 8 2 x 2 6 10 14 18 fx 14 60 10p 112 36 Mean =

Q15. The following table gives the age of doctors working in a hospital.   Age in years 25-30 30-35 35-40 40-45 45-50 No. of doctors 4 5 8 6 7 Change the above distribution to more than type and draw its ogive. Hence, find the median of the data.

Solution

Cumulative distribution of more than type is as follows: Age in years No. of doctors Age more than or equal to Cumulative Frequency  25 - 30  4  25  30 30 - 35  5  30  26  35 - 40  8  35  21  40 - 45  6  40  13  45 - 50  7  45  7   Total  30     Now, mark the lower class limits on x-axis and cumulative frequencies along Y-axis on suitable scales. Thus, we plot the points (25, 30) (30, 26) (35, 21) (40, 13) and (45, 7). By joining these points, we obtain the ogive as shown below:   Here, N = 30 Now, draw a line parallel to x-axis, from the point marked with frequency = 15, cutting the more than ogive at point P. Then, draw perpendicular PM from P on the x-axis. The x-coordinate of point M gives the median. Thus, M = 38.8 (approx).

Q16. Find the mode of the following data   Class 50 - 60 60- 70 70 - 80 80 - 90 90 - 100 Frequency 9 12 20 11 10    

Solution

Modal class = 70 - 80 l = 70, f_o = 12, f₁ = 20, f₂ = 11, h = 10 = = 70 + 4.7 = 74.7

Q17.

Solution

Q18. If the mode of some data is 7 and their mean is also 7, then their median is:

• 1) 9
• 2) 10
• 3) 7
• 4) 8

Solution

We know: Mode = 3 Median - 2 Mean 3 Median = Mode + 2 Mean 3 Median = 7 + 2 x 7 = 7 + 14 = 21 Thus, Median = 7

Q19. The following expenditure gives state wise teacher student ratio in higher secondary schools in India. Find the mode of this data. Number of students per teacher Number of states 15-20 3 20-25 8 25-30 9 30-35 10 35-40 3 40-45 0 45-50 0 50-55 2

Solution

Number of students per teacher Number of states 15-20 3 20-25 8 25-30 9 30-35 10  modal class 35-40 3 40-45 0 45-50 0 50-55 2  

Q20. The lower limit of the modal class of the following data is:   C.I. 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 Frequency 5 8 13 7 6  

• 1) 50
• 2) 20
• 3) 10
• 4) 30

Solution

Modal class is 20 - 30. Thus, the required lower limit of the modal class is 20.

Q21. The distribution below given the weight of 30 students of a class. Find the median weight of the students. Weight in kg 40 - 45 45 - 50 50 - 55 55 - 60 60 - 65 65 - 70 70 - 75 No. of students 2 3 8 6 6 3 2    

Solution

Weight (in kg) No. of students c.f. 40 - 45 2 2 45 - 50 3 5 50 - 55 8 13 55 - 60 6 19 60 - 65 6 25 65 - 70 3 28 70 - 75 2 30   Here, n = 30   The cumulative frequency just greater than n/2=15 is 19 and the corresponding class is 55 - 60. Hence, 55 - 60 is the median class. Thus, we have   l = 55, c.f. = 13, f = 6, h = 5 Median = l + = 55 + 1.67 = 56.67

Q22.

Solution

Mean of 10 numbers = 12 Thus, sum of 10 numbers = 12 x 10 = 120 Mean of 20 numbers = 9 Thus, sum of 20 numbers = 20 x 9 = 180 Now, Mean of 30 numbers =  

Q23.

Solution

Q24. The following table shows the heights of 50 boys: Height (cm) 120 121 122 123 124 Frequency 5 8 18 10 9 Find the mode of heights.

Solution

We observe that the height 122 cm has the highest frequency of 18. Hence, the modal value of height is 122 cm.

Q25. Find the median of the following data Class 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 Total Frequency 8 16 36 34 6 100

Solution

Class 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 Frequency (f) 8 16 36 34 6 c.f 8 24 60 94 100 Here, n = 100 Median class = 20 - 30 l = 20, f = 36, c.f. = 24 and h = 10

Q26. The mean of 6 numbers is 16. With the removal of a number the mean of remaining numbers is 17. The number removed is:

• 1) 2
• 2) 22
• 3) 11
• 4) 6

Solution

We know: Mean = Mean of 6 numbers = 16 Sum of the 6 observations =  Mean of 5 observations = 17 Sum of the 5 observations =  Thus, number which is removed = 96 - 85 = 11

Q27.

Solution

Q28. Find the mean of the following frequency distribution. C.I. 0-100 100-200 200-300 300-400 400-500 f 2 3 5 2 3

Solution

To calculate the mean, first obtain the column of mid value and then multiply the corresponding values of frequency and mid value. C.I. f Mid value (x) fx 0-100 2 50 100 100-200 3 150 450 200-300 5 250 1250 300-400 2 350 700 400-500 3 450 1350 15 3850 Here and , so the mean is given as

Q29. The mode of the following distribution is 17.03.Find the value of p. Class interval     Frequency 0-5 6 5-10 11 10-15 p 15-20 24 20-25 17 25-30 13 30-35 5

Solution

Q30. Find the mean of the following data. x     0 12 14 16 18 f 2 3 1 6 3

Solution

     x_i           f_i       f_i x_i     0          2       0   12         3     36   14         1     14   16         6     96   18         3     54   200  

Q31. The mean of the following frequency distribution is 50. Find the value of p. Classes 0 - 20 20 - 40 40 - 60 60 - 80 80 - 100 Frequency 17 28 32 p 19

Solution

Classes x_i f_i x_if_i 0 - 20 10 17 170 20 - 40 30 28 840 40 - 60 50 32 1600 60 - 80 70 p 70p 80 - 100 90 19 1710 Total 96 + p 4320 + 70p

Q32. If the median class is 30-40, the frequency of the median class is 20, the cumulative frequency of the preceding class is 15 and the total frequency is 50, then find the median.

• 1) 15
• 2) 25
• 3) 35
• 4) 45

Solution

According to the question, l = 30, h = 10, f = 20, cf= 15, n = 50, n/2 = 25

Q33. An incomplete distribution is given as follows:  Variable 0-10 10-20 20-30 30-40 40-50 50-60 60-70 Frequency 10 20 ? 40 ? 25 15 The median value is 35 and the sum of all the frequencies is 170. Using the median formula, fill up the missing frequencies. 

Solution

Let the frequency of class 20 – 30 be f₁ and that of class 40 – 50 be f₂. The total frequency is 170. Class interval f cf 0 -10 10 10 10 – 20 20 30 20 – 30 f₁ 30 + f₁ 30 – 40 40 70 + f₁ 40 -50 f₂ 70 + f₁+ f₂ 50 - 60 25 95 + f₁+ f₂ 60 -70 15 110 + f₁+ f₂    

Q34. 200 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in English alphabets in the surnames was obtained as follows: No. of letters 1 - 5 5 - 10 10 - 15 15 - 20 20 - 25 No. of surnames 20 60 80 32 8 Find the median.

Solution

  C.I 1 - 5 5 - 10 10 - 15 15 - 20 25 - 25 f 20 60 80 32 8 c.f. 20 80 160 192 200 Median class = 10 - 15 l = 10, f = 80, c.f. = 80 and h = 5 Thus, Median = 10 + 1.25 = 11.25

Q35. Find the mode from the following data: 2-4 4-6 6-8 8-10 10-12 12-14 5 8 10 6 7 2  

• 1) 6.5
• 2) 7
• 3) 10
• 4) 6

Solution

2­-4 4-6 6-8 8-10 10-12 12-14 5 8 = f₀ 9 = f₁ 6 = f₂ 7 2 8 = f₀, 9 = f₁, 6 = f₂, h = 2, l = 6 

Q36. Find the mean of the following frequency distribution, using step deviation method. Classes 100 - 150 150 - 200 200 - 250 250 - 300 300 - 350 Frequency 4 5 12 2 2

Solution

C.I. f x d = x - 225 d' = fd' 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 25 -7 Here, A = 225

Q37. Find the values of a, b, c and d in the following frequency distribution: Class interval Frequency Cumulative Frequency 0 – 20 5 a 20 – 40 10 15 40 - 60 b 23 60 - 80 c 30 80 - 100 4 d

Solution

Q38. Draw a 'less than ogive' for the given data. Marks 10-20 20-30 30-40 40-50 50-60 No. of students 2 4 7 6 1

Solution

 

Q39. The mean () of a frequency distribution is 45. If the value of = 20 then the value of is:

• 1) 900
• 2) 800
• 3) 2.25
• 4)

Solution

We know:

Q40. Change the following frequency distribution to more than type distribution and draw its ogive. Hence, find its median. Classes 0 - 5 5 - 10 10 - 15 15 - 20 20 - 25 Frequency (f) 6 8 10 6 4  

Solution

  Lower class limits c.f. More than 0 34 More than 5 28 More than 10 20 More than 15 10 More than 20 4   Plot the points (0, 34), (5, 28), (10, 20), (15, 10), (20, 4) to obtain the required ogive. The median will be corresponding x-coordinate of the point whose ordinate is .

Q41. A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if the policies are given only to persons having age between 18 and 60 years. Age in years Number of policy holders Below 20 2 Below 25 6 Below 30 24 Below 35 45 Below 40 78 Below 45 89 Below 50 92 Below 55 98 Below 60 100

Solution

Class interval f cf 15-20 2 2 20-25 4 6 25-30 18 24 30-35 21 45 35-40 33 78 40-45 11 89 45-50 3 92 50-55 6 98 55-60 2 100  

Q42.

Solution

Q43. A survey regarding the heights (in cm) of 50 girls of class X of a school was conducted and the following data was obtained: Height (in cm) 120 - 130 130 - 140 140 - 150 150 – 160 160 – 170 Total Number of girls 2 8 12 20 8 50 Find the mean, median and mode of the above data.

Solution

  Height in cm Number of girls Cumulative Frequency 120 – 130 2 2 130 – 140 8 10 140 – 150 12 22 150 – 160 20 42 160 – 170 8 50 Total 50      

Q44. In the distribution given below 50% of the observations is more than 14.4. Find the values of x and y, if the total frequency is 20. Class Interval 0-6 6-12 12-18 18-24 24-30 Frequency 4 x 5 y 1

Solution

  Class Interval 0-6 6-12 12-18 18-24 24-30 Frequency 4 x 5 y 1 Cumulative frequency 4 4+x 9+x 9+x+y 10+x+y It is given that total frequency N is 20 So, 10+x+y = 20 i.e. x + y = 10 Given 50% of the observations are greater than 14.4 so median = 14.4, which lies in the class interval 12-18. = 12, cf = 4 + x, h = 6, f = 5, N = 20 Median = 14.4 = 12 + x 6 14.4 - 12 = x6 = 6 - x Þx =4 Now using equation, 10 + x + y = 20, we get y = 6. Hence x = 4 and y = 6.

Q45. Write the following distribution as 'less than' type cumulative frequency distribution:   C.I 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60 60 - 70 70 - 80 Frequency 5 3 4 3 3 4 7 9  

Solution

Less than type cumulative frequency distribution is as follows:

Q46.

Solution

Q47. Compute the median for the following data: Marks        (more than or equal to) Number of students 80 150 90 141 100 124 110 105 120 60 130 27 140 12 150 0

Solution

First prepare the cumulative frequency table:     Marks No. of students Cumulative frequency 80 - 90 150 – 141 = 9 9 90 – 100 141 – 124 = 17 9 +17 = 26 100 - 110 124 – 105 = 19 26 +19 = 45 110 - 120 105 – 60 = 45 45 + 45 = 90 120 -130 60 – 27 = 33 90 + 33 = 123 130 – 140 27 – 12 = 15 123 + 15 = 138 140 -150 12- 0 = 12 138 + 12 =150  

Q48. Compute the median for the following data: Class interval Less than 20 Less than 30 Less than 40 Less than 50 Less than 60 Less than 70 Less than 80 Less than 90 Less than 100 Cumulative Frequency 0 4 16 30 46 66 82 92 100  

Solution

C.I. f c.f. 20 - 30 4 4 30 - 40 12 16 40 - 50 14 30 50 - 60 16 46 60 - 70 20 66 70 - 80 16 82 80 - 90 10 92 90 - 100 8 100   100   Here, N = 100   The cumulative frequency just greater than N/2=50 is 66 and the corresponding class is 60 - 70   Hence, 60 - 70 is the median class.   Thus, we have l = 60, c.f. = 46, f = 20, h = 10 Median = = = 60 + 2 = 62

Q49. The mean of the following frequency distribution is 25. Determine the value of p Classes 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 Frequency 5 18 15 P 6

Solution

Classes Frequency (f) Mid-value (x) fx 0 - 10 5 5 25 10 - 20 18 15 270 20 - 30 15 25 375 30 - 40 P 35 35p 40 - 50 6 45 270 44 + p 940 + 35p Mean = 940 + 35 p = 1100 + 25p 10p = 160 p = 16

Q50. Find the median wages for the following frequency distribution: Wages per day (in Rs.) Number of workers 61 – 70 5 71 – 80 15 81 – 90 20 91 – 100 30 101 – 110 20 111 - 120 8

Solution

Here the classes are in discontinuous form. So, convert the classes into continuous form. Adjustment factor = 1/2 (lower limit of one class – upper limit of previous class                               = 1/2(71 – 70) =.5   To convert discontinuous classes, subtract the adjustment factor from each lower limit and add the adjustment factor to each upper limit.   Wages per day f cf 60.5 – 70.5   5    5 70.5 - 80.5 15 20 80.5 – 90.5 20 40 90.5 – 100.5 30 70 100.5 – 110.5  20 90 110.5 – 120.5    8 98

Q51. Find the mean of the following data. x 0-2 2-4 4-6 6-8 8-10 10-12 12-14 f 1 2 1 5 6 2 3

Solution

Class Intervals       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   20   162

Q52. The distribution given below shows the weights of 30 students of a class. Find the median weight of the students Weight (in kg) 40-45 45-50 50-55 55-60 60-65 65-70 70-75 No. of students 2 3 8 6 6 3 2

Solution

Class interval f cf 40-45 2 2 45-50 3 5 50-55 8 13 55-60 6 19 60-65 6 25 65-70 3 28 70-75 2 30  

Q53. The arithmetic mean of the following data is 25. Find the value of p.   Classes: 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 Frequency, f: 3 P 3 6 2  

Solution

Classes: 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 Frequency, f: 3 p 3 6 2 x: 5 15 25 35 45 fx: 15 15p 75 210 90  

Q54. Construct the cumulative frequency distribution of the following distribution and find the median class.   Class 12.5 - 17.5 17.5 - 22.5 22.5 - 27.5 27.5 - 32.5 32.5 - 37.5 Frequency 2 22 19 14 13  

Solution

C.I. f c.f. 12.5 - 17.5 2 2 17.5 - 22.5 22 24 22.5 - 27.5 19 43 27.5 - 32.5 14 57 32.5 - 37.5 13 70 Here, N = 70 _{The cumulative frequency just greater than N/2=35 is 43 and the corresponding class is 22.5 - 27.5.} Median Class is 22.5 - 27.5

Q55. The average height of 30 boys is 150 cm. It was detected later that one value of 165 cm was wrongly copied as 135 cm for the computation of the mean. Find the correct mean.

Solution

Q56. The mean and median of same data are 24 and 26 respectively. The value of mode is :

• 1) 25
• 2) 30
• 3) 26
• 4) 23

Solution

We know: Mode = 3 Median - 2 Mean = 3 x 26 - 2 x 24 = 78 - 48 = 30

Q57. The mean of 5 observations x, x + 2, x + 4, x + 6 and x + 8 is 11, then the value of x is:

• 1) 4
• 2) 6
• 3) 11
• 4) 7

Solution

Mean =

Q58. Change the following frequency distribution to less than type distribution and draw it's ogive. Hence, obtain the median value. Classes 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60 Frequency f 2 5 7 12 8 6

Solution

Upper class limits c.f. Less than 10 2 Less than 20 7 Less than 30 14 Less than 40 26 Less than 50 34 Less than 60 40 Plot the points (10, 2), (20, 7), (30, 14), (40, 26), (50, 34), (60, 40) to obtain the required ogive. Here, N = 40 N/2 = 20 So, mark the point whose ordinate is 20, its x-coordinate is 32. Median = 32 (Approx.)

Q59. The mean of 25 observations is 36. If the mean of first 13 observations is 32 and that of the last 13 observations is 39, find the 13^th observation.

Solution

Q60. For the following data, find mode. Class 1 - 3 3 - 5 5 - 7 7 - 9 9 - 11 Frequency 14 16 4 4 2

Solution

Modal class = 3 - 5 l = 3, f₁ = 16, f₀ = 14, f₂ = 4, h = 2 Mode = 3 + Mode = 3.3 (approximately)

Q61. Find the median of the following data: Marks 0 - 10 10 - 30 30 - 60 60 - 80 80 - 100 Frequency 5 15 30 8 2  

Solution

  Marks f cf 0 - 10 5 5 10 - 30 15 20 30 - 60 30 50 60 - 80 8 58 80 - 100 2 60 N= Here, N = 60 So, N/2 = 30 The cumulative frequency is just greater than N/2 = 30 is 50 and the corresponding class is 30-60. Hence, 30-60 is the median class. Therefore, l = 30, f = 30, cf = 20, h = 30 Thus, Median is given by Thus, Median = 40 

Q62. Change the given distribution to more than type distribution and draw its ogive. Classes 20 - 25 25 - 30 30 - 35 35 - 40 40 - 45 45 - 50 Frequency 3 6 12 9 6 4

Solution

More than type distribution: Classes Frequency More than 20 40 More than 25 37 More than 30 31 More than 35 19 More than 40 10 More than 45 4

Q63. Find the median of the following data: Classes 500 - 600 600 - 700 700 - 800 800 - 900 900 - 1000 Frequency 40 28 35 22 25  

Solution

  Classes f cf 500 - 600 40 40 600 - 700 28 68 700 - 800 35 103 800 - 900 22 125 900 - 1000 25 150 Here, N = 150   The cumulative frequency just greater than N/2=75 is 103 and the corresponding class is 700 - 800.   Hence, 700 - 800 is the median class.   Thus, we have l = 700, c.f. = 68, f = 35, h = 100 Thus, Median = 

Q64.

Solution

Q65. Convert the following data to a less than type distribution and draw its Ogive. Also find the median from the graph.   Class 100 - 120 120 - 140 140 - 160 160 - 180 180 -200 Frequency 12 14 8 6 10    

Solution

Frequency distribution table of less than type is as follows:             Daily income      (in Rs)             (upper class limits) Cumulative frequency Less than 120 12 Less than 140 12 + 14 = 26 Less than 160 26 + 8 = 34 Less than 180 34 + 6 = 40 Less than 200 40 + 10 = 50   Now taking upper class limits of class intervals on x-axis and their respective frequencies on y-axis, we can draw its ogive as follows: Here, N = 50 N/2 = 25 Now, mark the point on curve whose y-coordinate is 25, its corresponding x-coordinate is 138.5. So median of this data is 138.5 (approximately). 

Q66.

Solution

Q67. Find the mean of the following data using step deviation method:   Classes 25 - 30 30 - 35 35 - 40 40 - 45 45 - 50 50 - 55 55 - 60 Frequency 14 22 16 6 5 3 4    

Solution

Class x_i f_i d_i = xi - a f_i u_i 25 - 30 27.5 14 -15 -3 -42 30 - 35 32.5 22 -10 -2 -44 35 - 40 37.5 16 -5 -1 -16 40 - 45 42.5 = a 6 0 0 0 45 - 50 47.5 5 5 1 5 50 - 55 52.5 3 10 2 6 55 - 60 57.5 4 15 3 12     = -79                     a = 42.5, h = 5 = 42.5 + 5 = 42.5 - 5.64 = 36.86

Q68. Find the mean of the following frequency distribution using assumed mean method: Classes 2 - 8 8 - 14 14 - 20 20 - 26 26 - 32 Frequency 6 3 12 11 8

Solution

Classes f_i x_i d_i = x_i - A d_i f_i 2 - 8 6 5 -12 -72 8 - 14 3 11 -6 -18 14 - 20 12 (17) = A 0 0 20 - 26 11 23 6 66 26 - 32 8 29 12 96 Here, A = 17 = 17 + 1.8 = 18.8

Q69. The ages of employees in a factory areas follows: Age in years 17 - 23 23 - 29 29 - 35 35 - 41 41 - 47 47 - 53 No. of employees 2 5 6 4 2 1 Find the median age of the employees.

Solution

  Age 17 - 23 23 - 29 29 - 35 35 - 41 41 - 47 47 - 53 No. of employees (f) 2 5 6 4 2 1 c.f. 2 7 13 17 19 20 _{Here, N = 20}  The cumulative frequency just greater than N/2=10 is 13 and the corresponding class is 29-35. Hence, Median class is 29-35.   Thus, we have   l = 29, c.f. = 7, f = 6, h = 6 Thus, Median is given by Median = 

Q70. Find the median of the following data: Class 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 Total Frequency 8 16 36 34 6 100

Solution

  Class Frequency (f) c.f. 0 - 10 8 8 10 - 20 16 24 20 - 30 36 60 30 - 40 34 94 40 - 50 6 100 Here, n = 100   The cumulative frequency just greater than N/2=50 is 60 and the corresponding class is 20 - 30. Hence, 20 - 30 is the median class. Thus, we have   l = 20, c.f. = 24, f = 36, h = 10 Thus, Median = 27.2

Q71. Find unknown entries a, b, c, d, e, f in the following distribution of heights of students in a class if the total number of students in the class in 50. Height in cm 150 - 155 155 - 160 160 - 165 165 - 170 170 - 175 175 - 180 Frequency 12 b 10 d e 2 Cumulative Frequency a 25 c 43 48 f

Solution

a = 12 a + b = 25 b = 13 c = 25 + 10 = 35 c + d = 43 d = 43 - 35 = 8 43 + e = 48 e = 5 f = 48 + 2 = 50

Q72. Find the mode of the following distribution of marks obtained by 50 students. Marks 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 No. of students 4 8 10 20 8

Solution

Marks 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 No. of students 4 8 10 20 8 f₀ f₁ f₂ Maximum frequency = 20 (f₁) Modal class = 30 - 40 = 34.55

Q73.

Solution

       

Q74. Write any two merits and demerits of arithmetic mean.

Solution

Merits: (1) It is based on all observations (2)It is simple to understand and calculate Demerits: (1) It is affected by extreme values (2) It cannot be determined graphically.

Q75. Find the median of the following data: Class 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 Total Frequency 8 16 36 34 6 100  

Solution

  Class 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 Frequency 8 16 36 34 6 c.f. 8 24 60 94 100 n = 100 = 50 Thus, Median class = 20 - 30, We have, l = 20, h = 10, c.f. = 24, f = 36 Median=l + = 20 + 7.2 = 27.2

Q76. The following distribution shows the daily pocket money of children of a locality. The mean pocket allowance is Rs. 18. Find the missing frequency f. Daily pocket money (in Rs.) No. of children 11-13 7 13-15 6 15-17 9 17-19 13 19-21 f 21-23 5 23-25 4

Solution

Daily pocket money (in Rs.)    x_i             f_i             f_i x_i             11-13   12            7         84             13-15   14            6         84             15-17   16            9                  144                     17-19   18            13          234             19-21   20            f          20f             21-23   22            5          110             23-25   24            4           96     44+f  

Q77. The following distribution gives the daily wages of workers of a factory. Daily wages (in Rs) 20 - 40 40 - 60 60 - 80 80 - 100 100- 120 120 - 140 140 - 160 Number of workers 4 6 10 16 12 7 3 Convert the above distribution into a less than type cumulative frequency distribution. Draw its ogive and find the median.

Solution

We find the cumulative frequency distribution of daily wages of 50 workers of a factory as follows: Daily wages (in Rs) No. of Workers Wages less than Cumulative Frequency 20 - 40 40 - 60 60 - 80 80 - 100 100 - 120 120 - 140 140 - 160 4 6 10 16 12 7 3 40 60 80 100 120 140 160 4 10 20 36 48 55 58 Now we mark the upper class limits along x-axis on a suitable scale and the cumulative frequencies along y-axis on a suitable scale. Plot the points (40, 4), (60, 10), (80, 20), (100, 36), (120, 48), (140, 55) and (160, 58). Now, join the plotted points by line segments as shown. N = 58 The median is corresponding to the frequency 29 In order to find the median, we first locate the point corresponding to 29 on the y-axis let the point be P. From this point, draw a line parallel to the x-axis cutting the curve at Q. From this point Q draw a line parallel to y-axis and meeting x-axis at point M. The x-coordinate of M is the median. Thus, median = 91 (approx).

Q78. The following distribution shows the heights of students of a certain school. Find the modal height. Height (in cm) No. of students 160-162 15 163-165 118 166-168 142 169-171 127 172-174 18

Solution

Here the classes are in discontinuous form. So, first convert the classes into continuous form.   Now, subtract the adjustment factor from each lower limit and add to each upper limit.   Height (in cm) No. of students 159.5-162.5 15 162.5-165.5 118 165.5-168.5 142 168.5-171.5 127 171.5-174.5 18  

Q79. Find the mean age of the following data. Age (yrs) 10 - 30 30 - 50 50 - 70 70 - 90 No. of persons 15 12 18 5

Solution

x_i (Mid-value) 20 40 60 80 f_i 15 12 18 5 x_i f_i 300 480 1080 400

Q80. Arithmetic mean of a set of 40 values is 65. If each of the 40 values is increased by 5, what will be the mean of the set of new values?

Solution

Q81. Find the mode of the given data:   Class 3 - 6 6 - 9 9 - 12 12 - 15 15 - 18 18 - 21 21 - 24 Frequency 2 5 10 23 21 12 3  

Solution

Modal class = 12 - 15 l = 12, f₁ = 23, f_o = 10, f₂ = 21, h = 3

Q82. Construct a cumulative frequency distribution table of the following distribution: C.I. 12.5-17.5 17.5-22.5 22.5-27.5 27.5-32.5 32.5-37.5  F 2 22 19 14 13   C.I. 12.5-17.5 17.5-22.5 22.5-27.5 27.5-32.5 32.5-37.5  F 2 22 19 14 13

Solution

C.I. F C.F. 12.5-17.5 2 2 17.5-22.5 22 24 22.5-27.5 19 43 27.5-32.5 14 57 32.5-37.5 13 70

Q83. The following in the daily pocket money spent by students.   Pocket money () 0 - 15 15 - 30 30 - 45 45 - 60 60 - 75 No. of students 8 15 7 4 6   Find the mode of the above data.

Solution

Modal class is 15 - 30 Thus, we have l = 15, f₁ = 15, f₀ = 8, f₁ = 7, h = 15 Now, mode is given by

Q84. Find the mode of given distribution: Class Interval 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 Frequency 2 12 22 8 6

Solution

Modal class = 20 - 30 Here, l = 20, f₀ = 12, f₁ = 22, f₂ = 8, h = 10 Mode = l + Mode = 20 +

Q85. The mean weight of 150 students in a class is 60 kg. The mean weight of boys is 70 kg and that of girls is 55 kg. Find the number of boys and girls.

Solution

Q86. If the mean of the following frequency distribution is 54, find the value of p. Classes 0 - 20 20 - 40 40 - 60 60 - 80 80 - 100 Frequency (f) 7 p 10 9 13

Solution

Classes f_i x_i f_ix_i 0 - 20 7 10 70 20 - 40 P 30 30p 40 - 60 10 50 500 60 - 80 9 70 630 80 - 100 13 90 1170 = 39 + p = 2370 + 30p Mean =

Q87. Find the mode of the following data. x 10 12 14 16 18 20 f 5 3 10 3 2 1

Solution

We observe that the value 14 has the maximum frequency i.e. 10. Hence, the modal value is 14.

Q88.

Solution

Q89. In the following distribution: Monthly income range (In Rs.) No. of families Income more than Rs 10000 100 Income more than Rs 13000 85 Income more than Rs 16000 69 Income more than Rs 19000 50 Income more than Rs 22000 33 Income more than Rs 25000 15   Find the number of families having income range (in Rs.) 16000-19000?

Solution

Monthly income range (In Rs.) No. of families 10000-13000 15 13000-16000 16 16000-19000 19 19000-22000 17 22000-25000 18 25000-28000 15 No. of families having income range (in Rs.) 16000-19000 is 19.