## NCERT Solutions Class 10 |

Class Intervals = Number of Plants | Frequency (f_{i}) = Number of houses | Mid-point (x_{i}) = (U.L + L.L)/2 | Product of f_{i} and 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 |

∑ f_{i} = 20 | ∑ f_{i}x_{i} = 162 |

The values of f_{i} and x_{i} are numerically small, so we will use the direct method to calculate the mean.

Mean = x? =∑ f_{i}x_{i}/∑ f_{i}

= 162/20 = 8.1

Mean number of plant per house = 8.1

**2. Consider the following distribution of daily wages of 50 workers of a factory.**

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

**SOLUTION**

Class Intervals = Daily wages | Frequency (f_{i}) = Number of workers | Mid-point (x_{i}) = (U.L + L.L)/2 | u_{i} = (x_{i} - 150)/20 | Product of f_{i} and u_{i} |

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

120 - 140 | 14 | 130 | -1 | -14 |

140 - 160 | 8 | 150 | 0 | 0 |

160 - 180 | 6 | 170 | 1 | 6 |

180 - 200 | 10 | 190 | 2 | 20 |

∑ f_{i} = 50 | ∑ f_{i}u_{i} = -12 |

Let us assume the value of mean to be A = 150.

Class interval = h = U.L - L.L = 20

True Mean = x? = A + h∑ f_{i}u_{i}/∑ f_{i}

= 150 + 20(-12)/50

= 145.2

Mean daily wage of the workers = Rs 145.2

**3. The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is Rs 18. Find the missing frequency f.**

**SOLUTION**

Class Intervals = Daily pocket allowance (in Rs) | Frequency (f_{i}) = Number of children | Mid-point (x_{i}) = (U.L + L.L)/2 | Product of f_{i} and x_{i} |

11 - 13 | 7 | 12 | 84 |

13 - 15 | 6 | 14 | 84 |

15 - 17 | 9 | 16 | 144 |

17 - 19 | 13 | 18 | 234 |

19 - 21 | f | 20 | 20f |

21 - 23 | 5 | 22 | 110 |

23 - 25 | 4 | 24 | 96 |

∑ f_{i} = f + 44 | ∑ f_{i}x_{i} = 752 + 20f |

Mean = x? = ∑f_{i}x_{i}/∑f_{i} = (752 + 20f)/(f + 44)

We know that x? = 18. Therefore

18 = (752 + 20f)/(f + 44)

18(f + 44) = 752 + 20f

18f + 792 = 752 + 20f

40 = 2f

**f = 20**

**4. Thirty women were examined in a hospital by a doctor and the number of heartbeats per minute were recorded and summarised as follows. Find the mean heartbeats per minute for these women, choosing a suitable method.**

**SOLUTION**

Class Intervals = Number of heartbeats per minutes | Frequency (f_{i}) = Number of women | Mid-point (x_{i}) = (U.L + L.L)/2 | u_{i} = (x_{i} - 75.5)/3 | Product of f_{i} and u_{i} |

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

68 - 71 | 4 | 69.5 | -2 | -8 |

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

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

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

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

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

∑ f_{i} = 30 | ∑ f_{i}u_{i} = 4 |

Let us assume the mean to be A = 75.5

Class height = U.L - L.L = 3

True Mean = x? = A + h∑ f_{i}u_{i}/∑ f_{i}

= 75.5 + 3(4)/10

= 75.9

Mean heart beats per minute of the women is 75.9

**5. In a retail market, fruit vendors were selling mangoes kept in packing boxes. These boxes contained varying number of mangoes. The following was the distribution of mangoes according to the number of boxes.**

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

**SOLUTION**

The given data does not have continuous class intervals. Therefore, we will add 0.5 to all the Upper Limit values and subtract 0.5 from all the Lower Limit values.

The given values for f_{i} are numerically big, so we will use the step-deviation method.

Class Intervals = Number of mangoes | Frequency (f_{i}) = Number of boxes | Mid-point (x_{i}) = (U.L + L.L)/2 | d_{i} = x_{i} - A | Product of f_{i} and d_{i} |

49.5 - 52.5 | 15 | 51 | -6 | 90 |

52.5 - 55.5 | 110 | 54 | -3 | -330 |

55.5 - 58.5 | 135 | 57 | 0 | 0 |

58.5 - 61.5 | 115 | 60 | 3 | 345 |

61.5 - 64.5 | 25 | 63 | 6 | 150 |

∑ f_{i} = 400 | ∑ f_{i}d_{i} = 75 |

Let us assume the mean to be A = 57

Class height = h = U.L - L.L = 3

True Mean = x? = A + h∑ f_{i}d_{i}/∑ f_{i}

= 57 + 0.1875

= 57.19

Mean number of mangoes kept in one box is 57.19

**6. The table below shows the daily expenditure on food of 25 households in a locality.**

**Find the mean daily expenditure on food by a suitable method.**

**SOLUTION**

Class Intervals = Daily expenditure | Frequency (f_{i}) = Number of households | Mid-point (x_{i}) = (U.L + L.L)/2 | u_{i} = (x_{i} - 150)/50 | Product of f_{i} and u_{i} |

100 - 150 | 4 | 125 | -2 | -8 |

150 - 200 | 5 | 175 | -1 | -5 |

200 - 250 | 12 | 225 | 0 | 0 |

250 - 300 | 2 | 275 | 1 | 2 |

300 - 350 | 2 | 325 | 2 | 4 |

∑ f_{i} = 25 | ∑ f_{i}u_{i} = -7 |

Let us assume the mean to be A = 225

Class height = h = U.L - L.L = 50

True Mean = x? = A + h∑ f_{i}u_{i}/∑ f_{i}

= 225 + 50(-7)/25

= 211

Mean daily expenditure on food is Rs 211.

**7. To find out the concentration of SO _{2} in the air (in parts per million, i.e., ppm), the data was collected for 30 localities in a certain city and is presented below:**

**Find the mean concentration of SO _{2} in the air.**

**SOLUTION**

Class Intervals = Concentration of SO_{2} in ppm | Frequency (f_{i}) | Mid-point (x_{i}) = (U.L + L.L)/2 | Product of f_{i} and x_{i} |

0.00 - 0.04 | 4 | 0.02 | 0.08 |

0.04 - 0.08 | 9 | 0.06 | 0.54 |

0.08 - 0.12 | 9 | 0.1 | 0.9 |

0.12 - 0.16 | 2 | 0.14 | 0.28 |

0.16 - 0.20 | 4 | 0.18 | 0.72 |

0.2 - 0.24 | 2 | 0.22 | 0.44 |

∑ f_{i} = 30 | ∑ f_{i}x_{i} = 2.96 |

Mean = x? = ∑ f_{i}x_{i}/∑ f_{i}

= 2.96/30

= 0.0986

Mean concentration of SO_{2} in the air is 0.0986 ppm.

**8. 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.**

**SOLUTION**

Class Intervals = Number of Days | Frequency (f_{i}) = Number of students | Mid-point (x_{i}) = (U.L + L.L)/2 | Product of f_{i} and x_{i} |

0 - 6 | 11 | 3 | 33 |

6 - 10 | 10 | 8 | 80 |

10 - 14 | 7 | 12 | 84 |

14 - 20 | 4 | 17 | 68 |

20 - 28 | 4 | 24 | 96 |

28 - 38 | 3 | 33 | 99 |

38 - 40 | 1 | 39 | 39 |

∑ f_{i} = 40 | ∑ f_{i}x_{i} = 499 |

Mean = x? = ∑ f_{i}x_{i}/∑ f_{i}

= 499/40

= 12.48

Mean number of days a student was absent is 12.48

**9. The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.**

**SOLUTION**

Class Intervals = Literacy rate (in %) | Frequency (f_{i}) = Number of cities | Mid-point (x_{i}) = (U.L + L.L)/2 | u_{i} = (x_{i} - 70)/10 | Product of f_{i} and u_{i} |

45 - 55 | 3 | 50 | -2 | -6 |

55 - 65 | 10 | 60 | -1 | -10 |

65 - 75 | 11 | 70 | 0 | 0 |

75 - 85 | 8 | 80 | 1 | 8 |

85 - 95 | 3 | 90 | 2 | 6 |

∑ f_{i} = 35 | ∑ f_{i}u_{i} = -2 |

Let us assume the mean to be A = 70

Class height = U.L - L.L = 10

True Mean = x? = A + h∑ f_{i}u_{i}/∑ f_{i}

= 70 + 10(-2)/35

= 69.428

Mean literacy rate of a city is 69.428 %

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

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

**SOLUTION**

The highest frequency in the given data is 23. Therefore, the modal class will be 35 - 45.

l = 35

f_{1} = 23

f_{2} = 14

f_{0} = 21

Class width = h = 10

Mode = 36.81 years

Class Intervals = Age (in years) | Frequency (f_{i}) = Number of patients | Mid-point (x_{i}) = (U.L + L.L)/2 | Product of f_{i} and x_{i} |

5 - 15 | 6 | 10 | 60 |

15 - 25 | 11 | 20 | 220 |

25 - 35 | 21 | 30 | 630 |

35 - 45 | 23 | 40 | 920 |

45 - 55 | 14 | 50 | 700 |

55 - 65 | 5 | 60 | 300 |

∑ f_{i} = 80 | ∑ f_{i}x_{i} = 2830 |

Mean = x? = ∑ f_{i}x_{i}/∑ f_{i}

= 2830/80

= 35.375 years

Mean age of a patient is 35.375 years

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

**Determine the modal lifetimes of the components.**

**SOLUTION**

The highest frequency in the given data is 61. Therefore, the modal class will be 60 - 80.

l = 60

f_{1} = 61

f_{2} = 38

f_{0} = 52

Class width = h = 20

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

**SOLUTION**

The highest frequency in the given data is 40. Therefore, the modal class will be 1500 - 2000.

l = 1500

f_{1} = 40

f_{2} = 33

f_{0} = 24

Class width = h = 500

Class Intervals = Expenditure | Frequency (f_{i}) = Number of families | Mid-point (x_{i}) = (U.L + L.L)/2 | u_{i} = (x_{i} - 2750)/500 | Product of f_{i} and u_{i} |

1000 - 1500 | 24 | 1250 | -3 | -72 |

1500 - 2000 | 40 | 1750 | -2 | -80 |

2000 - 2500 | 33 | 2250 | -1 | -33 |

2500 - 3000 | 28 | 2750 | 0 | 0 |

3000 - 3500 | 30 | 3250 | 1 | 30 |

3500 - 4000 | 22 | 3750 | 2 | 44 |

4000 - 4500 | 16 | 4250 | 3 | 48 |

4500 - 5000 | 7 | 4750 | 4 | 28 |

∑ f_{i} = 200 | ∑ f_{i}u_{i} = -35 |

Let us assume the mean to be A = 2750

True mean = x? = A + h∑ f_{i}u_{i}/∑ f_{i}

= 2750 + 500(-35)/200

= 2662.5

Mean monthly expenditure of the families = Rs 2662.5

**4. 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**

The highest frequency in the given data is 10. Therefore, the modal class will be 30 - 35.

l = 30

f_{1} = 10

f_{2} = 3

f_{0} = 9

Class width = h = 5

Class Intervals = Age (in years) | Frequency (f_{i}) = Number of patients | Mid-point (x_{i}) = (U.L + L.L)/2 | Product of f_{i} and x_{i} |

15 - 20 | 3 | 17.5 | 52.5 |

20 - 25 | 8 | 22.5 | 180 |

25 - 30 | 9 | 27.5 | 247.5 |

30 - 35 | 10 | 32.5 | 325 |

35 - 40 | 3 | 37.5 | 112.5 |

40 - 45 | 0 | 42.5 | 0 |

45 - 50 | 0 | 47.5 | 0 |

50 - 55 | 2 | 52.7 | 105.5 |

∑ f_{i} = 35 | ∑ f_{i}x_{i} = 1022.5 |

Mean = x? = ∑f_{i}x_{i}/∑f_{i}

= 1022.5/35

= 29.214

Mean student-teacher ratio of the States/U.T.s = 29.214

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

**Find the mode of the data.**

**SOLUTION**

The highest frequency in the given data is 18. Therefore, the modal class will be 4000 - 5000.

l = 4000

f_{1} = 18

f_{2} = 9

f_{0} = 4

Class width = h = 1000

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

**SOLUTION**

The highest frequency in the given data is 20. Therefore, the modal class will be 40 - 50.

l = 40

f_{1} = 20

f_{2} = 11

f_{0} = 12

Class width = h = 10

**1. 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.**

**SOLUTION**

Class Interval = Monthly consumption | Frequency (f_{i}) = Number of consumers | Cumulative frequency |

65 - 85 | 4 | 4 |

85 - 105 | 5 | 9 |

105 - 125 | 13 | 22 |

125 - 145 | 20 | 42 |

145 - 165 | 14 | 56 |

165 - 185 | 8 | 64 |

185 - 205 | 4 | 68 |

n = 68 |

n = 68

n/2 = 34

Therefore, the median class will be 125 - 145 and cumulative frequency is 42

l = 125

f = 20

cf = 22

Class height = h = 20

Median monthly consumption = 137 units

The highest frequency in the given data is 20. Therefore, the modal class will be 125 - 145.

l = 125

f_{1} = 20

f_{2} = 14

f_{0} = 13

Class width = h = 20

Mode = 103.823

Class Intervals = Monthly Consumption | Frequency (f_{i}) = Number of consumers | Mid-point (x_{i}) = (U.L + L.L)/2 | u_{i} = (x_{i} - 135)/20 | Product of f_{i} and u_{i} |

65 - 85 | 4 | 75 | -3 | -12 |

85 - 105 | 5 | 95 | -2 | -10 |

105 - 125 | 13 | 115 | -1 | -13 |

125 - 145 | 20 | 135 | 0 | 0 |

145 - 165 | 14 | 155 | 1 | 14 |

165 - 185 | 8 | 175 | 2 | 16 |

185 - 205 | 4 | 195 | 3 | 12 |

∑ f_{i} = 68 | ∑ f_{i}u_{i} = 7 |

Let us assume the mean to be A = 135

True Mean = x? = A + h∑ f_{i}u_{i}/∑ f_{i}

= 135 + 20(7)/68

= 137.06

Mean monthly consumption is 137.06 units.

On comparison of mean, median and mode of the data, we can conclude that they are approximately equal to each other.

**2. If the median of the distribution given below is 28.5, find the values of x and y.**

**SOLUTION**

Class Interval | Frequency (f_{i}) | Cumulative frequency |

0 - 10 | 5 | 5 |

10 - 20 | x | 5 + x |

20 - 30 | 20 | 25 + x |

30 - 40 | 15 | 40 + x |

40 - 50 | y | 40 + x + y |

50 - 60 | 5 | 45 + x + y |

n = 60 |

n = 60

n/2 = 30

Therefore, the median class will be 20 - 30 with cumulative frequency 25 + x

l = 20

f = 20

cf = 5 + x

Class height = h = 10

∑ f_{i} = 45 + x + y = n

45 + 8 + y = 60

53 + y = 60

y = 7

Hence, the values of x and y are 8 and 7 respectively.

**3. A life insurance agent found the following data for distribution of ages of 100 policy holders. Calculate the median age, if policies are given only to persons having age 18 years onwards but less than 60 year.**

**SOLUTION**

Class Interval | Frequency (f_{i}) | Cumulative frequency |

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 |

n = 100 |

n = 100

n/2 = 50

Therefore, the median class will be 35 - 40 with cumulative frequency = 78

l = 35

f = 33

cf = 45

Class height = h = 5

Median age = 35.75 years

**4. The lengths of 40 leaves of a plant are measured correct to the nearest millimetre, and the data obtained is represented in the following table :**

**Find the median length of the leaves.**

**(Hint : The data needs to be converted to continuous classes for finding the median, since the formula assumes continuous classes. The classes then change to 117.5 - 126.5, 126.5 - 135.5, . . ., 171.5 - 180.5.)**

**SOLUTION**

Since the data isn't continuous, we will add 0.5 to all the Upper Limit values and subtract 0.5 from all the Lower Limit values.

Class Interval | Frequency (f_{i}) | Cumulative frequency |

117.5 - 126.5 | 3 | 3 |

126.5 - 135.5 | 5 | 8 |

135.5 - 144.5 | 9 | 17 |

144.5 - 153.5 | 12 | 29 |

153.5 - 162.5 | 5 | 34 |

162.5 - 171.5 | 4 | 38 |

171.5 - 180.5 | 2 | 40 |

n = 40 |

n = 40

n/2 = 20

Therefore, the median class will be 144.5 - 153.5 with cumulative frequency = 29

l = 144.5

f = 12

cf = 17

Class height = h = 9

Median length of the leaves = 146.25 mm.

**5. The following table gives the distribution of the life time of 400 neon lamps :**

**Find the median life time of a lamp.**

**SOLUTION**

Class Interval | Frequency (f_{i}) | Cumulative frequency |

1500 - 2000 | 14 | 14 |

2000 - 2500 | 56 | 70 |

2500 - 3000 | 60 | 130 |

3000 - 3500 | 86 | 216 |

3500 - 4000 | 74 | 290 |

4000 - 4500 | 62 | 352 |

4500 - 5000 | 48 | 400 |

n = 400 |

n = 400

n/2 = 200

Therefore, the median class will be 3000 - 3500 with cumulative frequency = 216

l = 3000

f = 86

cf = 130

Class height = h = 500

Median life time of a lamp = 3406.976 hours

**6. 100 surnames were randomly picked up from a local telephone directory and the frequency distribution of the number of letters in the English alphabets in the surnames was obtained as follows:**

**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**

Class Interval | Frequency (f_{i}) | Cumulative frequency |

1 - 4 | 6 | 6 |

4 - 7 | 30 | 36 |

7 - 10 | 40 | 76 |

10 - 13 | 16 | 92 |

13 - 16 | 4 | 96 |

16 - 19 | 4 | 100 |

n = 100 |

n = 100

n/2 = 50

Therefore, the median class will be 7 - 10 with cumulative frequency = 76

l = 7

f = 40

cf = 36

Class height = h = 3

Median number of letters = 8.05

Class Intervals = Number of letters | Frequency (f_{i}) = Number of surnames | Mid-point (x_{i}) = (U.L + L.L)/2 | Product of f_{i} and x_{i} |

1 - 4 | 6 | 2.5 | 15 |

4 - 7 | 30 | 5.5 | 165 |

7 - 10 | 40 | 8.5 | 340 |

10 - 13 | 16 | 11.5 | 184 |

13 - 16 | 4 | 14.5 | 51 |

16 - 19 | 4 | 17.5 | 70 |

∑ f_{i} = 100 | ∑ f_{i}x_{i} = 825 |

Mean = x? = ∑ f_{i}x_{i}/∑ f_{i}

= 825/100 = 8.25

Mean number of letters = 8.25

The highest frequency in the given data is 40. Therefore, the modal class will be 7 - 10.

l = 7

f_{1} = 40

f_{2} = 16

f_{0} = 30

Class width = h = 3

Modal number of letters = 7.882

**7. The distribution below gives the weights of 30 students of a class. Find the median weight of the students.**

**SOLUTION**

Class Interval | Frequency (f_{i}) | Cumulative frequency |

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 |

n = 30 |

n = 30

n/2 = 15

Therefore, the median class will be 55 - 60 with cumulative frequency = 19

l = 55

f = 6

cf = 13

Class height = h = 5

Median weight of the students = 56.667 kg

**1. The following distribution gives the daily income of 50 workers of a factory.**

**Convert the distribution above to a less than type cumulative frequency distribution, and draw its ogive.**

**SOLUTION**

We need to convert the data into less than type distribution and find cumulative frequencies.

Daily Income | Frequency (f_{i}) = Number of workers | Cumulative frequency |

Less than 120 | 12 | 12 |

Less than 140 | 14 | 26 |

Less than 160 | 8 | 34 |

Less than 180 | 6 | 40 |

Less than 200 | 10 | 50 |

n = 50 |

Using the upper limits and the corresponding cumulative frequencies, we can obtain the points (120, 12), (140, 26), (160, 34), (180, 40), and (200, 50). We will plot a graph with these coordinates and join the points to obtain the less than type ogive curve.

**2. During the medical check-up of 35 students of a class, their weights were recorded as follows:**

**Draw a less than type ogive for the given data. Hence obtain the median weight from the graph and verify the result by using the formula.**

**SOLUTION**

Using the upper limits and the corresponding cumulative frequencies, we can obtain the points (38, 0), (40, 3), (42, 5), (44, 9), (46, 14), (48, 28), (50, 32) and (52, 35). We will plot a graph with these coordinates and join the points to obtain the less than type ogive curve.

Sum of frequencies is 35. Therefore, we will locate the point 35/2 = 17.5 on the y-axis or the cumulative frequency axis.

Produce a line parallel to x-axis through the point (0, 17.5) until it meets the ogive curve.

Draw the perpendicular at the point at which this line intersects the ogive curve.

The coordinates of this point are found out to be (46.5, 17.5). Therefore, 46.5 is the median weight of the students from the given data.

Verification :

We need to find the actual frequencies.

Weight | Frequency (f_{i}) = Number of students | Cumulative frequency |

Less than 38 | 0 | 0 |

Less than 40 | 3 - 0 = 3 | 3 |

Less than 42 | 5 - 3 = 2 | 5 |

Less than 44 | 9 - 5 = 4 | 9 |

Less than 46 | 14 - 9 = 5 | 14 |

Less than 48 | 28 - 14 = 14 | 28 |

Less than 50 | 32 - 28 = 4 | 32 |

Less than 52 | 35 - 32 = 3 | 35 |

n = 35 |

n = 35

n/2 = 17.5

Therefore, the median class will be 46 - 48 with cumulative frequency = 28

l = 46

f = 14

cf = 14

Class height = h = 2

Median weight of the students = 46.5 kg

Hence, the median weight of students has been verified.

**3. The following table gives production yield per hectare of wheat of 100 farms of a village.**

**Change the distribution to a more than type distribution, and draw its ogive.**

**SOLUTION**

We need to convert the given data to more than type distribution and find out the cumulative frequencies.

Production yeild | Frequency (f_{i}) = Number of farms | Cumulative frequency |

More than 50 | 2 | 100 |

More than 55 | 8 | 100 - 2 = 98 |

More than 60 | 12 | 98 - 8 = 90 |

More than 65 | 24 | 90 - 12 = 78 |

More than 70 | 38 | 78 - 24 = 54 |

More than 75 | 16 | 54 - 38 = 16 |

n = 50 |

Using the upper limits and the corresponding cumulative frequencies, we can obtain the points (50, 100), (55, 90), (60, 90), (65, 78), (70, 54) and (75, 16). We will plot a graph with these coordinates and join the points to obtain the more than type ogive curve.

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