It can be observed that the maximum class frequency is 10, belonging to 30 - 35. Therefore, Modal class = 30 - 35 Class size (h) = 5 Lower limit (l) of modal class = 30 Frequency (f₁) of modal class = 10 Frequency (f₀) of class preceding the modal class = 9 Frequency (f₂) of class succeeding the modRead more
It can be observed that the maximum class frequency is 10, belonging to 30 – 35.
Therefore, Modal class = 30 – 35
Class size (h) = 5
Lower limit (l) of modal class = 30
Frequency (f₁) of modal class = 10
Frequency (f₀) of class preceding the modal class = 9
Frequency (f₂) of class succeeding the modal class 3
Mode = l + ((f₁ – f₀)/(2f₁ – f₀ – f₂)) × h = 30 + ((10 – 9)/(2 × 10 – 9 – 3)) × 5 = 30 + 1/8 × 5 = 30 + 0.625 = 30.625
It represent that most of the states/U.T have a teacher-student ratio as 30.6.
To find the class marks, the following relation is used.
xᵢ = (Upper Limit + Lower Limit)/2
Taking 32.5 as assumed mean (a), dᵢ, uᵢ and fᵢuᵢ are calculated as follows.
From the table, we obtain
∑fᵢ = 35, ∑fᵢuᵢ = -23, a = 32.5 and h = 5
mean (X̄) = a + (∑fᵢuᵢ / ∑fᵢ)h = 32.5 + (-23/35) × 5 = 32.5 – 23/7 = 32.5 – 3.28 = 29.22
Therefore, mean of the data is 29.2.
It represents that on an average, teacher-student ration was 29.2
To find the class marks (xᵢ), the following relation is used. xᵢ = (Upper Limit + Lower Limit)/2 Taking 30 as assumed mean (a), dᵢ and fᵢdᵢ are calculated as follows. From the table, we obtain ∑fᵢ = 80, ∑fᵢdᵢ = 430 and a = 30 mean (X̄) = a + (∑fᵢdᵢ / ∑fᵢ) = 30 + (430/80) = 30 + 5.375 = 35.375 = 35.3Read more
To find the class marks (xᵢ), the following relation is used.
xᵢ = (Upper Limit + Lower Limit)/2
Taking 30 as assumed mean (a), dᵢ and fᵢdᵢ are calculated as follows.
From the table, we obtain
∑fᵢ = 80, ∑fᵢdᵢ = 430 and a = 30
mean (X̄) = a + (∑fᵢdᵢ / ∑fᵢ) = 30 + (430/80) = 30 + 5.375 = 35.375 = 35.38
Mean of this data is 35.38. It represents that on an average, the age of a patient admitted to hospital was 35.38 years. It can be observed that the maximum class frequency is 23 belonging to class interval 35 -45.
Modal class = 35 – 45
Lower limit (l) of modal class 35
Frequency (f₁) of modal class = 23
Class size (h) = 10
Frequency (f₀) of class preceding the modal class = 21
Frequency (f₂) of class succeeding the modal class 14
Mode = l + ((f₁ – f₀)/(2f₁ – f₀ – f₂)) × h = 35 + ((23 – 21)/(2 × 23 – 21 – 14)) × 10 = 35 + 2/11 × 10 = 35 + 1.81 = 36.81
Mode is 36.8. It represents that the age of maximum number of patients admitted in hospital was 36.8 years.
Here is the video explanation of the above question😃
From the data given above, it can be observed that the maximum class frequency is 61, belonging to class interval 60 - 80. Therefore, Modal class = 60 - 80 Lower limit (l) of modal class = 60 Frequency (f₁) of modal class = 61 Frequency (f₀) of class preceding the modal class = 52 Frequency (f₂) ofRead more
From the data given above, it can be observed that the maximum class frequency is 61, belonging to class interval 60 – 80.
Therefore, Modal class = 60 – 80
Lower limit (l) of modal class = 60
Frequency (f₁) of modal class = 61
Frequency (f₀) of class preceding the modal class = 52
Frequency (f₂) of class succeeding the modal class 38
Class size (h) = 20
Mode = l + ((f₁ – f₀)/(2f₁ – f₀ – f₂)) × h = 60 + ((61 – 52)/(2 × 61 – 52 – 38)) × 20 = 60 + 9/32 × 20 = 60 + 5.625 = 65.625
Therefore, modal lifeline of electrical components is 65.625 hours.
It can be observed that the maximum class frequency is 40, belonging to 1500 - 2000 intervals. Therefore, Modal class = 1500 - 2000 Lower limit (l) of modal class = 1500 Frequency (f₁) of modal class = 40 Frequency (f₀) of class preceding the modal class = 24 Frequency (f₂) of class succeeding the mRead more
It can be observed that the maximum class frequency is 40, belonging to 1500 – 2000 intervals.
Therefore, Modal class = 1500 – 2000
Lower limit (l) of modal class = 1500
Frequency (f₁) of modal class = 40
Frequency (f₀) of class preceding the modal class = 24
Frequency (f₂) of class succeeding the modal class 33
Class size (h) = 500
Mode = l + ((f₁ – f₀)/(2f₁ – f₀ – f₂)) × h = 500 + ((40 – 24)/(2 × 40 – 24 – 33)) × 500 = 1500 + 16/23 × 500 = 1500 + 347.826 = 1847.83
Therefore, modal monthly expenditure was ₹1847.83
To find the class mark, the following relation is used.
xᵢ = (Upper Limit + Lower Limit)/2
Class size (h) of the given data = 500.
Taking 2750 as assumed mean (a), dᵢ, uᵢ and fᵢuᵢ are calculated as follows.
From the table, we obtain
∑fᵢ = 200, ∑fᵢuᵢ = -35, a = 2750 and h = 500
mean (X̄) = a + (∑fᵢuᵢ / ∑fᵢ)h = 2750 + (-35/200) × 500 = 2750 – 87.5 = 2662.50
We know that each angle of equilateral triangle is 60°. Area of sector OCDE = 60°/360° × πr² = 1/6 × π(6)² = 1/6 × 22/7 × 6 × 6 = 132/7 cm² Area of equilateral triangle OAB = √3/4 (12)² = 36√3 cm² Area of circle = πr² = π(6)² = 22/7 × 6 × 6 = 792/7 cm² Area of shaded region = Area of circle + Area oRead more
We know that each angle of equilateral triangle is 60°.
Area of sector OCDE
= 60°/360° × πr² = 1/6 × π(6)² = 1/6 × 22/7 × 6 × 6 = 132/7 cm²
Area of equilateral triangle OAB
= √3/4 (12)² = 36√3 cm²
Area of circle
= πr² = π(6)² = 22/7 × 6 × 6 = 792/7 cm²
Area of shaded region
= Area of circle + Area of triangle OAB- Area of sector OCDE
= (792/7 + 36√3 – 132/7) cm²
= (36√3 + 660/7) cm²
The following distribution gives the state-wise teacher-student ratio in higher secondary schools of India.
It can be observed that the maximum class frequency is 10, belonging to 30 - 35. Therefore, Modal class = 30 - 35 Class size (h) = 5 Lower limit (l) of modal class = 30 Frequency (f₁) of modal class = 10 Frequency (f₀) of class preceding the modal class = 9 Frequency (f₂) of class succeeding the modRead more
It can be observed that the maximum class frequency is 10, belonging to 30 – 35.
Therefore, Modal class = 30 – 35
Class size (h) = 5
Lower limit (l) of modal class = 30
Frequency (f₁) of modal class = 10
Frequency (f₀) of class preceding the modal class = 9
Frequency (f₂) of class succeeding the modal class 3
Mode = l + ((f₁ – f₀)/(2f₁ – f₀ – f₂)) × h = 30 + ((10 – 9)/(2 × 10 – 9 – 3)) × 5 = 30 + 1/8 × 5 = 30 + 0.625 = 30.625
It represent that most of the states/U.T have a teacher-student ratio as 30.6.
To find the class marks, the following relation is used.
xᵢ = (Upper Limit + Lower Limit)/2
Taking 32.5 as assumed mean (a), dᵢ, uᵢ and fᵢuᵢ are calculated as follows.
From the table, we obtain
See less∑fᵢ = 35, ∑fᵢuᵢ = -23, a = 32.5 and h = 5
mean (X̄) = a + (∑fᵢuᵢ / ∑fᵢ)h = 32.5 + (-23/35) × 5 = 32.5 – 23/7 = 32.5 – 3.28 = 29.22
Therefore, mean of the data is 29.2.
It represents that on an average, teacher-student ration was 29.2
The following table shows the ages of the patients admitted in a hospital during a year:
To find the class marks (xᵢ), the following relation is used. xᵢ = (Upper Limit + Lower Limit)/2 Taking 30 as assumed mean (a), dᵢ and fᵢdᵢ are calculated as follows. From the table, we obtain ∑fᵢ = 80, ∑fᵢdᵢ = 430 and a = 30 mean (X̄) = a + (∑fᵢdᵢ / ∑fᵢ) = 30 + (430/80) = 30 + 5.375 = 35.375 = 35.3Read more
To find the class marks (xᵢ), the following relation is used.
xᵢ = (Upper Limit + Lower Limit)/2
Taking 30 as assumed mean (a), dᵢ and fᵢdᵢ are calculated as follows.
From the table, we obtain
∑fᵢ = 80, ∑fᵢdᵢ = 430 and a = 30
mean (X̄) = a + (∑fᵢdᵢ / ∑fᵢ) = 30 + (430/80) = 30 + 5.375 = 35.375 = 35.38
Mean of this data is 35.38. It represents that on an average, the age of a patient admitted to hospital was 35.38 years. It can be observed that the maximum class frequency is 23 belonging to class interval 35 -45.
Modal class = 35 – 45
Lower limit (l) of modal class 35
Frequency (f₁) of modal class = 23
Class size (h) = 10
Frequency (f₀) of class preceding the modal class = 21
Frequency (f₂) of class succeeding the modal class 14
Mode = l + ((f₁ – f₀)/(2f₁ – f₀ – f₂)) × h = 35 + ((23 – 21)/(2 × 23 – 21 – 14)) × 10 = 35 + 2/11 × 10 = 35 + 1.81 = 36.81
Mode is 36.8. It represents that the age of maximum number of patients admitted in hospital was 36.8 years.
Here is the video explanation of the above question😃
See lessThe following data gives the information on the observed lifetimes (in hours) of 225 electrical components:
From the data given above, it can be observed that the maximum class frequency is 61, belonging to class interval 60 - 80. Therefore, Modal class = 60 - 80 Lower limit (l) of modal class = 60 Frequency (f₁) of modal class = 61 Frequency (f₀) of class preceding the modal class = 52 Frequency (f₂) ofRead more
From the data given above, it can be observed that the maximum class frequency is 61, belonging to class interval 60 – 80.
See lessTherefore, Modal class = 60 – 80
Lower limit (l) of modal class = 60
Frequency (f₁) of modal class = 61
Frequency (f₀) of class preceding the modal class = 52
Frequency (f₂) of class succeeding the modal class 38
Class size (h) = 20
Mode = l + ((f₁ – f₀)/(2f₁ – f₀ – f₂)) × h = 60 + ((61 – 52)/(2 × 61 – 52 – 38)) × 20 = 60 + 9/32 × 20 = 60 + 5.625 = 65.625
Therefore, modal lifeline of electrical components is 65.625 hours.
The following data gives the distribution of total monthly household expenditure of 200 families of a village.
It can be observed that the maximum class frequency is 40, belonging to 1500 - 2000 intervals. Therefore, Modal class = 1500 - 2000 Lower limit (l) of modal class = 1500 Frequency (f₁) of modal class = 40 Frequency (f₀) of class preceding the modal class = 24 Frequency (f₂) of class succeeding the mRead more
It can be observed that the maximum class frequency is 40, belonging to 1500 – 2000 intervals.
Therefore, Modal class = 1500 – 2000
Lower limit (l) of modal class = 1500
Frequency (f₁) of modal class = 40
Frequency (f₀) of class preceding the modal class = 24
Frequency (f₂) of class succeeding the modal class 33
Class size (h) = 500
Mode = l + ((f₁ – f₀)/(2f₁ – f₀ – f₂)) × h = 500 + ((40 – 24)/(2 × 40 – 24 – 33)) × 500 = 1500 + 16/23 × 500 = 1500 + 347.826 = 1847.83
Therefore, modal monthly expenditure was ₹1847.83
To find the class mark, the following relation is used.
xᵢ = (Upper Limit + Lower Limit)/2
Class size (h) of the given data = 500.
Taking 2750 as assumed mean (a), dᵢ, uᵢ and fᵢuᵢ are calculated as follows.
From the table, we obtain
See less∑fᵢ = 200, ∑fᵢuᵢ = -35, a = 2750 and h = 500
mean (X̄) = a + (∑fᵢuᵢ / ∑fᵢ)h = 2750 + (-35/200) × 500 = 2750 – 87.5 = 2662.50
Find the area of the shaded region in Figure.
We know that each angle of equilateral triangle is 60°. Area of sector OCDE = 60°/360° × πr² = 1/6 × π(6)² = 1/6 × 22/7 × 6 × 6 = 132/7 cm² Area of equilateral triangle OAB = √3/4 (12)² = 36√3 cm² Area of circle = πr² = π(6)² = 22/7 × 6 × 6 = 792/7 cm² Area of shaded region = Area of circle + Area oRead more
We know that each angle of equilateral triangle is 60°.
Area of sector OCDE
= 60°/360° × πr² = 1/6 × π(6)² = 1/6 × 22/7 × 6 × 6 = 132/7 cm²
Area of equilateral triangle OAB
= √3/4 (12)² = 36√3 cm²
Area of circle
= πr² = π(6)² = 22/7 × 6 × 6 = 792/7 cm²
Area of shaded region
= Area of circle + Area of triangle OAB- Area of sector OCDE
= (792/7 + 36√3 – 132/7) cm²
= (36√3 + 660/7) cm²
Here is the video explanation 😃🙌
See less