The cumulative frequency with their respective class intervals are as follows. Cumulative frequency just greater than n/2 (i.e. 30/2 = 15) is 19, belonging to class interval 55 - 60. Median class = 55 - 60 Lower limit (l) of median class = 55 Frequency (f) of median class = 6 Cumulative frequency (cRead more
The cumulative frequency with their respective class intervals are as follows.
Cumulative frequency just greater than n/2 (i.e. 30/2 = 15) is 19, belonging to class interval 55 – 60.
Median class = 55 – 60
Lower limit (l) of median class = 55
Frequency (f) of median class = 6
Cumulative frequency (cf) of median class = 13
Class size (h) = 5
Median = l + ((n/2 – cf)/f) × h = 55 + ((15 – 13)/6) × 5 = 55 + 10/6 = 55 + 1.67 = 56.67.
Therefore, median weight is 56.67 kg.
The cumulative frequency for the given data is calculated as follows. From the table, it can be observed that n = 60 45 + x + y = 60 or x + y = 15 ...(1) Median of thr data is given as 28.5 which lies in interval 20 - 30. Therefore, median class = 20 - 30 Lower limit (l) of median class = 20 CumulatRead more
The cumulative frequency for the given data is calculated as follows.
From the table, it can be observed that n = 60
45 + x + y = 60 or x + y = 15 …(1)
Median of thr data is given as 28.5 which lies in interval 20 – 30.
Therefore, median class = 20 – 30
Lower limit (l) of median class = 20
Cumulative frequency (cf) of class preceding the median class = 5 + x
Frequency (f) of median class = 20
Class size (h) = 10
Median = l + ((n/2 – cf)/(f)) × h
⇒ 28.5 = 20 + {(60/2 – (5 + x))/20} × 10
⇒ 8.5 = (25 – x)/2
⇒ 17 = 25 – x
⇒ x = 8
From equation (1), 8 + y = 15 ⇒ y = 7
Hence, the value of x and y are 8 and 7 respectively.
The given data does not have continuous class intervals. It can be observed that the difference between two class intervals is 1. Therefore, 1/2 = 0.5 has to be added and subtracted to upper class limits and lower-class limits respectively. Continuous class intervals with respective cumulative frequRead more
The given data does not have continuous class intervals. It can be observed that the difference between two class intervals is 1. Therefore, 1/2 = 0.5 has to be added and subtracted to upper class limits and lower-class limits respectively.
Continuous class intervals with respective cumulative frequencies he represented follows.
From the table, it can be observed that the cumulative frequency just greater than n/2 (i.e. 40/2 = 20) is 29, belonging to class interval 144.5 – 153.5.
Median class = 144.5 – 153.5
Lower Limit (l) of median class = 144.5 and class size (h) = 9
Frequency (f) of median class 12
Cumulative frequency (cf) of class preceding median class = 17
Median = l + (n/2 – cf)/f) × h = 144.5 + ((20 – 17)/12) × 9 = 144.5 + 2.25 = 146.75
Therefore, median length of leaves is 146.75mm.
See this explanation video for better understanding✌😃
The cumulative frequencies with their respective class intervals are as follows. It can be observed that the cumulative frequency just greater than n/2 (i.e.400/2 = 200) is 216, belonging to class interval 3000 - 3500. Median class = 3000 - 3500 Lower limt (l) of median class = 3000 Frequency (f) ofRead more
The cumulative frequencies with their respective class intervals are as follows.
It can be observed that the cumulative frequency just greater than n/2 (i.e.400/2 = 200) is 216, belonging to class interval 3000 – 3500.
Median class = 3000 – 3500
Lower limt (l) of median class = 3000
Frequency (f) of median class = 86
Cumulative frequency (cf) of class preceding median class = 130
Class size = 500
Median = l + ((n/2 – cf)/f) × h = 3000 + ((200 – 130)/86) × 500 = 3000 + 35000/86 = 3000 + 406.978 + 3406.978
Therefore, median lifetime of lamps is 3406.98 hours.
To find the class marks, the following relation is used. xᵢ = (Upper limit + Lower limit)/2 Taking 135 as assumed mean (a). dᵢ, uᵢ, fᵢuᵢ are calculated according to step deviation method as follows. From the table, we obtain. ∑fᵢ = 68, ∑fᵢuᵢ = 7, a = 135 and h = 20 mean (X̄) = a + (∑fᵢuᵢ / ∑fᵢ)h = 1Read more
To find the class marks, the following relation is used.
xᵢ = (Upper limit + Lower limit)/2
Taking 135 as assumed mean (a). dᵢ, uᵢ, fᵢuᵢ are calculated according to step deviation method as follows.
From the table, we obtain.
∑fᵢ = 68, ∑fᵢuᵢ = 7, a = 135 and h = 20
mean (X̄) = a + (∑fᵢuᵢ / ∑fᵢ)h = 135 + (7/68) × 20 = 135 + 2.058 = 137.058
From the table, it can be observed that the maximum class frequency is 20, belonging to class interval 125 – 145.
Modal class 125 – 145
Lower limit (l) of modal class = 125
Class size (h) = 20
Frequency (f₁) of modal class = 20
Frequency (f₀) of class preceding modal class = 13
Frequency (f₂) of class succeeding the modal class = 14
Mode = l + (f₁ – f₀ / 2f₁ – f₀ – f₂)h = 125 × (20 – 13 / 2 × 20 – 13 – 14) × 20 = 125 + 7/13 × 20 = 125 + 10.76 = 135.76
To find the median of the given data, cumulative frequency is calculated as follows.
Monthly consumption (in units) Number of consumers Cumulative frequency
65 – 85 4 4
85 – 105 5 4 + 5 = 9
105 – 125 13 9 + 13 = 22
125 – 145 20 22 + 20 = 42
145 – 165 14 42 + 14 = 56
165 – 185 8 56 + 8 = 64
185 – 205 4 64 + 4 = 68
From the table, we obtain n = 18
Cumulative frequency (cf) just greater than n/2 (i.e, 68/2 = 34) is 42, belonging to interval 125 – 145.
Therefore, median class = 125 – 145
Lower limit (l) of median class = 125 and class size (h) = 20
Frequency (f) of median class = 20
Cumulative frequency (cf) of class preceding median class = 22
Median = l + ((n/2 – cf)/f) × h = 125 + (34 – 22 / 20) × 20 = 125 + 12 = 137
Therefore, median, mode, mean of the given data is 137, 135.76, and 137.05 respectively.
The three measures are approximately the same in this case.
The distribution below gives the weights of 30 students of a class.
The cumulative frequency with their respective class intervals are as follows. Cumulative frequency just greater than n/2 (i.e. 30/2 = 15) is 19, belonging to class interval 55 - 60. Median class = 55 - 60 Lower limit (l) of median class = 55 Frequency (f) of median class = 6 Cumulative frequency (cRead more
The cumulative frequency with their respective class intervals are as follows.
Cumulative frequency just greater than n/2 (i.e. 30/2 = 15) is 19, belonging to class interval 55 – 60.
See lessMedian class = 55 – 60
Lower limit (l) of median class = 55
Frequency (f) of median class = 6
Cumulative frequency (cf) of median class = 13
Class size (h) = 5
Median = l + ((n/2 – cf)/f) × h = 55 + ((15 – 13)/6) × 5 = 55 + 10/6 = 55 + 1.67 = 56.67.
Therefore, median weight is 56.67 kg.
If the median of the distribution given below is 28.5, find the values of x and y.
The cumulative frequency for the given data is calculated as follows. From the table, it can be observed that n = 60 45 + x + y = 60 or x + y = 15 ...(1) Median of thr data is given as 28.5 which lies in interval 20 - 30. Therefore, median class = 20 - 30 Lower limit (l) of median class = 20 CumulatRead more
The cumulative frequency for the given data is calculated as follows.
From the table, it can be observed that n = 60
See less45 + x + y = 60 or x + y = 15 …(1)
Median of thr data is given as 28.5 which lies in interval 20 – 30.
Therefore, median class = 20 – 30
Lower limit (l) of median class = 20
Cumulative frequency (cf) of class preceding the median class = 5 + x
Frequency (f) of median class = 20
Class size (h) = 10
Median = l + ((n/2 – cf)/(f)) × h
⇒ 28.5 = 20 + {(60/2 – (5 + x))/20} × 10
⇒ 8.5 = (25 – x)/2
⇒ 17 = 25 – x
⇒ x = 8
From equation (1), 8 + y = 15 ⇒ y = 7
Hence, the value of x and y are 8 and 7 respectively.
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 :
The given data does not have continuous class intervals. It can be observed that the difference between two class intervals is 1. Therefore, 1/2 = 0.5 has to be added and subtracted to upper class limits and lower-class limits respectively. Continuous class intervals with respective cumulative frequRead more
The given data does not have continuous class intervals. It can be observed that the difference between two class intervals is 1. Therefore, 1/2 = 0.5 has to be added and subtracted to upper class limits and lower-class limits respectively.
Continuous class intervals with respective cumulative frequencies he represented follows.
From the table, it can be observed that the cumulative frequency just greater than n/2 (i.e. 40/2 = 20) is 29, belonging to class interval 144.5 – 153.5.
Median class = 144.5 – 153.5
Lower Limit (l) of median class = 144.5 and class size (h) = 9
Frequency (f) of median class 12
Cumulative frequency (cf) of class preceding median class = 17
Median = l + (n/2 – cf)/f) × h = 144.5 + ((20 – 17)/12) × 9 = 144.5 + 2.25 = 146.75
Therefore, median length of leaves is 146.75mm.
See this explanation video for better understanding✌😃
See lessThe following table gives the distribution of the life time of 400 neon lamps:
The cumulative frequencies with their respective class intervals are as follows. It can be observed that the cumulative frequency just greater than n/2 (i.e.400/2 = 200) is 216, belonging to class interval 3000 - 3500. Median class = 3000 - 3500 Lower limt (l) of median class = 3000 Frequency (f) ofRead more
The cumulative frequencies with their respective class intervals are as follows.
It can be observed that the cumulative frequency just greater than n/2 (i.e.400/2 = 200) is 216, belonging to class interval 3000 – 3500.
See lessMedian class = 3000 – 3500
Lower limt (l) of median class = 3000
Frequency (f) of median class = 86
Cumulative frequency (cf) of class preceding median class = 130
Class size = 500
Median = l + ((n/2 – cf)/f) × h = 3000 + ((200 – 130)/86) × 500 = 3000 + 35000/86 = 3000 + 406.978 + 3406.978
Therefore, median lifetime of lamps is 3406.98 hours.
The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality.
To find the class marks, the following relation is used. xᵢ = (Upper limit + Lower limit)/2 Taking 135 as assumed mean (a). dᵢ, uᵢ, fᵢuᵢ are calculated according to step deviation method as follows. From the table, we obtain. ∑fᵢ = 68, ∑fᵢuᵢ = 7, a = 135 and h = 20 mean (X̄) = a + (∑fᵢuᵢ / ∑fᵢ)h = 1Read more
To find the class marks, the following relation is used.
xᵢ = (Upper limit + Lower limit)/2
Taking 135 as assumed mean (a). dᵢ, uᵢ, fᵢuᵢ are calculated according to step deviation method as follows.
From the table, we obtain.
See less∑fᵢ = 68, ∑fᵢuᵢ = 7, a = 135 and h = 20
mean (X̄) = a + (∑fᵢuᵢ / ∑fᵢ)h = 135 + (7/68) × 20 = 135 + 2.058 = 137.058
From the table, it can be observed that the maximum class frequency is 20, belonging to class interval 125 – 145.
Modal class 125 – 145
Lower limit (l) of modal class = 125
Class size (h) = 20
Frequency (f₁) of modal class = 20
Frequency (f₀) of class preceding modal class = 13
Frequency (f₂) of class succeeding the modal class = 14
Mode = l + (f₁ – f₀ / 2f₁ – f₀ – f₂)h = 125 × (20 – 13 / 2 × 20 – 13 – 14) × 20 = 125 + 7/13 × 20 = 125 + 10.76 = 135.76
To find the median of the given data, cumulative frequency is calculated as follows.
Monthly consumption (in units) Number of consumers Cumulative frequency
65 – 85 4 4
85 – 105 5 4 + 5 = 9
105 – 125 13 9 + 13 = 22
125 – 145 20 22 + 20 = 42
145 – 165 14 42 + 14 = 56
165 – 185 8 56 + 8 = 64
185 – 205 4 64 + 4 = 68
From the table, we obtain n = 18
Cumulative frequency (cf) just greater than n/2 (i.e, 68/2 = 34) is 42, belonging to interval 125 – 145.
Therefore, median class = 125 – 145
Lower limit (l) of median class = 125 and class size (h) = 20
Frequency (f) of median class = 20
Cumulative frequency (cf) of class preceding median class = 22
Median = l + ((n/2 – cf)/f) × h = 125 + (34 – 22 / 20) × 20 = 125 + 12 = 137
Therefore, median, mode, mean of the given data is 137, 135.76, and 137.05 respectively.
The three measures are approximately the same in this case.