| Monthly consumption(in units) | No. of customers |
|---|---|
| 65-85 | 4 |
| 85-105 | 5 |
| 105-125 | 13 |
| 125- 145 | 20 |
| 145-165 | 14 |
| 165-185 | 8 |
| 185- 205 | 4 |
Find the cumulative frequency of the given data as follows:
| Class Interval | Frequency | 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 |
From the table, it is observed that, N = 68 and hence $\dfrac{N}{2}$=34
Hence, the median class is 125-145 with cumulative frequency = 42
Where, l = 125, N = 68, cf = 22, f = 20, h = 20
Median is calculated as follows:
| Median =$l+\dfrac{(\dfrac{N}{2})-cf}{f}\times h$ |
= 125 + $[\dfrac{(34 − 22)}{20}]$ × 20
= 125 + 12
= 137
Therefore, median = 137
To calculate the mode:
Modal class = 125-145,
$f_m$ or $f_1$ = 20, $f_0$ = 13, $f_2$ = 14 & h = 20
Mode formula:
| Mode = $l+ [\dfrac{(f_1 – f_0)}{(2f_1 – f_0 – f_2)}] × h$ |
Mode = 125 + $[\dfrac{(20 – 13)}{(40 – 13 – 14)}]$ × 20
= 125 + $(\dfrac{140}{13})$
= 125 + 10.77
= 135.77
Therefore, mode = 135.77
Calculate the Mean:
| Class Interval | $f_i$ | $x_i$ | $d_i=x_i-a$ | $u_i=\dfrac {d_i}{h}$ | $f_iu_i$ |
|---|---|---|---|---|---|
| 65-85 | 4 | 75 | -60 | -3 | - 12 |
| 85-105 | 5 | 95 | -40 | -2 | -10 |
| 105- 125 | 13 | 115 | -20 | -1 | -13 |
| 125-145 | 20 | 135 = a | 0 | 0 | 0 |
| 145-165 | 14 | 155 | 20 | 1 | 14 |
| 165 -185 | 8 | 175 | 40 | 2 | 16 |
| 185- 205 | 4 | 195 | 60 | 3 | 12 |
| Sum$ f_i$ = 68 | Sum $ f_iu_i$= 7 |
| $\overline{x} = a + h (\dfrac{∑ f_iu_i }{∑f_i})$ |
= 135 + 20 $(\dfrac{7}{68})$
Mean = 137.05
In this case, mean, median and mode are more/less equal in this distribution.
| Class Interval | Frequency |
|---|---|
| 0- 10 | 5 |
| 10-20 | x |
| 20-30 | 20 |
| 30-40 | 15 |
| 40- 50 | y |
| 50-60 | 5 |
| Total | 60 |
Given data, n = 60
Median of the given data = 28.5
| CI | 0-10 | 10-20 | 20-30 | 30-40 | 40-50 | 50- 60 |
| Frequency | 5 | x | 20 | 15 | y | 5 |
| Cumulati ve frequency | 5 | 5+x | 25+x | 40+x | 40+x+y | 45+x+y |
Where, $\dfrac{N}{2}$ = 30
Median class is 20 – 30 with a cumulative frequency = 25 + x
Lower limit of median class, l = 20,
cf = 5 + x,
f = 20 & h = 10
| Median =$l+\dfrac{(\dfrac{N}{2})-cf}{f}\times h$ |
Substitute the values
28.5 = 20 + $[\dfrac{(30 − 5 − x)}{20}]$ × 10
8.5 = $\dfrac{(25 – x)}{2}$
17 = 25 – x
Therefore, x = 8
Now, from cumulative frequency, we can identify the value of x + y as follows:
Since,
60 = 45 + x + y
Now, substitute the value of x, to find y
60 = 45 + 8 + y
y = 60 – 53
y = 7
Therefore, the value of x = 8 and y = 7
| Age (in years) | Number of policy holder |
|---|---|
| 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 |
| Class interval | Frequency | 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 |
Given data: N = 100 and $\dfrac{N}{2}$ = 50
Median class = 35-40
Then, l = 35, cf = 45, f = 33 & h = 5
| Median =$l+\dfrac{(\dfrac{N}{2})-cf}{f}\times h$ |
Median = 35 + $[\dfrac{(50 – 45)}{33}]$ × 5
= 35 + $(\dfrac{25}{33})$
= 35.76
Therefore, the median age = 35.76 years.
| Length (in mm) | Number of leaves |
|---|---|
| 118-126 | 3 |
| 127-135 | 5 |
| 136-144 | 9 |
| 145- 153 | 12 |
| 154-162 | 5 |
| 163-171 | 4 |
| 172- 180 | 2 |
(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.)
Since the data are not continuous reduce 0.5 in the lower limit and add 0.5 in the upper limit.
| Class Interval | Frequency | 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 |
So, the data obtained are:
N = 40 and $\dfrac{N}{2}$ = 20
Median class = 144.5-153.5
then, l = 144.5,
cf = 17, f = 12 & h = 9
| Median =$l+\dfrac{(\dfrac{N}{2})-cf}{f}\times h$ |
Median = 144.5 +$ [\dfrac{(20 – 17)}{12}]$ × 9
= 144.5 + $(\dfrac{9}{4})$
= 146.75 mm
Therefore, the median length of the leaves = 146.75 mm.
| Lifetime (in hours) | Number of lamps |
|---|---|
| 1500-2000 | 14 |
| 2000-2500 | 56 |
| 2500- 3000 | 60 |
| 3000-3500 | 86 |
| 3500-4000 | 74 |
| 4000- 4500 | 62 |
| 4500-5000 | 48 |
| Class Interval | Frequency | Cumulative |
|---|---|---|
| 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 |
Data:
N = 400 & $\dfrac{N}{2}$ = 200
Median class = 3000 – 3500
Therefore, l = 3000, cf = 130,
f = 86 & h = 500
| Median =$l+\dfrac{(\dfrac{N}{2})-cf}{f}\times h$ |
Median = 3000 + $[\dfrac{(200 – 130)}{86}]$ × 500
= 3000 + $(\dfrac{35000}{86})$
= 3000 + 406.98
= 3406.98
Therefore, the median lifetime of the lamps = 3406.98 hours
| Number of letters | 1-4 | 4-7 | 7-10 | 10- 13 | 13-16 | 16-19 |
| Number of surnames | 6 | 30 | 40 | 16 | 4 | 4 |
To calculate median:
| Class Interval | Frequency | 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 |
Given:
N = 100 & $\dfrac{N}{2}$ = 50
Median class = 7-10
Therefore, l = 7, cf = 36, f = 40 & h = 3
| Median =$l+\dfrac{(\dfrac{N}{2})-cf}{f}\times h$ |
Median = 7 + $[\dfrac{(50 – 36)}{40}]$ × 3
Median = 7 + (\dfrac{42}{40})
Median = 8.05
Calculate the Mode:
Modal class = 7-10,
Where, l = 7, $f_1$ = 40, $f_0$ = 30, $f_2$ = 16 & h = 3
Mode = 7 + $[\dfrac{(40 – 30)}{(2 × 40 – 30 – 16)}]$ × 3
= 7 + $(\dfrac{30}{34})$
= 7.88
Therefore mode = 7.88
Calculate the Mean:
| Class Interval | $f_i$ | $x_i$ | $f_ix_i$ |
|---|---|---|---|
| 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 | 58 |
| 16-19 | 4 | 17.5 | 70 |
| Sum $f_i$ = 100 | Sum $f_ix_i$ = 832 |
Mean = $\overline{x} = \dfrac{∑f_i x_i}{∑f_i}$
Mean = $\dfrac{832}{100}$ = 8.32
Therefore, mean = 8.32
| Weight(in kg) | 40-45 | 45-50 | 50-55 | 55- 60 | 60-65 | 65-70 | 70-75 |
| Number of students | 2 | 3 | 8 | 6 | 6 | 3 | 2 |
| Class Interval | Frequency | 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 |
Given: N = 30 and $\dfrac{N}{2}$= 15
Median class = 55-60
l = 55, Cf = 13, f = 6 & h = 5
| Median =$l+\dfrac{(\dfrac{N}{2})-cf}{f}\times h$ |
Median = 55 + $[\dfrac{(15 – 13)}{6}]$ × 5
= 55 + $(\dfrac{10}{6})$
= 55 + 1.666
= 56.67
Therefore, the median weight of the students = 56.67