| Age (in years) | 5-15 | 15-25 | 25-35 | 35- 45 | 45-55 | 55-65 |
| Number of patients | 6 | 11 | 21 | 23 | 14 | 5 |
To find out the modal class, let us the consider the class interval with high frequency.
Here, the greatest frequency = 23, so the modal class = 35 – 45,
Lower limit of modal class = l = 35,
class width (h) = 10,
$f_m$ = 23,
$f_{1}$ = 21 and $f_{2}$ = 14
The formula to find the mode is
| Mode = $l + [ \dfrac{(f_m – f_{1})}{(2f_m – f_{1} – f_{2})}] × h$ |
Substitute the values in the formula, we get
Mode = 35+$[\dfrac{(23-21)}{(46-21-14)}]$×10
= 35 + $\dfrac{20}{11}$
= 35 + 1.8
= 36.8 years
So the mode of the given data = 36.8 years
Calculation of Mean:
First find the midpoint using the formula, $x_{i} = \dfrac{(upper \: limit + lower \: limit)}{2}$
| Class Interval | Frequency $(f_{i})$ | Mid-point $(x_{i})$ | $f_{i}x_{i}$ |
|---|---|---|---|
| 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 |
| Sum $f_i$ = 80 | Sum $f_{i}x_{i}$ = 2830 |
The mean formula is
| Mean =$ \overline{x} = \dfrac{∑f_{i}x_{i} }{∑f_{i}}$ |
= $\dfrac{2830}{80}$
= 35.375 years
Therefore, the mean of the given data = 35.375 years
| Lifetime (in hours) | 0-20 | 20-40 | 40-60 | 60- 80 | 80-100 | 100-120 |
| Frequency | 10 | 35 | 52 | 61 | 38 | 29 |
From the given data the modal class is 60–80.
Lower limit of modal class = l = 60,
The frequencies are:
$f_m$ = 61,$ f_{1}$ = 52, $f_{2}$ = 38 and h = 20
The formula to find the mode is
| Mode =$ l+ [\dfrac{(f_m – f_{1})}{(2f_m – f_{1} – f_{2})}] × h$ |
Substitute the values in the formula, we get
Mode = 60 + $[\dfrac{(61 – 52)}{(122 – 52 – 38)}]$ × 20
Mode = 60 + $[\dfrac{(9 × 20)}{32}]$
Mode = 60 + $(\dfrac{45}{8})$ = 60 + 5.625
Therefore, modal lifetime of the components = 65.625 hours.
| Expenditure (in Rs.) | Number of families |
|---|---|
| 1000-1500 | 24 |
| 1500-2000 | 40 |
| 2000- 2500 | 33 |
| 2500-3000 | 28 |
| 3000-3500 | 30 |
| 3500- 4000 | 22 |
| 4000-4500 | 16 |
| 4500-5000 | 7 |
Given data:
Modal class = 1500-2000,
l = 1500,
Frequencies:
$f_m$ = 40 $f_{1}$ = 24, $f_{2}$ = 33 and
h = 500
Mode formula:
| Mode = $l + [\dfrac{(f_m – f_{1})}{(2f_m – f_{1} – f_{2})}] × h$ |
Substitute the values in the formula, we get
Mode = 1500 + $[\dfrac{(40 – 24)}{(80 – 24 – 33)}]$ × 500
Mode = 1500 + $[\dfrac{(16 × 500)}{23}]$
Mode = 1500 + $\dfrac{8000}{23}$ = 1500 + 347.83
Therefore, modal monthly expenditure of the families = Rupees 1847.83
Calculation for mean:
First find the midpoint using the formula, $x_{i} =\dfrac{(upper \: limit + lower \: limit)}{2}$
Let us assume a mean, (a) be 2750.
| Class Interval | $f_i$ | $x_i$ | $d_{i} = x_{i} – a$ | $u_{i} =\dfrac{ d_{i}}{h}$ | $f_{i}x_{i}$ |
|---|---|---|---|---|---|
| 1000-1500 | 24 | 1250 | -1500 | -3 | -72 |
| 1500-2000 | 40 | 1750 | -1000 | -2 | -80 |
| 2000-2500 | 33 | 2250 | -500 | -1 | -33 |
| 2500-3000 | 28 | 2750 = a | 0 | 0 | 0 |
| 3000-3500 | 30 | 3250 | 500 | 1 | 30 |
| 3500-4000 | 22 | 3750 | 1000 | 2 | 44 |
| 4000-4500 | 16 | 4250 | 1500 | 3 | 48 |
| 4500-5000 | 7 | 4750 | 2000 | 4 | 28 |
| $f_i$=200 | $f_iu_i$ = -35 |
The formula to calculate the mean,
| Mean = $\overline{x} = a +(\dfrac{∑f_iu_i}{∑f_i}) × h$ |
Substitute the values in the given formula
= 2750 + $(\dfrac{-35}{200})$ × 500
= 2750 – 87.50
= 2662.50
So, the mean monthly expenditure of the families = Rs. 2662.50
| No of students per teacher | $\dfrac{Number of states} {U.T}$ |
|---|---|
| 15-20 | 3 |
| 20-25 | 8 |
| 25- 30 | 9 |
| 30-35 | 10 |
| 35-40 | 3 |
| 40-45 | 0 |
| 45- 50 | 0 |
| 50-55 | 2 |
Given data:
Modal class = 30 – 35,
l = 30,
Class width (h) = 5,
$f_m$ = 10, $f_{1}$ = 9 and $f_{2}$ = 3
Mode Formula:| Mode = $l + [\dfrac{(f_m – f_{1})}{(2f_m – f_{1} – f_{2})}] × h$ |
Substitute the values in the given formula
Mode = 30 + $[\dfrac{(10 – 9)}{(20 – 9 – 3)}]$ × 5
= 30 + $(\dfrac{5}{8})$
= 30 + 0.625
= 30.625
Therefore, the mode of the given data = 30.625
Calculation of mean:
Find the midpoint using the formula, $x_{i} =\dfrac{(upper \: limit + lower \: limit)}{2}$
| Class Interval | Frequency $(f_{i})$ | Mid-point $(x_{i})$ | $f_{i}x_{i}$ |
|---|---|---|---|
| 15-20 | 3 | 17.5 | 52.5 |
| 20-25 | 8 | 22.5 | 180.0 |
| 25-30 | 9 | 27.5 | 247.5 |
| 30-35 | 10 | 32.5 | 325.0 |
| 35-40 | 3 | 37.5 | 112.5 |
| 40-45 | 0 | 42.5 | 0 |
| 45-50 | 0 | 47.5 | 0 |
| 50-55 | 2 | 52.5 | 105.0 |
| Sum $f_i$ = 35 | Sum $f_{i}x_{i}$=1022.5 |
| Mean = $\overline{x} = \dfrac{∑f_{i}x_{i}}{∑f_{i}}$ |
= $\dfrac{1022.5}{35}$
= 29.2 (approx)
Therefore, mean = 29.2
| Run Scored | Number of Batsman |
|---|---|
| 3000-4000 | 4 |
| 4000-5000 | 18 |
| 5000- 6000 | 9 |
| 6000-7000 | 7 |
| 7000-8000 | 6 |
| 8000- 9000 | 3 |
| 9000-10000 | 1 |
| 10000-11000 | 1 |
Given data:
Modal class = 4000 – 5000,
l = 4000,
class width (h) = 1000,
$f_m$ = 18, $f_{1}$ = 4 and $f_{2}$ = 9
Mode Formula:
| Mode = $l + [\dfrac{(f_m – f_{1})}{(2f_m – f_{1} – f_{2})}] × h$ |
Substitute the values
Mode = 4000 + $[\dfrac{(18 – 4)}{(36 – 4 – 9)}]$ × 1000
= 4000 + $(\dfrac{14000}{23})$
= 4000 + 608.695
= 4608.695
= 4608.7 (approximately)
Thus, the mode of the given data is 4608.7 runs.
| Number of cars | Frequency |
|---|---|
| 0 -10 | 7 |
| 10-20 | 14 |
| 20-30 | 13 |
| 30- 40 | 12 |
| 40-50 | 20 |
| 50-60 | 11 |
| 60- 70 | 15 |
| 70-80 | 8 |
Given Data:
Modal class = 40 – 50, l = 40,
Class width (h) = 10, $f_m$ = 20, $f_{1}$ = 12 and $f_{2}$ = 11
| Mode =$ l + [\dfrac{(f_m – f_{1})}{(2f_m – f_{1} – f_{2})}] × h$ |
Substitute the values
Mode = 40 + $[\dfrac{(20 – 12)}{(40 – 12 – 11)}]$ × 10
= 40 + $(\dfrac{80}{17})$
= 40 + 4.7
= 44.7
Thus, the mode of the given data is 44.7 cars.