CBSE 10th Maths - Statistics- Exercise 14.2

Solution:

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

Solution:

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.

Solution:

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

Solution:

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

Solution:

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.

Solution:

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.