Book Online Demo

Check Your IQ

Try Test

Courses

Dropper NEET Course Dropper JEE Course Class - 12 NEET Course Class - 12 JEE Course Class - 11 NEET Course Class - 11 JEE Course Class - 10 Foundation NEET Course Class - 10 Foundation JEE Course Class - 10 CBSE Course Class - 9 Foundation NEET Course Class - 9 Foundation JEE Course Class -9 CBSE Course Class - 8 CBSE Course Class - 7 CBSE Course Class - 6 CBSE Course

Offline Centres

Q.

The following distribution gives the state-wise teacher student ratio in higher secondary schools of India. What is the mode and mean of this data?

see full answer

Start JEE / NEET / Foundation preparation at rupees 99/day !!

21% of IItians & 23% of AIIMS delhi doctors are from Sri Chaitanya institute !!

An Intiative by Sri Chaitanya

a

Mode = 31.7, Mean = 28.2

b

Mode = 33.6, Mean = 25.3

c

Mode = 35.7, Mean = 26.3

d

Mode = 30.6, Mean = 29.2

answer is D.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

It is given the state-wise teacher student ratio in higher secondary schools of India.
Question Image

Question Image

We convert the data in terms of the following expression.

x_{i} = \frac{lower limit + upper limit}{2}

Number of students	$x_{i}$	Number of state $(f_{i})$	${(x}_{i}) \times (f_{i}) = (x_{i} f_{i})$
15 - 20	17.5	3	52.5
20 - 25	22.5	8	180
25 - 30	27.5	9	247.5
30 - 35	32.5	10	325
35 - 40	37.5	3	112.5
40 - 45	42.5	0	0
45 - 50	47.5	0	0
50 - 55	52.5	2	105
		$Σ f_{i} = 35$	$Σ f_{i} x_{i} = 1022.5$

From the table, we get

Σ f_{i} = 35

Σ f_{i} x_{i} = 1022.5

We know that,

Mean = \frac{Σ x_{i} f_{i}}{Σ f_{i}}

\Rightarrow Mean = \frac{1022.5}{35}

\Rightarrow Mean = 29.2

We also know that,
Mode

= l + \frac{f - f_{1}}{2 f - f_{1} - f_{2}} \times h

Where:

l \to

lower limit of the modal class

f \to

frequency of the modal class

f_{1} \to

frequency of the class preceding the modal class

f_{2} \to

frequency of the class succeeding the modal class

h \to

class width
From the given table, we get that the maximum frequency is

10

and the corresponding class is

30 - 35

. Therefore,

30 - 35

is the modal class.
So, we have

l = 30, h = 5, f = 10, f_{1} = 9

and

f_{2} = 3

Putting the values in the formula, we get

Mode = 30 + \frac{10 - 9}{20 - 9 - 3} \times 5

\Rightarrow Mode = 30 + \frac{1}{8} \times 5

\Rightarrow Mode = 30 + \frac{5}{8}

\Rightarrow Mode = 30 + 0.625

\Rightarrow Mode = 32.625

Therefore, the mode of the given frequency distribution is 32.625 and the mean is 29.2.
Hence, the correct option is 4.

Watch 3-min video & get full concept clarity

Related Blogs

Get ₹ 1 crore* worth scholarship for JEE / NEET / Foundation

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET

How to Join IIT after 10th

Get Expert Academic Guidance – Connect with a Counselor Today!

Related Blogs

Get ₹ 1 crore* worth scholarship for JEE / NEET / Foundation

Result Producing online coaching for JEE/NEET of south India

Largest All India Test Series for JEE/NEET

Best Books for IIT JEE

Best Books for NEET

How to Join IIT after 10th

Not sure what to do in the future? Don’t worry! We have a FREE career guidance session just for you!