Frequency distribution table

(i) Mode is 19.
(ii) If 19 is changed to 1 6. then frequency of 16 is 5 and frequency of 19 becomes 1i.e.,16 has maximum frequency
\(\Rightarrow\) 16 is the mode.

Median class : 40 - 50 \(\Rightarrow \) Lower Limit =40
Modal class   : 40 - 50  \(\Rightarrow \) Upper Limit = 50
Their sum = 40+50=90

Given, \( \sum { f_{ i }u_{ i }=30 } \), \(\sum { f_{ i }=40 } \) and \(u_{ i }=\frac { x_{ i }-20 }{ 10 } \)
We know that, \(u_{ i }=\frac { x_{ i }-a }{ h } \)
On comparing, we get a=20, h=10
\(\therefore \quad \overline { x } =a+\left\{ \frac { \sum { f_{ i }u_{ i } } }{ \sum { f_{ i } } } \right\} \times h=20+\left\{ \frac { 30 }{ 40 } \right\} \times 10\\ =20+\frac { 30 }{ 4 } =\frac { 80+30 }{ 4 } =\frac { 110 }{ 4 } =27.5\)

Here, number of terms, N=4n, whch is an even number.
Median=Mean of \(\left( \frac { n }{ 2 } \right) \) th term and \(\left( \frac { n }{ 2 } +1 \right) \) th term
i.e. Median=Mean of (2n)th term and (2n+1)th term.

Let a = 413,h = 25

\(\bar{x}=a+{ {\sum { { f }_{ i } }{u}_{i} }\over{ \sum { { f }_{ i } } } }\times h=413+{ { (-79)}\over{ 50} }\times25=413-{ { 79}\over{ 2} }=413-39.5\) = 373.5

\(\overset { - }{ x } =a+\frac { \Sigma f_{ i }u_{ i } }{ \Sigma f_{ i } } \times h\)
\(=35+\frac { -44 }{ 65 } \times 10=35-6.76=28.24\)

Here, maximum frequency = 9, hence modal class is 60 - 90
Mode= L+h  \(\left( \frac { f_{ 1 }-f_{ 0 } }{ 2f_{ 1 }-f_{ 0 }-f_{ 2 } } \right) \)
Here, - 60,f₁ = 9,f₀ = 6,f₂= 6 and h = 30.
Mode = 60+30 \(\left( \frac { 9-6 }{ 2\times 9-6-6 } \right) \)
= \(60+\frac { 30\times 3 }{ 6 } =60+15\)
=75

We construct a cumulative frequency distribution of the less than type as follows:

Taking upper class limits of class intervals on x-axis and their respective frequencies on y-axis, its ogive can be drawn as follows.

Ans.138.57 cm

Cumulative frequency distribution of the less than type
Here, the number of students who have scored marks less than 10 are 5. The number of students who have scored marks less than 20 includes the number of students who have scored marks from 0-10 as well as the number of students who have scored marks from 10-20.
Thus, the total number of students with marks less than 20 is 5 + 7, i.e. 12. So, the cumulative frequency of the class 10-20 is 12.
Similarly, on computing the cumulative frequencies of the other classes, i.e. the number of students with marks less than 30, less than 40, ... less than 100, we get the distribution which is called the cumulative frequency distribution of the less than type.

Here, 10,20,30, ... , 100 are the upper limits of the respective class intervals.
Cumulative frequency distribution of the more than type
For this type of distribution, we make the table for the number of students with scores, more than or equal to 0, more than or equal to 10, more than or equal to 20 and so on. From the question, we observed that all 58 students have scored marks more than or equal to O.
There are 5 students scoring marks in the interval 0-10, it shows that there are 58 - 5 = 53 students getting more than or equal to 10 marks. In the same manner, the number of students scoring 20 marks or above = 53 - 7 = 46 students, scoring 30 or above = 46 - 4 = 42 students and so on.
Similarly, computing the cumulative frequencies of the other classes, i.e. the number of students with marks more than or equal to 40, more than or equal to 50, ... , we get the distribution which is called the cumulative frequency distribution of the more than type.

Here, 0, 10,20,30, ... ,90 are the lower limits of the respective class intervals.

Benefits of regular medical check-up are
(t) It enables us to take pre-medical action on time.
(it) It reduces the chances of getting sick.

Ans. 50

 Here, \(\frac { n }{ 2 } =\frac { 25 }{ 2 } =12.5\)
Since, cumulative frequency just greater than 12.5 and corresponding class is 60-70.
Median class=60-70
Since, highest frequency is 9 and corresponding class is 50-60
Modal class=50-60

(a) Modal class is the class interval in a frequency distribution table which has the highest frequency. According to the given data, the highest frequency is 132 and the class interval corresponding to it is 20-24.
Thus, the modal class of the above data is 20-24.
Hence, lower limit of the modal class is 20.
(b) (i) Convert the given data as continuous classes and calculate the cumulative frequency.

Here, N = 365 (odd)
\(\frac{N}{2}=\frac{365}{2}=182.5\)
Cumulative frequency greater than and nearer to 182.5 is 194 which belongs to the class interval 19.5-24.5
\(\therefore\) The median class of the above data is 19.5 - 24.5.
(ii) Number of participants of age less than 50 yr who undergo vocational training
= 62 + 132 + 96 + 37 + 13 + 11 + 10 = 361
Therefore, required number of participants is 361.
(c) The empirical relationship between mean, median and mode is given by
Mode = 3 Median - 2 Mean

(i)  (a)
In \(\triangle A B C \text { and } \triangle A D E,\) 
\(\angle A=\angle A\) (common)
\(\angle B=\angle D\)  (coorresponding angles)
\(\therefore \quad \Delta A B C \sim \Delta A D E\)  (By AA similarity criteria)
(ii) (b)
We have, \(A B=2 B C \Rightarrow B C=\frac{20}{2}=10 \mathrm{~m}\) 
So,height of ship = 10m
(iii) (c)
We have, AD=12 BC
\(\Rightarrow A D=12 \times 10=120 \mathrm{~m}\) 
\(\therefore \triangle A B C \sim \triangle A D E\) 
\(\therefore \frac{A B}{A D}=\frac{B C}{D E}\) 
\(\Rightarrow \frac{B C}{D E}=\frac{20}{120}=\frac{1}{6}\) 
So,ratio of height of ship and light house is 1:6
(iv) (a) 
Since,\(\triangle A B C \sim \triangle A D E\) 
\(\Rightarrow \frac{A B}{A D}=\frac{A C}{A E}\) 
\(\Rightarrow \frac{A E}{A C}-1=6-1 \Rightarrow \frac{A E-A C}{A C}=5 \Rightarrow \frac{E C}{A C}=5\) 
 \(\Rightarrow \frac{A C}{E C}=\frac{1}{5}\) 

\(\therefore\) Required ratio = 1 :5
(v) (b) 
Height of lighthouse = DE
Now, \(\frac{B C}{D E}=\frac{1}{6}\) 
\(\Rightarrow D E=6 B C\) 
= 6 X 10 = 60m

(i) (c): 

\(\text { Mean }=\frac{1720}{40}\)
= 43
(ii) (c): 60
(iii) (b): Median
(iv) (c): Median class 40 - 60, Modal class = 40 - 60
Sum of lower limits of median class and modal class = 40 + 40 = 80
(v) (c): Number of students are = 8 + 10 + 13
= 31

(i) (d): We know that,
\(\text { Class mark }=\frac{\text { Lower limit }+\text { Upper limit }}{2} \)
\(\Rightarrow m=\frac{\text { Lower limit }+b}{2} \Rightarrow \text { Lower limit }=2 m-b\)
(ii) (a):

\(\begin{aligned} &\therefore \quad \text { Average lifetime of a packet }\\ &=A+\frac{\sum f_{i} d_{i}}{\sum f_{i}}=325+\frac{6400}{400}=341 \mathrm{hrs} \end{aligned}\)

(iii) (b) : \(\text { Here, } N=400 \Rightarrow \frac{N}{2}=200\)
Also, cumulative frequency for the given distribution are 14, 70, 130,216,290,352,400
\(\therefore\) c.f just greater than 200 is 216, which is
corresponding to the interval 300-350.
l= 300, f=86, c.f = 130, h = 50
\(\therefore \quad \text { Median }=l+\left(\frac{\frac{N}{2}-c . f .}{f}\right) \times h=300+\left(\frac{200-130}{86}\right) \times 50\)
= 300 + 40.697 = 340.697 ""340 hrs (approx.)
(iv) (a) : We know that Mode = 3 Median - 2 Mean
= 3(340.697) -2(341)
= 1022.091 - 682 = 340.091 ""340 hrs
(v) (c): Since, minimum of mean, median and mode is approximately 340 hrs. So, manufacturer should claim that lifetime of a packet is 340 hrs.

We have the following table:

(i) (c): Here, it is given that total frequency = 100
\(\therefore\) 76 + x + y = 100 \(\Rightarrow\) x + y = 24
(ii) (c): Here \(\frac{N}{2}=\frac{100}{2}=50\)
Also, median = 525
\(\therefore\) Median class is 500-600.
\(\text { Now, median }=l+\left(\frac{N / 2-c . f .}{f}\right) \times h \)
\(\Rightarrow 525=500+\left(\frac{50-(36+x)}{20}\right) \times 100 \)
\(\Rightarrow 5=50-36-x \Rightarrow x=9\)
(iii) (b) : Since, maximum frequency is 20, so modal class is 500 - 600. Hence, upper limit of modal class is 600.
(iv) (b) : Since, x + y = 24 \(\Rightarrow\) y = 24 - 9 = 15
Required average consumption
\(\begin{aligned} & 50 \times 2+150 \times 5+250 \times 9+350 \times 12+450 \times 17 \\ =& \frac{+550 \times 20+650 \times 15+750 \times 9+850 \times 7+950 \times 4}{100} \\ =& \frac{52200}{100}=522 \mathrm{kwh} \end{aligned}\)
(v) (d)