Kerala Plus One Answer Key 2025 Statistics

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ANSWER KEY
FIRT YEAR HIGHER SECONDARY EXAMINATION - MARCH 2025
P 4 <.T I I, SUBJECT: STATISTICS CoDE No: FY 332
Qn Sub Total
Answer key/ value points Score
No Ons Score
1 (c) ISI 1 1

2 (b) I is correct and II is not correct 1 1

3 (b) \4ec!ian I I
4 (!|l only I I
5 (a)Positively skewed 1 I
6 (c) Both cost and time 1 I
7 At least two relevant points 2xl=2 2
8 Any 4 methods 4xl=2 2
Class 0-10 10 - 20 20-30 30-40 40-50 Total
Frequency J 7 l0 2 3 25
9 ,7 2 2
Percentage 1x100=12 l0 2
1x100=12 100
Frequency 25 25x100=28 25x100=40 6x100=8 25

l0 Histogram, Ogives 2xl=2 2
t+%
ll Fr=4=3
p; 2
The distribution is meso lortic %
t2 Any two nonprobability sampling methods 2xl=2 2
l3
Job ofparents
Category Self Govemment Private sector Tolal
employed Employees employee
Boy 6 94 30 130 3 J
Girl 180 40 270 490
Total 186 134 300 620
(Give.full score.for correct
OR for correct structure q[lgble give 2 score)
t4 (a) (ii) 2 only OR (iil) and2 I 1
(b) Total weight of all the boys : 100 x 52 = 5200 %
Corrected total weight : 5200-15 + 50 = 5235 3
%
5235
Corrected averase
- - 100 =52.35 L
15 (a) (iv) None of these 1

(b)
0, = 15, Qt=50 %
J
gp=Q,-Q,=50-15=r7.5
-22 7+%

t6 8t-q =15, Qr+Q, =35, Qr=29 1

S* =Qr+-Q,-2Q, -35-2x20 = -0.33 J
" Qr-Q, 15
l+1

t7 (a) (iii) 2 only 1
(b) of Equally likely events
D efi n ition/Explanation/Examp I e 1 J
Definition/Explanatior/Example of Mutually exclusive events L

I

l8 Definition /Explanation of simple random sampling 2
J
Lottery method, Random number table method Yz+ %
t9 (a) (i) Enumerator I
(b) 4
Any 3 difference between questionnaire and schedule 3xl=3
20 Drawing Histogram 4
(Drawing histogram with space between bars - give 2 score) 4
2t Ar - A white ball is transferred from first to second box
Az - A black ball is transferred from first to second box
A - A white ball is drawn from the second box.

P(A) =
%, r<,n l = %, r(%,)= %r, r(/r)= %, 2

PQa)xP 4
,(^,/^)=
P(A,)x P A/ + P(,\)x P I
/A, Al
4/ -5
== -'6^--13 - =I=0.625
"'"--
2,A"9,ir*+,u"11;,, 8
1

22
SD
rl00= 0'0134x100=0.14
For StudentA,Cl, =Me.an 9.8 1+l
For Student B, CV =o'o592xloo=0.6 4
9.822
CV is less for student A. So student A is more stable in measuring

( a) (ii) GM I
(b) I _;1-l
\
HM L; I
t t(t l\ 0.05+0.025 0.075
_, _i_ t-_
HM 2\20 40) 2 2
I
4
1
.'.HM= ' =26.67 1
0.07s
The average speed: 26.67 krn/hr
(Give full score forJinding HM using any other
formula or procedure)
(Give I scoreforfinding AM)
24 Wages (t) Below 250 - 500 - 750 - 1000- 1250 -
250 500 750 1000 1250 I 500
l0 30 40 25 20 l5 I
Cum. l0 40 80 105 125 140

r/ = 140,
I:ro Median class is 500 - 750 t/
/2

N 4
L
2
Median = I + I
"f
(lo- qo\"zso
r
-(QQrr =68?.5 7+t/z
40

2

(a)
(u) _I
25 1

"26
(b) P(AU B) - r (t)+ r(B)- P (Ar B) I
0.7 = P(A)+r(,1)-o.t
%
P(A) = 0.4
%
6
(c) 35 I
l8c,oo 816 on o.o+:
P(All are blue): 'C,

8C'x3C x7C' J-
P(All are different colour) -
18C, 6p 204 oR 0.0343
1

P(2 red and I white balls):
t!,."'q oR l-oR 0.103 I
lScr 68
26 (a)
nr=25,it =120, nr=30, x, =135 I

- nr1, + nri. I
\+n"
25x120+30x135 I
= 128.18
55
(b) 6
ModalClass is 24-26 %

c
Mode = I I
z.fr-.fo- f,
22-13 x2
:24+0.857 =24.86 l+%
2x22-13-10
,

27 (a) (1) 24 I
(b)
x f fx f,'
5 5 25 125

l5 8 120 1 800

25 t2 300 7500
2
35 9 315 I 1025

45 6 270 12150 6

T 40 1030 32600

SD(o) =
N
Jx-
(+)' I

'32600
__t_ / t o:o
l+l
40 [+o )' = Jrsr.s3zs =1233.

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