Kerala Plus Two Answer Key 2025 Statistics

ANWSER KEY
SECOND YEAR IIIGHER SECONDARY EXAMINATION - MARCH 2025
CODE No: SY 432 PART III SUBJECT: STATISTICS SCORES: 60

Qn Sub Total
No Answer key/ value points Score
Ons Score
I (d) Income and Expenditure OR ( c) and distance I I
2 (b) Nature I I
3 (a\ 2 I I
4 (d) AII of the above OR (a) consistent OR(b) sufficient OR(c) efticeint I I
5 (b) '-" * c I I
-11
6 (b) (ii) istrue I I
7 i+ /= -3 (l)

(l)r (2) = 6F= -le I =- -t9 I
= 6 2

(2\-s x(l)+ -6V=-t - Y:T
1
I
8 dY
dx
=6xs+r44x2+24 2 2

9 I
ZP@) =t
0.2+0.3+ k+ 0.3 = 1 2
= k =0.2 I
2 score for of k without
the value of
l0 (i) -(2), (ii)- (3), (iii)-(4), (iv)- (l) 4"i 2

ll Normality, Homogeneity, Additivity, Independence 4"t 2

t2 The moment estimate ofpopulation mean = Sample Mean I
2
35+45+40+ 42+39+55+63 2
= 45.57 _l
7 I2
(Estimating Population mean without using formula may also consider and sive 2 score)
l3
!x = 105,1 y =173,2*' =2575,2y' =729l,>,xy = 34g5 I
,lw-(l*XIy)
I

5x3485-105x173
3
zszs -(ros)' zzsr - (r zr)'
/(s, ), (s, )
I
Cov(x' y)
(Also consider the formula , -
6r\oy
Cov(x,y)=-29.6, o, =8.6, o"=l6.2and r=4)1)
14 3xl
1 score each for each definition 3
r5
Source I d,f s.s M.S.S F F o.or
Between 16 466;8 29.18 4.63 3.62
Error l4 88.2 6.3 6"* 3
Total 30 555
score value for atleast one correct

1

l6 1
Zpo=Zo8,ltr=227
: 227 t+|+f 3
Srmple Aggregate Index Number x 100 = xl00 = 109.13
* 208
t7 x I 2 3 4 5 6
0.1 0.3 0.1 0.2 0.1
0.1 0.4 0.5 0.7 0.8 1
3
(Give score 3

18 (l P(-1.2<Z <1.2) =2x0.3M9:0.7698 I
(ii) P(0< z <2.3)= 0.4893 I 3

(ili) P(z < -1.6) = 0.5 - 0.4452 = 0.0548 I

t9 V :76, s :12 ,n : 100 .L
2

99% confidence interval for the population mean is given by

(i-*r,,,
.,ln ,i+|r,,): 1ta-$x2.58 )7, xz.ss) l+l
tln {100 ,76aJ100 3

: (72.904, 79.096) I
2
Note: Ta6ng stmdard deviation as s or o can be admissible
Finding C.I by taking table value as 1.96 give 3 score
20 Year 200s 2005 2007 2008 2009 2010 2011 2012 2013
sales 2480 L594 3706 1531 2il2 3L24 t234 3531
5 yr mov
L2062 L2876 133s8
total
2+2 4
5 yr mov
.4

(5 yr moving are not compulsory)
2t lrL
To find ld2 =58
6\ a'
6x 58
P=l- n"?-n =l- l0'-10 0.6s 1+1 4

l-
correlation is positive 2

22 (i) (c) less than 30
- I
(ir)
2x-5y+33=0 - - ----(l) 30x-9y-108=0 -------(2)
Let us assume that eqn (1) be the regression line ofY on X and eqn (?) be that ofX
on Y.

5y = 2r+ 33 + , =?- ** :.br, =? I
4
9 I
Srmilarly eqn (2) becomes,3 0x = 9 y+ 108 +' = fr , * $ b,y
30
2 9 18
r" x b,r, =-x---<I
b I
5 30 1s0 )
So our assumption is right. Hence the regression line of Y on X is
2x -5y +33 = 0 I
2

2

23
:T:H Ir =l:2.96.
14.8 R:2 =0.29
: In I
2
tn5m
LCL = V- ,l.rF = 2.96 - 0.729 x0.28 = 2.76 7

UCL =V+ ,l,rn = 2.96 + 0.729 x0.28 = 3.16 1

:5 4
IrcL
LCL 1
:5

o
12345
Seglc No !2
The process is out ofcontrol
(proper qrplanation of conclusion without diasram eive 1 score)
24

Hs: Performance and age groups are independent
I
2
H1: Theyare not independent

The observed values are 180,220,270,330
l
*, 4
rhe test statistic * r' =Z(ry)=ror.ro
I
Here 72 =107.74> )t|=3.8q 2
So we reject Ho. Hence the performance is depends on age grolrys. I
2

25

P(X <35)=6.97 +P 2.35- P I
= 0.07
o 2

=35-P=-1.48
o
+ p-1.48o = 35------(1) I

= r(2.ry)=0.8e
4
P(X < 63)=0.89 I
2

63- P
= o
=r.23 = tt +7.23o = 63- - - - - -(2) I
Solving (l) an (2) we get, It=50.29, o = 10.33 I
26
I oo%= 844,I ponr :1073, I prQo =9ll,ln,nr= 1154 2

(i) Laspeyre', rN:&-x100 = 9l I I 'l
)' no% "100 =107.94 844 2'2
(ii) paache's IN : &.x '-'""
100 =
I 154
too : 107.55
"'""
,, lrl
5

lnoq 1073 za,
L'1
(iii) Fisher's IN = JZ,, P = 07.94x 107.55 =107 .74
zfT

3

27 (i) (b) Seasonal variation I
(ii) True I
(iir)
SlNo I 2 J 4 5 6 7 8 9 l0
Total
Samples (3,4) (4,6)
I
ample 3.5 4 4.5 6 4.s 5 6.5 5.5 7 7.5 54
5

E(7): uean of sample means : = t.o
I
# 2

1+4+5+6+9 27 t.O
p = Population mean: I
=; = ,
I
E (7\ = p, So Sample mean is unbiased for Population mean 2
(i)
(a) E(x)= l.rp(x)=s r]+ e*l+ 1,1
28
15 15
or!+*l+ 2rL=!
t5 15 15 15 = z.o, 2'2

(b) E 6\ =\x'1 p(x\ = s' * *+ 6',fr + o',I +t',L + 2'. f, = ff = o.z r l,l
2'2
(c) rr (X) = E(X') - (z<x))' =t0.2 - (z.ot)' = 5.92 I'l
2t 2

(iD I
np=9, npq=6 = q=*=1, O=l,r=r, 2
5
plx\ = nC"p'qn-' ,x =0,1,...,n I
2

fZ)'",x = 0,t,...,27
= 27 c.f1)'
-'"'[sJ Ig./ I
2

P(x = 2) = 27 cz, (i)' , (i)" I
2

oR give 2 score ror correct pmr, p(x) = r7 c,(:)(;f ,x = 0,1,...,27

Scheme prepared
No Name & Desipation Signature
Dr. Biju G V (155642), G.VHSS Vattiyoorkavs Thiruvananthapuram (01144) Mob 9447584301

^y
1

2 Darsana Kumari D (157904),NSS HSS Chathannoor, Kollam. (02050) , Mob:

3 Smitha M 5(157531), SN HSS Poochakkal, Alappuzha (M046), Mob:
,r*-F
4 Vidya Ramachanran (1 5 67 02),TD HSS Thuravoor, Alappuzha (04027),Mob:
9,b
5

6

7
Ambily A (195690),St. Peters HSS Kolenchery, Emakulam (07089),Mob:

Dr. Sajish Kumar M (194802),MNKM HSS Chittilamchery, Palakkad (09050), Mob:9,M7380150

Dr. Vidhla G Nair (449871),Govt. HSS Vazhakkad, Malapuram (l1013),Mob:
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4