Kerala PSC Previous Years Question Paper & Answer

Title : HIGHER SECONDARY SCHOOL TEACHER (STATISTICSE)
Question Code :

Page:5


Below are the scanned copy of Kerala Public Service Commission (KPSC) Question Paper with answer keys of Exam Name 'HIGHER SECONDARY SCHOOL TEACHER (STATISTICSE)' And exam conducted in the year 2023. And Question paper code was '044/2023/OL'. Medium of question paper was in Malayalam or English . Booklet Alphacode was ''. Answer keys are given at the bottom, but we suggest you to try answering the questions yourself and compare the key along wih to check your performance. Because we would like you to do and practice by yourself.

page: 5 out of 14
Excerpt of Question Code: 044/2023/OL

Correct Answer:- Option-D
Question32:-For the sequence of random variables *{X_(n)}* such that *P[X_(n)=+-2%(n)]=(1)/(2),n=1, 2, ...°
A:-WLLN holds
B:-WLLN does not hold
C:-SLLN holds
D:-None of these
Correct Answer:- Option-B
Question33:-Let X be a random variable having t-distribution with n degrees of freedom. Then the distribution of
1/2) 6
A:-F distribution with (n, 1) degrees of freedom
B:-F distribution with (1, n) degrees of freedom
C:-chi*(2)* distribution with n degrees of freedom
D:-‘chi*(2)° distribution with n — 1 degrees of freedom
Correct Answer:- Option-A
Question34:-The joint density function of *X_(1)° and *X_(2)* is
f(x_(1),x_(2))=e*(-(x_(1)+x_(2)), *x_(1)>0, x_(2)>0)
The distribution of *(X_(1))/(X_(1)+X_(2))* will be
A:-Gamma distribution with parameters 1 and 2
B:-Exponential with parameter 2
C:-Uniform in the interval (0, 1)
D:-None of these
Correct Answer:- Option-C
Question35:-The mode of F-distribution with *(n_(1),n_(2))* degrees of freedom is

)2 - 71012
)2 + 7( مر
)2 - 111( 112

8.11 012 + 2)
N, (Nz + 2)
ಯ್‌ 71, (01, 7 2)
72 (701 + 2)

p. 704 (702 )
Correct Answer:- Option-B
Question36:-The joint probability distribution of two random variables X and Y are given by:

P{X=—1, Y=0}=P{X=0, Y=0}=P{X=0, Y=1} = P{X=1, Y=0}="(1)/(8)° and P{X=-1, Y= X=1, Y=-1}="(1)/(4)°.
Obtain the conditional probability distribution of X given Y = 1.
0 | -1 ] ٤اا‎ 1
f(x/y=1) 2 2 | ¢ |
Kes 6 3
% -1 | 0 | 1
f(x/y=1) 3 1 ட்‌
3 3 3
B:-
0 | -1 | 0 | 3
f(x/y=1) 2 1 | ० |
ಯ. ತಿ ತಿ
೫ -1 | 0 |
5 f(x/y=1) 1 | 0 | | 0 |

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