A lamp - exp(A = 1/4 per hour), hence, on average, 1 failure per 4 hours. What is ?the probability that the lamp lasts between 4 to 5 hours

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A lamp - exp() = 1/4 per hour), hence, on average, 1 failure per 4 hours. What is
?the probability that the lamp lasts between 4 to 5 hours
%3D
e"a*
.a
F(x)=
x= 0,1,.
x!
otherwise
p(x)=
1-0
[0,
E(X) = a = V(X)
ANSWER: P(4 s x s 5) = 0.818
.b O
le*, x20
(0,
F(x =
0, elsewhere
r<0
[ ie*d=1-e*, x20
EX) = 1/, V(X) = 1/2
ANSWER: P(4 s xs 5) = 0.081
ANSWER: None of these
.c
.d O
p°q"*, x=0,1,2,., n
p(x) =-
|0,
otherwise
Е(x) — пр, Ү(X) - пра
%3D
ANSWER: P(4sxs 5) = 0.743
Transcribed Image Text:A lamp - exp() = 1/4 per hour), hence, on average, 1 failure per 4 hours. What is ?the probability that the lamp lasts between 4 to 5 hours %3D e"a* .a F(x)= x= 0,1,. x! otherwise p(x)= 1-0 [0, E(X) = a = V(X) ANSWER: P(4 s x s 5) = 0.818 .b O le*, x20 (0, F(x = 0, elsewhere r<0 [ ie*d=1-e*, x20 EX) = 1/, V(X) = 1/2 ANSWER: P(4 s xs 5) = 0.081 ANSWER: None of these .c .d O p°q"*, x=0,1,2,., n p(x) =- |0, otherwise Е(x) — пр, Ү(X) - пра %3D ANSWER: P(4sxs 5) = 0.743
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