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Continous Probability
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Find the value of $\lambda$ such that function f(x) is valid probability density function
$f(x)=\lambda (x1)(2x)$ for $1 \leq x \leq 2$
$=0$ otherwise
My $\lambda$ is coming to be $ \frac{6}{5}$
Am I correct?
probability
randomvariable
asked
Nov 15, 2018
in
Probability
by
Ayush Upadhyaya
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I m getting 6...
0
Yeah 6 is given in the key.
0
Integrate from 1 to 2 which evaluates to 1 ,Therefore lambda=6
0
akash.dinkar12
plzz add answer .. i am also getting diff.
+2
see this....
+1
I am getting 6...i think you have done some calculation mistake.
0
yes..
0
Yeah I was doing calculation mistake. Answer is 6.
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