in Mathematical Logic
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in Mathematical Logic
450 views

2 Comments

a= .79

b= .98
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plz chk now
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1 Answer

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$a)$ $P\left ( X\leq a \right )=\int_{\alpha}^{ a}f(x)dx$

$=\int_{0}^{a }3x^{2}dx$

$=\left [ \frac{-3a^{3}}{3} \right ]=a^{3}$..........................i

 

$P\left ( X>a \right )$

$=\int_{a}^{\alpha }3x^{2}dx$

$=\int_{a}^{1 }3x^{2}dx$

$=\left [ x^{3} \right ]_{a}^{1}$

$=1-a^{3}$............................................ii

 

Now, according to question

$a^{3}=1-a^{3}$

$2a^{3}=1$

$a=\frac{1}{\sqrt[3]{2}}$

 

 

$b)$ $P(X>b)=0.05$

$\Rightarrow \int_{b}^{\alpha }3x^{2}dx=0.05$

$1-b^{3}=0.05$

$b^{3}=0.95$

$b=\sqrt[3]{0.95}$

4 Comments

but in (i) part a can be +ve or -ve so how can we take its range from 0 to a??
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range is between 0 to 1

So, it cannot be -ve
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@Saumya

what is ans given?
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