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Suppose $X$ is $\mathrm{N}(10,1) .$ Find (i) $P[X>10.5]$, (ii) $P[9.5<X<11]$, (iii) $x$ such that $P[X<x]=0.95$. You will need to use Standard Normal tables.
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Let $Z$ denote a $N(0,1)$ random variable from now on and let $\Phi$ denote its $c d f$.
(i) $P[X>10.5]=P[X-10>0.5]=1-\Phi(0.5)=0.3085$;
(ii) $P[9.5<X<11]=\Phi(1)-\Phi(-0.5)=$
$0.5328$;
(iii) $P[X<x]=P[Z<x-10]=0.95$ when $x-10=1.645$ i.e. $x=11.645$.
by Diamond (74,866 points)

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