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The mean length of 600 stainless steel sticks is 181mm and the standard deviation is 60mm.Assuming that the length is normally distributed,

1) find the probability that a randomly chosen stick is between 150 and 190mm in length.

2)Given that the length of a particular stick is more than 195mm, find the conditional probability that is actual length exceeds 210mm.

I already solve part (1). For part 2 i dont know. Could someone help me out

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Bayes' theorem? – You Sep 19 '12 at 16:58
not sure. It says conditional probability – David Sep 19 '12 at 16:59
up vote 3 down vote accepted

Let $A$ be the event "greater than $195$" and let $B$ be the event "greater than $210$." We want $\Pr(B|A)$. By a standard formula, $$\Pr(B|A)=\frac{\Pr(A\cap B)}{\Pr(A)}.$$

Note that $\Pr(B\cap A)$ is just $\Pr(B)$. If you solved the first problem then you know how to find $\Pr(A)$ and $\Pr(B)$.

Remark: Intuitively, $\Pr(A)$ is the area under a certain "tail" of the normal. Our answer $\dfrac{\Pr(B)}{\Pr(A)}$ is just the ratio of the area past $210$ to the area past $195$.

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how about Pr(A)?? – David Sep 19 '12 at 17:12
So this is my steps : P(X>210|X>195)= P(Z>0.483|Z>0.233)=$$\frac{P(Z>0.483)}{P(Z>0.233)}$$.And i got the answer 0.7712. Is my answer right?? – David Sep 19 '12 at 17:26
$\Pr(A)$ is the probability of being more than $14$ above the mean, so it is $\Pr(Z\gt 14/60$. Now use normal tables or software. – André Nicolas Sep 19 '12 at 17:27
yeah andre i got my Pr(A) as above i posted.Just want to clarify with you tht is my working above is correct – David Sep 19 '12 at 17:29
Yes, the method is right, and a glance at a normal table shows that your numbers look OK. – André Nicolas Sep 19 '12 at 17:32

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