Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
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 $195$.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.