The weight in pounds of people in a certain population has a normal distribution, with a mean of 150 and standard deviation of 40. Of the people who weigh over 170 lbs, what percent weigh more than 200? (using conditional probability)

To do this, I calculated the probability P(x>170) ==> .3085 (using normal distribution) Using the conditional probability formula P( x>200 | x>170) = P( x>200 AND x>170) / P(x>170)

*Here it seems like P(x>200 and x>170) is equal to P(x>200) -- is this wrong to assume?

I'm not sure how to come up with P(x>200 and x>170) -- sticking to formulas, are these events dependent?

Any insights are greatly appreciated. Thanks for your time!


P(x>200 and x>170) = P(x>200) .

I wouldn't call it an assumption. I'd call it a fact. Any x which is > 200 is necessarily greater than 170. I.e., In the event "x>200 and x>170", "and x > 170" is redundant.

  • 1
    $\begingroup$ Thank you! This is obvious, not sure why I was confused by it:) Thanks for your help $\endgroup$ – Stats_anon Jan 3 '18 at 0:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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